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## Vertex cover might be hard to approximate to within $$2 - \varepsilon$$.(English)Zbl 1133.68061

Summary: Based on a conjecture regarding the power of unique 2-prover-1-round games presented in [S. Khot, “On the power of unique 2-prover 1-round games”, in: Proc. 34th ACM Symp. on Theory of Computing, STOC, May 2002, 767–775 (2002)], we show that vertex cover is hard to approximate within any constant factor better than 2. We actually show a stronger result, namely, based on the same conjecture, vertex cover on $$k$$-uniform hypergraphs is hard to approximate within any constant factor better than $$k$$.

### MSC:

 68R10 Graph theory (including graph drawing) in computer science 68Q17 Computational difficulty of problems (lower bounds, completeness, difficulty of approximation, etc.) 68W25 Approximation algorithms
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### References:

 [1] S. Arora, B. Bollobás, L. Lovász, Proving integrality gaps without knowing the linear program, in: Proceedings of the 43rd Annual Symposium on Foundations of Computer Science, FOCS 2002, Vancouver, Canada, Nov. 2002, pp. 313-322 [2] Arora, S.; Lund, C.; Motwani, R.; Sudan, M.; Szegedy, M., Proof verification and the hardness of approximation problems, J. ACM, 45, 3, 501-555, (1998) · Zbl 1065.68570 [3] Arora, S.; Safra, S., Probabilistic checking of proofs: A new characterization of NP, J. ACM, 45, 1, 70-122, (1998) · Zbl 0903.68076 [4] Bellare, M.; Goldreich, O.; Sudan, M., Free bits, PCPs, and nonapproximability—towards tight results, SIAM J. comput., 27, 3, 804-915, (1998) · Zbl 0912.68041 [5] M. Charikar, K. Makarychev, Y. Makarychev, Near-optimal algorithms for unique games, in: Proceedings of the 38th ACM Symposium on Theory of Computing, STOC 2006 · Zbl 1301.68267 [6] S. Chawla, R. Krauthgamer, R. Kumar, Y. Rabani, D. Sivakumar, On the hardness of approximating sparsest cut and multicut, in: Proc. of 20th IEEE Annual Conference on Computational Complexity, CCC 2005, pp. 144-153 · Zbl 1132.68418 [7] I. Dinur, V. Guruswami, S. Khot, Vertex cover on k-uniform hypergraphs is hard to approximate within factor (k−3−&z.epsiv;), in: Electronic Colloquium on Computational Complexity, Technical Report TR02-027, 2002 [8] Dinur, I.; Guruswami, V.; Khot, S.; Regev, O., A new multilayered PCP and the hardness of hypergraph vertex cover, SIAM J. comput., 34, 5, 1129-1146, (2005) · Zbl 1079.68039 [9] I. Dinur, E. Mossel, O. Regev, Conditional hardness for approximate coloring, in: Proc. 38th ACM Symp. on Theory of Computing, STOC 2006 · Zbl 1301.68143 [10] Dinur, I.; Safra, S., On the hardness of approximating minimum vertex cover, Ann. of math., 162, 1, (2005), preliminary version in STOC 2002 · Zbl 1084.68051 [11] U. Feige, S. Goldwasser, L. Lovász, S. Safra, M. Szegedy, Approximating clique is almost NP-complete, in: Proc. 32nd IEEE Symp. on Foundations of Computer Science, FOCS 1991, pp. 2-12 [12] Friedgut, E., Boolean functions with low average sensitivity depend on few coordinates, Combinatorica, 18, 1, 27-35, (1998) · Zbl 0909.06008 [13] O. Goldreich, Using the FGLSS-reduction to prove inapproximability results for minimum vertex cover in hypergraphs, ECCC Technical Report TR01-102, 2001 · Zbl 1343.68094 [14] A. Gupta, K. Talwar, Approximating unique games, in: Proceedings of Annual Symposium on Discrete Algorithms, SODA 2006, pp. 99-106 · Zbl 1192.91012 [15] Guruswami, V.; Håstad, J.; Sudan, M., Hardness of approximate hypergraph coloring, SIAM J. comput., 31, 6, 1663-1686, (2002) · Zbl 1008.68052 [16] Halperin, E., Improved approximation algorithms for the vertex cover problem in graphs and hypergraphs, SIAM J. comput., 31, 5, 1608-1623, (2002) · Zbl 1041.68130 [17] Håstad, J., Clique is hard to approximate within $$n^{1 - \operatorname{\&z.epsiv;}}$$, Acta math., 182, 1, 105-142, (1999) · Zbl 0989.68060 [18] Håstad, J., Some optimal inapproximability results, J. ACM, 48, 4, 798-859, (2001) · Zbl 1127.68405 [19] J. Holmerin, Improved inapproximability results for vertex cover on k-regular hyper-graphs, in: International Colloquium on Automata, Languages, and Programming, ICALP 2002, pp. 1005-1016 · Zbl 1057.68079 [20] J. Holmerin, Vertex cover on 4-regular hyper-graphs is hard to approximate within $$2 - \operatorname{\&z.epsiv;}$$, in: Proc. 34th ACM Symp. on Theory of Computing, STOC 2002, pp. 544-552 · Zbl 1192.68323 [21] G. Karakostas, A better approximation ratio for the vertex cover problem, in: Proceedings of the 32nd International Colloquium on Automata, Languages and Programming, ICALP, Lisboa, Portugal, July 2005, pp. 1043-1050 · Zbl 1085.68112 [22] S. Khot, On the power of unique 2-Prover 1-Round games, in: Proc. 34th ACM Symp. on Theory of Computing, STOC, May 2002, pp. 767-775 · Zbl 1192.68367 [23] S. Khot, G. Kindler, E. Mossel, R. O’Donnell, Optimal inapproximability results for MAX-CUT and other 2-variable CSPs?, in: Proc. 45th IEEE Symp. on Foundations of Computer Science, FOCS 2004, pp. 146-154 · Zbl 1135.68019 [24] S. Khot, N. Vishnoi, The unique games conjecture, integrality gap for cut problems and embeddability of negative type metrics into $$\ell_1$$, in: Proc. 46th IEEE Symp. on Foundations of Computer Science, FOCS 2005, pp. 53-62 · Zbl 1321.68316 [25] Margulis, G., Probabilistic characteristics of graphs with large connectivity, Prob. peredachi inform., 10, 101-108, (1974) · Zbl 0322.05147 [26] E. Mossel, R. O’Donnell, K. Oleszkiewicz, Noise stability of functions with low influences, invariance and optimality, in: Proc. 46th IEEE Symp. on Foundations of Computer Science, FOCS 2005, pp. 21-30 [27] Raz, R., A parallel repetition theorem, SIAM J. comput., 27, 3, 763-803, (1998) · Zbl 0911.68082 [28] Russo, L., An approximate zero-one law, Z. wahrsch. verw. gebiete, 61, 1, 129-139, (1982) · Zbl 0501.60043 [29] L. Trevisan, Non-approximability results for optimization problems on bounded degree instances, in: Proc. 33rd ACM Symp. on Theory of Computing, STOC 2001, pp. 453-461 · Zbl 1323.90059 [30] L. Trevisan, Approximation algorithms for unique games, in: Proceedings of the 46th Annual Symposium on Foundations of Computer Science, FOCS 2005, 2005, pp. 197-205
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


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# Physics Predictions/Thoughts (1 Viewer)

#### TeheeCat

##### Active Member
Woah, you memorised his equation?? for schrodinger just know how he introduced quantum ideas and the shift from a deterministic view on subatomic interactions to a probabilistic view, it uncovered the probabilistic nature of quantum particles. Also, explain how his model was also an improvement of de Broglie’s, where he defined the electron waves as a probability density of finding a particular electron within a specific orbitals. I’d also mention the different s p d f orbitals if you do chemistry and explain their shapes. Also talk about his wavefunction and the ‘collapse’ of the significance of the collapse of a wavefunction in the probability of finding an electron within an orbital. If you can, also mention the work of Dirac in using Schrodinger‘s ideas coupled with special relativity to discover the existence of anti-particles, and the work of Heissenberg and his uncertainty principle, and how it further supported the probabilistic nature of quantum physics
Haha no no, I didn't word that right. I didn't memorise the actual equation, but rather the fact he derived an equation and what it was used for. Thank you so much btw, you've been very helpful to all of us in this thread.

#### ABBAS38

##### Member
I’d say that is reasonable as a modern measurement, in fact I was also thinking of using interferometry as a modern technique
Ye coz hertz in the end still found the speed of light and in my tutoring book methods similar to hertz which were practically the same was under the subheading of resonance and interferometry and also how the accuracy increased through the modern time as the frequency and wavelength measurements became more accurate and in turn the speed of light became more accurate (v=f times wavelength)

#### lplsz2000

##### New Member
Woah, you memorised his equation?? for schrodinger just know how he introduced quantum ideas and the shift from a deterministic view on subatomic interactions to a probabilistic view, it uncovered the probabilistic nature of quantum particles. Also, explain how his model was also an improvement of de Broglie’s, where he defined the electron waves as a probability density of finding a particular electron within a specific orbitals. I’d also mention the different s p d f orbitals if you do chemistry and explain their shapes. Also talk about his wavefunction and the significance of the collapse of a wavefunction in the probability of finding an electron within an orbital. If you can, also mention the work of Dirac in using Schrodinger‘s ideas coupled with special relativity to discover the existence of anti-particles, and the work of Heissenberg and his uncertainty principle, and how it further supported the probabilistic nature of quantum physics
Man, you are getting high band 6 for sure. You are incredible. May I ask to what degree should be learn about the astronomy way of measuring the speed of light. I was just going to mention that the person measure the time it takes for Jupiter's moon to be eclipsed by Jupiter and concluded that it take light 22 minutes to travel the diameter of Earth's orbit. And I noticed that different textbook has different revolution path of stars and the mass that differentiate them is different, is it a big deal? Which one should we stick to? Thanks in advance

#### Arrowshaft

##### Well-Known Member
Man, you are getting high band 6 for sure. You are incredible. May I ask to what degree should be learn about the astronomy way of measuring the speed of light. I was just going to mention that the person measure the time it takes for Jupiter's moon to be eclipsed by Jupiter and concluded that it take light 22 minutes to travel the diameter of Earth's orbit. And I noticed that different textbook has different revolution path of stars and the mass that differentiate them is different, is it a big deal? Which one should we stick to? Thanks in advance
For older measurements I’d mention the work of Galileo and how his assistant uncovered their lanterns and used a sand/water based clock to determine the time took for the travel, and I’d memorise another technique - you can use Ole Romer (like you said, to measure the successive appearances and disappearances of Jupiter’s moon Io). As for the modern measurements I’d mention James Bradley’s work on stellar abberration to determine the speed of light, Fizeau’s cog tooth experiment (or Foucalt’s improvement on his method by using rotating mirrors). Just memorising 2 from each in detail should be good enough!

#### Hscbuzman

##### Active Member
@Arrowshaft what about this one? I feel I can only ever get 8/9 when this man marks ahahaha

#### Attachments

• 2.7 MB Views: 31

#### Balajanovski

##### Member
@Arrowshaft what about this one? I feel I can only ever get 8/9 when this man marks ahahaha
Honestly, the difference between an 8 & 9 is how well you express scientific ideas, not your actual understandings of them.
Its pretty obvious that you understand your physics, but you do use terminology which is not strictly scientific.
Little things like calling Bohr's orbits "orbit states".
"Fall into the nucleus" could be more scientifically described as "orbital decay".

I know its petty, but from briefly skimming your answer and looking at the marker's feedback, this seems to be your problem.

#### Hscbuzman

##### Active Member
Honestly, the difference between an 8 & 9 is how well you express scientific ideas, not your actual understandings of them.
Its pretty obvious that you understand your physics, but you do use terminology which is not strictly scientific.
Little things like calling Bohr's orbits "orbit states".
"Fall into the nucleus" could be more scientifically described as "orbital decay".

I know its petty, but from briefly skimming your answer and looking at the marker's feedback, this seems to be your problem.
Okay thanks for the feedback

#### Arrowshaft

##### Well-Known Member
Also you seem to have gotten some information wrong, you say that 1 in 8000 alpha particles were deflected directly back, however it was 1 in 8000 that were deflected back at an obtuse angle (>90 degrees). You also said something along the lines of the “the light shun” its “shone”, just little things like that. Overall, its a solid response however, but I do think your slightly informal tendencies in writing occasionally may steer the marker away from giving you a 9, but most likely - I’d say, that would be a 9.

#### Arrowshaft

##### Well-Known Member
Another point, as @Balajanovski said, with orbital decay I’d explain as to why that happens. I would say, by Maxwell’s fourth equation (Ampere’s Circuitial Law), an accelerating charge (such as the electrons orbiting the nucleus, due to centripetal acceleration) will emit EMR, thus by the law of conservation of energy; the kinetic energy of the electrons will be lost causing them to spiral inward through orbital decay.

#### Balajanovski

##### Member
Just a dump on what I know about Planck's hypothesis. Could you verify if my understanding is correct @Arrowshaft?

Q: Describe how Planck's explanation of black body radiation changed the direction of scientific thinking in the early twentieth century. (3 marks)

Prior to Max Planck's hypothesis on the nature of black body radiation, the accepted model of light was Maxwell's continuous wave theory. However, Maxwell's theory was unable to account for the features of a black body curve. It predicted that vibrational modes of higher frequencies in cavity radiators were more likely than lower frequencies, which resulted in the predicted curve, according to Rayleigh-Jeans law, having no peak emission, and approaching infinite intensity at higher frequencies, violating the law of conservation of energy. This was known as the UV catastrophe. Physicists, such as Wien, attempted to account for the UV catastrophe through classical arguments, however essentially introduced arbitrary constants to fit a polynomial with no real underlying model.

Max Planck's hypothesis however had the extraordinary insight to approach the black body problem through a discrete, not continuous model. He modified Wien's law to remove the arbitrary constants and built a mathematical model with the underlying assumption that the atoms in the walls of a cavity radiator acted to transmit and receive only certain modes of oscillation based on their transitions between quantised thermal energy states. This resulted in the electromagnetic energy a black body could emit and receive being made up of discrete quanta E = hf.

Since most atoms in a black body would be at a given temperature, certain thermal energy transitions would be more likely than others, explaining the peak wavelength. The thermal energies required for atoms to undertake the large energy transitions required to release high frequency radiation were furthermore, very unlikely in the black bodies studied, explaining the UV catastrophe.

The immense success of Planck's quantised model in explained the black body curved hinted to physicists that there were shortcomings to the continuous, classical view of the world. Planck's hypothesis would go on to change the course of physics from the classical to the quantum, with his quantisation of electromagnetic energy being fully realised in Einstein's quantised photon model of light.

#### Arrowshaft

##### Well-Known Member
Just a dump on what I know about Planck's hypothesis. Could you verify if my understanding is correct @Arrowshaft?

Q: Describe how Planck's explanation of black body radiation changed the direction of scientific thinking in the early twentieth century. (3 marks)

Prior to Max Planck's hypothesis on the nature of black body radiation, the accepted model of light was Maxwell's continuous wave theory. However, Maxwell's theory was unable to account for the features of a black body curve. It predicted that vibrational modes of higher frequencies in cavity radiators were more likely than lower frequencies, which resulted in the predicted curve, according to Rayleigh-Jeans law, having no peak emission, and approaching infinite intensity at higher frequencies, violating the law of conservation of energy. This was known as the UV catastrophe. Physicists, such as Wien, attempted to account for the UV catastrophe through classical arguments, however essentially introduced arbitrary constants to fit a polynomial with no real underlying model.

Max Planck's hypothesis however had the extraordinary insight to approach the black body problem through a discrete, not continuous model. He modified Wien's law to remove the arbitrary constants and built a mathematical model with the underlying assumption that the atoms in the walls of a cavity radiator acted to transmit and receive only certain modes of oscillation based on their transitions between quantised thermal energy states. This resulted in the electromagnetic energy a black body could emit and receive being made up of discrete quanta E = hf.

Since most atoms in a black body would be at a given temperature, certain thermal energy transitions would be more likely than others, explaining the peak wavelength. The thermal energies required for atoms to undertake the large energy transitions required to release high frequency radiation were furthermore, very unlikely in the black bodies studied, explaining the UV catastrophe.

The immense success of Planck's quantised model in explained the black body curved hinted to physicists that there were shortcomings to the continuous, classical view of the world. Planck's hypothesis would go on to change the course of physics from the classical to the quantum, with his quantisation of electromagnetic energy being fully realised in Einstein's quantised photon model of light.
That’s a very thorough response, I loved your use of the idea of quantum oscillators (although I do think it is a tad overkill for high school physics). I’d also mention the similarity in both curves toward higher wavelengths but most importantly I’d define what a black body is, as, correct me if I am wrong, but you didn’t mention how blackbodies absorb all incident radiation. That being said, this is a definite 3/3.

#### Arrowshaft

##### Well-Known Member
Also, since you went super in depth, i’d also draw the comparison Planck made with standing waves to the vibrational modes, supporting quantisation

#### ABBAS38

##### Member
but lets be real that is too much for 3 marks and nesa wouldnt give that many lines either.
But never the less good depth on the topic and so a question with bigger marks, you would be totally fine!

#### Arrowshaft

##### Well-Known Member
but lets be real that is too much for 3 marks and nesa wouldnt give that many lines either
as i said, it is overkill, that would be the typical response for a 5+ marker from a state ranker haha.

#### ABBAS38

##### Member
as i said, it is overkill, that would be the typical response for a 5+ marker from a state ranker haha.
tru tru!

#### Hscbuzman

##### Active Member
Remember everyone, you can get extra writing booklets!

#### Arrowshaft

##### Well-Known Member
Hey guys, a question - for the applications of step up and step down transformers, how much detail have you gone to? Because I dont know whether it is worth memorising the structure of powerstations and use of pylons, etc. or just knowing the basic concept of stepping up voltage to reduce current and hence power loss by
$\bg_white P_{loss}=I^2R$

#### Hscbuzman

##### Active Member
Hey guys, a question - for the applications of step up and step down transformers, how much detail have you gone to? Because I dont know whether it is worth memorising the structure of powerstations and use of pylons, etc. or just knowing the basic concept of stepping up voltage to reduce current and hence power loss by
$\bg_white P_{loss}=I^2R$
Like i can't see more then 4 marks on that sort of question due to the fact it is now less important in this syllabus compared to the old one.

#### Skuxxgolfer

##### Member
Could someone clarify for me "why cathode rays were a mystery at the time/ an area of debate. Thanks

#### Arrowshaft

##### Well-Known Member
Could someone clarify for me "why cathode rays were a mystery at the time/ an area of debate. Thanks
In summary: cathode rays were argued to be either waves or particles.

Evidence for the particle nature:
- when a paddle wheel was placed inside an evacuated CRT, the paddle wheel spun - implying the cathode rays had momentum and thus mass.
- when cathode rays were passed through electric and magnetic fields, they deflected
- they travelled significantly slower than light.
- they left at right angles to the surface of the cathode

Evidence for wave nature:
- rectilinear propagation (deduced from the shadows produced from the rays, refer to the Maltese across experiment)
- they passed through thin metal foils without damaging them

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Main Content

## Conditional Variance Models

### General Conditional Variance Model Definition

Consider the time series

`${y}_{t}=\mu +{\epsilon }_{t},$`

where ${\epsilon }_{t}={\sigma }_{t}{z}_{t}$. Here, zt is an independent and identically distributed series of standardized random variables. Econometrics Toolbox™ supports standardized Gaussian and standardized Student’s t innovation distributions. The constant term, $\mu$, is a mean offset.

A conditional variance model specifies the dynamic evolution of the innovation variance,

`${\sigma }_{t}^{2}=Var\left({\epsilon }_{t}|{H}_{t-1}\right),$`

where Ht–1 is the history of the process. The history includes:

• Past variances, ${\sigma }_{1}^{2},{\sigma }_{2}^{2},\dots ,{\sigma }_{t-1}^{2}$

• Past innovations, ${\epsilon }_{1},{\epsilon }_{2},\dots ,{\epsilon }_{t-1}$

Conditional variance models are appropriate for time series that do not exhibit significant autocorrelation, but are serially dependent. The innovation series ${\epsilon }_{t}={\sigma }_{t}{z}_{t}$ is uncorrelated, because:

• E(εt) = 0.

• E(εtεt–h) = 0 for all t and $h\ne 0.$

However, if ${\sigma }_{t}^{2}$ depends on ${\sigma }_{t-1}^{2}$, for example, then εt depends on εt–1, even though they are uncorrelated. This kind of dependence exhibits itself as autocorrelation in the squared innovation series, ${\epsilon }_{t}^{2}.$

Tip

For modeling time series that are both autocorrelated and serially dependent, you can consider using a composite conditional mean and variance model.

Two characteristics of financial time series that conditional variance models address are:

• Volatility clustering. Volatility is the conditional standard deviation of a time series. Autocorrelation in the conditional variance process results in volatility clustering. The GARCH model and its variants model autoregression in the variance series.

• Leverage effects. The volatility of some time series responds more to large decreases than to large increases. This asymmetric clustering behavior is known as the leverage effect. The EGARCH and GJR models have leverage terms to model this asymmetry.

### GARCH Model

The generalized autoregressive conditional heteroscedastic (GARCH) model is an extension of Engle’s ARCH model for variance heteroscedasticity [1]. If a series exhibits volatility clustering, this suggests that past variances might be predictive of the current variance.

The GARCH(P,Q) model is an autoregressive moving average model for conditional variances, with P GARCH coefficients associated with lagged variances, and Q ARCH coefficients associated with lagged squared innovations. The form of the GARCH(P,Q) model in Econometrics Toolbox is

`${y}_{t}=\mu +{\epsilon }_{t},$`

where${\epsilon }_{t}={\sigma }_{t}{z}_{t}$ and

`${\sigma }_{t}^{2}=\kappa +{\gamma }_{1}{\sigma }_{t-1}^{2}+\dots +{\gamma }_{P}{\sigma }_{t-P}^{2}+{\alpha }_{1}{\epsilon }_{t-1}^{2}+\dots +{\alpha }_{Q}{\epsilon }_{t-Q}^{2}.$`

Note

The `Constant` property of a `garch` model corresponds to κ, and the `Offset` property corresponds to μ.

For stationarity and positivity, the GARCH model has the following constraints:

• $\kappa >0$

• ${\gamma }_{i}\ge 0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\alpha }_{j}\ge 0$

• ${\sum }_{i=1}^{P}{\gamma }_{i}+{\sum }_{j=1}^{Q}{\alpha }_{j}<1$

To specify Engle’s original ARCH(Q) model, use the equivalent GARCH(0,Q) specification.

### EGARCH Model

The exponential GARCH (EGARCH) model is a GARCH variant that models the logarithm of the conditional variance process. In addition to modeling the logarithm, the EGARCH model has additional leverage terms to capture asymmetry in volatility clustering.

The EGARCH(P,Q) model has P GARCH coefficients associated with lagged log variance terms, Q ARCH coefficients associated with the magnitude of lagged standardized innovations, and Q leverage coefficients associated with signed, lagged standardized innovations. The form of the EGARCH(P,Q) model in Econometrics Toolbox is

`${y}_{t}=\mu +{\epsilon }_{t},$`

where ${\epsilon }_{t}={\sigma }_{t}{z}_{t}$ and

`$\mathrm{log}{\sigma }_{t}^{2}=\kappa +\sum _{i=1}^{P}{\gamma }_{i}\mathrm{log}{\sigma }_{t-i}^{2}+\sum _{j=1}^{Q}{\alpha }_{j}\left[\frac{|{\epsilon }_{t-j}|}{{\sigma }_{t-j}}-E\left\{\frac{|{\epsilon }_{t-j}|}{{\sigma }_{t-j}}\right\}\right]+\sum _{j=1}^{Q}{\xi }_{j}\left(\frac{{\epsilon }_{t-j}}{{\sigma }_{t-j}}\right).$`

Note

The `Constant` property of an `egarch` model corresponds to κ, and the `Offset` property corresponds to μ.

The form of the expected value terms associated with ARCH coefficients in the EGARCH equation depends on the distribution of zt:

• If the innovation distribution is Gaussian, then

`$E\left\{\frac{|{\epsilon }_{t-j}|}{{\sigma }_{t-j}}\right\}=E\left\{|{z}_{t-j}|\right\}=\sqrt{\frac{2}{\pi }}.$`

• If the innovation distribution is Student’s t with ν > 2 degrees of freedom, then

`$E\left\{\frac{|{\epsilon }_{t-j}|}{{\sigma }_{t-j}}\right\}=E\left\{|{z}_{t-j}|\right\}=\sqrt{\frac{\nu -2}{\pi }}\frac{\Gamma \left(\frac{\nu -1}{2}\right)}{\Gamma \left(\frac{\nu }{2}\right)}.$`

The toolbox treats the EGARCH(P,Q) model as an ARMA model for$\mathrm{log}{\sigma }_{t}^{2}.$ Thus, to ensure stationarity, all roots of the GARCH coefficient polynomial,$\left(1-{\gamma }_{1}L-\dots -{\gamma }_{P}{L}^{P}\right)$, must lie outside the unit circle.

The EGARCH model is unique from the GARCH and GJR models because it models the logarithm of the variance. By modeling the logarithm, positivity constraints on the model parameters are relaxed. However, forecasts of conditional variances from an EGARCH model are biased, because by Jensen’s inequality,

`$E\left({\sigma }_{t}^{2}\right)\ge \mathrm{exp}\left\{E\left(\mathrm{log}{\sigma }_{t}^{2}\right)\right\}.$`

An EGARCH(1,1) specification will be complex enough for most applications. For an EGARCH(1,1) model, the GARCH and ARCH coefficients are expected to be positive, and the leverage coefficient is expected to be negative; large unanticipated downward shocks should increase the variance. If you get signs opposite to those expected, you might encounter difficulties inferring volatility sequences and forecasting (a negative ARCH coefficient can be particularly problematic). In this case, an EGARCH model might not be the best choice for your application.

### GJR Model

The GJR model is a GARCH variant that includes leverage terms for modeling asymmetric volatility clustering. In the GJR formulation, large negative changes are more likely to be clustered than positive changes. The GJR model is named for Glosten, Jagannathan, and Runkle [2]. Close similarities exist between the GJR model and the threshold GARCH (TGARCH) model—a GJR model is a recursive equation for the variance process, and a TGARCH is the same recursion applied to the standard deviation process.

The GJR(P,Q) model has P GARCH coefficients associated with lagged variances, Q ARCH coefficients associated with lagged squared innovations, and Q leverage coefficients associated with the square of negative lagged innovations. The form of the GJR(P,Q) model in Econometrics Toolbox is

`${y}_{t}=\mu +{\epsilon }_{t},$`

where${\epsilon }_{t}={\sigma }_{t}{z}_{t}$ and

`${\sigma }_{t}^{2}=\kappa +\sum _{i=1}^{P}{\gamma }_{i}{\sigma }_{t-i}^{2}+\sum _{j=1}^{Q}{\alpha }_{j}{\epsilon }_{t-j}^{2}+\sum _{j=1}^{Q}{\xi }_{j}I\left[{\epsilon }_{t-j}<0\right]{\epsilon }_{t-j}^{2}.$`

The indicator function $I\left[{\epsilon }_{t-j}<0\right]$ equals 1 if ${\epsilon }_{t-j}<0$, and 0 otherwise. Thus, the leverage coefficients are applied to negative innovations, giving negative changes additional weight.

Note

The `Constant` property of a `gjr` model corresponds to κ, and the `Offset` property corresponds to μ.

For stationarity and positivity, the GJR model has the following constraints:

• $\kappa >0$

• ${\gamma }_{i}\ge 0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\alpha }_{j}\ge 0$

• ${\alpha }_{j}+{\xi }_{j}\ge 0$

• ${\sum }_{i=1}^{P}{\gamma }_{i}+{\sum }_{j=1}^{Q}{\alpha }_{j}+\frac{1}{2}{\sum }_{j=1}^{Q}{\xi }_{j}<1$

The GARCH model is nested in the GJR model. If all leverage coefficients are zero, then the GJR model reduces to the GARCH model. This means you can test a GARCH model against a GJR model using the likelihood ratio test.

## References

[1] Engle, Robert F. “Autoregressive Conditional Heteroskedasticity with Estimates of the Variance of United Kingdom Inflation.” Econometrica. Vol. 50, 1982, pp. 987–1007.

[2] Glosten, L. R., R. Jagannathan, and D. E. Runkle. “On the Relation between the Expected Value and the Volatility of the Nominal Excess Return on Stocks.” The Journal of Finance. Vol. 48, No. 5, 1993, pp. 1779–1801.

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## Pages

### Graph y = (x - 1)(x - 4)

 Graph of y = (x - 1)(x - 4)
Comparing the equation with intercept form y = a(x - p)(x - q), p = 1 and q = 4. So the x-intercepts are (1, 0) and (4, 0).

Let the vertex be (h, k), then h = (p + q)/2 = (1 + 4)/2 = 5/2,

k = (5/2 - 1)(5/2 - 4)
k = (3/2)(-3/2)
k = -9/4

Therefore the vertex is (5/2, -9/4). Now plot the vertex and x-intercepts and join them with a free hand curve

### Graph y = 3(x - 6)(x - 3)

 Graph of y = 3(x - 6)(x - 3)
Comparing the equation with intercept form y = a(x - p)(x - q), p = 6 and q = 3. So the x-intercepts are (6, 0) and (3, 0).

Let the vertex be (h, k), then h = (p + q)/2 = (6 + 3)/2 = 9/2,

k = 3(9/2 - 6)(9/2 - 3)
k = 3(-3/2)(3/2)
k = -27/4

Therefore the vertex is (9/2, -27/4). Now plot the vertex and x-intercepts and join them with a free hand curve

### Graph y = -2(x - 4)(x - 5)

Comparing the equation with intercept form y = a(x - p)(x - q), p = 4 and q = 5. So the x-intercepts are (4, 0) and (5, 0).
 Graph y = -2(x - 4)(x - 5)

Let the vertex be (h, k), then h = (p + q)/2 = (4 + 5)/2 = 9/2,

k = -2(9/2 - 4)(9/2 - 5)
k = -2(1/2)(-1/2)
k = 1/2

Therefore the vertex is (9/2, 1/2). Get some more points on the parabola by making table of x-y values

 x Y = -2(x - 4)(x - 5) 2 -12 3 -4 6 -4 7 -12

Now plot the vertex and x-intercepts and join them with a free hand curve

### Graph y = (x + 2)(x + 3)

Comparing the equation with intercept form y = a(x - p)(x - q), p = -2 and q = -3. So the x-intercepts are (-2, 0) and (-3, 0).

Let the vertex be (h, k), then h = (p + q)/2 = (-2 + -3)/2 = -5/2,

k = (-5/2 + 2)(-5/2 + 3)
k = (-1/2)(1/2)
k = -1/4

Therefore the vertex is (-5/2, -1/4). Now plot the vertex and x-intercepts and join them with a free hand curve.  Get some more points on the parabola by making table of x-y values

 x y = (x + 2)(x + 3) -1 2 -4 2 0 6 -5 6
Now plot the vertex and x-intercepts and join them with a free hand curve
 Graph of y = (x + 2)(x + 3)


text:
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## GE Microwave Oven

Guías de desmontajes y reparación para hornos de microondas GE

Preguntas 320 Ver todo

# Replacing dead high voltage cap with similar of lower capacitance.

Hi,

The high voltage capacitor for my microwave oven has shorted. It is a 2100v, 86mfd capacitor.

I have a "spare" that is very similar: 2100v .76 mfd (so .76 instead of .86 mfd. Can I use it?

Thanks

Es esta una buena pregunta?

Puntuación 0

## 1 Respuesta

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Without knowing more, I'd say not to use it. That particular rating may be required for the proper operation of the magnetron (the thing that generates the microwaves). GE should be able to get you the right part.

Best of luck.

Fue útil esta respuesta?

Puntuación 0

Comentarios:

I had been hoping for an informed reply. It turns out, as I learned since I posted the question, that the high voltage capacitor is part o fhe voltage doubling circuit between the high voltage transformer and the magnetron. If the capacitance is a bit low, it reduces the final voltage - it does not quite "double" - to the magnetron, which results in a small reduction in the power the magnetron delivers, but with zero risk to the circuitry.

- de


text:
# MATLAB - read files from directory?

I wish to read files from a directory and iteratively perform an operation on each file. This operation does not require altering the file.

I understand that I should use a for loop for this. Thus far I have tried:

FILES = ls('path\to\folder');

for i = 1:size(FILES, 1);
end

The error returned here suggests to me, a novice, that listing a directory with ls() does not assign the contents to a data structure.

Secondly I tried creating a file containing on each line a path to a file, e.g.,

C:\Documents and Settings\My Documents\MATLAB\asd.pdb
C:\Documents and Settings\My Documents\MATLAB\asd.pdb

I then read this file using the following code:

fid = fopen('paths_to_files.txt');
FILES = textscan(fid, '%s');
FILES = FILES{1};
fclose(fid);

This code reads the file but creates a newline where a space exists in the pathway, i.e.

'C:\Documents'
'and'
'Setting\My'
'Documents\MATLAB\asd.pdb'

Ultimately, I then intended to use the for loop

for i = 1:size(FILES, 1)

to read each file but pdbread() throws an error proclaiming that the file is of incorrect format or does not exist.

Is this due to the newline separation of paths when the pathway file is read in?

Any help or suggestions greatly apppreciated.

Thanks, S :-)

-

First Get a list of all files matching your criteria:
( in this case pdb files in C:\My Documents\MATLAB )

matfiles = dir(fullfile('C:', 'My Documents', 'MATLAB', '*.pdb'))

Then read in a file as follows:
( Here i can vary from 1 to the number of files )

data = load(matfiles(i).name)

A simpler alternative if you can rename your files is as follows:-

First save the reqd. files as 1.pdb, 2.pdb, 3.pdb,... etc.

Then the code to read them iteratively in Matlab is as follows:

for i = 1:n
str = strcat('C:\My Documents\MATLAB', int2str(i),'.pdb');

% use our logic here
% before proceeding to the next file

end

-

I copy this from yahoo answers! It worked for me

% first, you have to find the folder folder = uigetdir; % check the help for uigetdir to see how to specify a starting path, which makes your life easier

% get the names of all files. dirListing is a struct array. dirListing = dir(folder);

% loop through the files and open. Note that dir also lists the directories, so you have to check for them. for d = 1:length(dirListing) if ~dirListing(1).isdir fileName = fullfile(folder,dirListing(d).name); % use full path because the folder may not be the active path

% open your file here fopen(fileName)

% do something

end % if-clause end % for-loop

-


text:
# Making the environment of A Cap B less thick

How do I decrease the thickness of the line encompassing A intersection B? This is my Code:

\documeentclass{book}
\usepackage{tikz}
\usetikzlibrary{patterns}
\begin{document}
\begin{venndiagram2sets}[tikzoptions={scale=1.7, line width = 0.2cm}, labelA= {}, labelB={}, showframe=false, shade={}]
\setpostvennhook{
\draw (labelA) node[above=0.2cm, left=1.1cm]{\LARGE $A$};
\draw (labelB) node[above=0.2cm, right=1.1cm]{\LARGE $B$};}
\begin{scope}[pattern= horizontal lines]
\fillACapB
\end{scope}
\begin{scope}[pattern= north east lines]
\fillACapB
\end{scope}
\end{venndiagram2sets}
\end{document}

I want it to be something like this:

• Maybe have a look at: tex.stackexchange.com/questions/9681/… it might help. Jun 18 at 12:37
• Jun 18 at 12:44
• Worst case you have to draw the diagram manually. Should not be very hard, figure out some way to draw circle section, fill a region, draw an arc etc. Jun 18 at 13:01

\documentclass[tikz, border=1cm]{standalone}
\usetikzlibrary{patterns}
\begin{document}
\begin{tikzpicture}
\end{tikzpicture}
\end{document}

\documentclass[tikz, border=1cm]{standalone}
\usetikzlibrary{patterns}
\begin{document}
\begin{tikzpicture}
\end{tikzpicture}
\end{document}

• Please can you also help me with the one of three sets A, B and C shaded in A intersection B intersection C. Please. Thank you so much for this. Jun 21 at 11:45
• See edit. You should normally start a new question for addition questions. -remember there is no limit on how many new question you can ask, and if you show some effort, all question will attract helpfull people. Jun 21 at 11:56
• Thank you. I will note that. Jun 21 at 12:15

Just to play a bit (if you want to do more complex diagrams, you'll have to calculate the angles more accurately).

\documentclass[tikz,border=3.14mm]{standalone}
\usetikzlibrary{intersections}

\begin{document}
\begin{tikzpicture}
\path[name path=circleA] (0,0) circle [radius=3];
\path[name path=circleB] (4,0) circle [radius=3];
\path[name intersections={of=circleA and circleB}] ;
\draw[red,fill=orange] (intersection-1) arc(132:228:3) arc(-48:48:3);
\draw[red,line width=3pt] (intersection-1) arc(48:312:3) arc(-132:132:3) -- cycle;
\end{tikzpicture}
\end{document}

## EDIT

A second approach (which I don't really like because of the white fill which doesn't allow to have something under the picture, but anyway it's quite unlikely) is the following where the lines seem better to my eyes.

\documentclass[tikz,border=3.14mm]{standalone}
\usetikzlibrary{intersections}

\begin{document}
\begin{tikzpicture}
\path[name path=circleA] (0,0) circle [radius=3];
\path[name path=circleB] (4,0) circle [radius=3];
\path[name intersections={of=circleA and circleB}] ;

\draw[red,line width=15pt] (intersection-1) arc(48:312:3) arc(-132:132:3) -- cycle;
\fill[white] (intersection-1) arc(48:312:3) arc(-132:132:3) -- cycle;
\draw[red,fill=orange] (intersection-1) arc(132:228:3) arc(-48:48:3);
\end{tikzpicture}
\end{document}


text:
# zbMATH — the first resource for mathematics

Values of Bernoulli polynomials and Hurwitz’s zeta function at rational points. (English) Zbl 0731.11014
Let $$B_ n(t)$$ denote the Bernoulli polynomial of degree n. It is shown that $$k^ n\{B_ n(h/k)-B_ n(0)\}$$ is an integer for all positive integers h and k.

##### MSC:
 11B68 Bernoulli and Euler numbers and polynomials 11M35 Hurwitz and Lerch zeta functions
##### Keywords:
integer values; Bernoulli polynomial


text:
# Revision history [back]

Your price formula can be simplified to :

=IFNA(VLOOKUP(A21,$'Material Stock'.A$3:E$1197,5,0),"")  Instead of selecting an item number, which could be difficult to remember without scrolling down your list, it might be better to select the serial number using the Validity cell range, or add a more descriptive column to your materials list for the purpose of selection. If the column for selection is not the first column of your table you would need something like the following instead of VLOOKUP. This searches column B and chooses value from column A. =IFNA(INDEX($'Material Stock'.A$3:A$1197,MATCH(A21,$'Material Stock'.B$3:B$1197,0)),"")  If you have experience with macros you might create a dialog to perform the selection but If you have a large list of material it might be better to start using a database. Your price formula can be simplified to : =IFNA(VLOOKUP(A21,$'Material Stock'.A$3:E$1197,5,0),"")

Instead of selecting an item number, which could be difficult to remember without scrolling down your list, it might be better to select the serial number using the Validity cell range, or add a more descriptive column to your materials list for the purpose of selection.

If the column for selection is not the first column of your table you would need something like the following instead of VLOOKUP. This searches column B and chooses value from column A.

=IFNA(INDEX($'Material Stock'.A$3:A$1197,MATCH(A21,$'Material Stock'.B$3:B$1197,0)),"")

If you have experience with macros you might create a dialog to perform the selection but If you have a large list of material it might be better to start using a database.

Edit: I think to allow either item number or Serial number to be entered and search for the other from the list would require a macro.

Attached example. Untitled 21.ods

I tried to base this on your description but it may need some modifications. The macro included is as follows and needs to be assigned to the sheet event "Content changed" on the workorder sheet.

If column A is changed then the formula to search is entered in column B. If column B is changed the formula is entered in column A.

REM  *****  BASIC  *****

Option Explicit

Sub ContentChangedEvent(oEvent As Object)
'   Assign this macro to the Sheet event "Content changed" or workorder sheet'

'   Only if single cell was modified'

'   If row of cell changed was greater than 20 and less than 26'
If Len(oEvent.Formula) > 0 Then
'   Ignore formula to avoid recursion of event'
If Left(oEvent.Formula, 1) <> "=" Then
Case 0  '   Column A'
"=IFNA(INDEX($'Material Stock'.B$3:B$1197;MATCH(A" & Format(oRngAd.StartRow + 1,"0") & ";$'Material Stock'.A$3:A$1197;0));"")"
Case 1  '   Column B'
"=IFNA(INDEX($'Material Stock'.A$3:A$1197;MATCH(B" & Format(oRngAd.StartRow + 1,"0") & ";$'Material Stock'.B$3:B$1197;0));"")"
Case Else
'   Do Nothing'
End Select
End If
End If
End If
End Sub

Your price formula can be simplified to :

=IFNA(VLOOKUP(A21,$'Material Stock'.A$3:E$1197,5,0),"")  Instead of selecting an item number, which could be difficult to remember without scrolling down your list, it might be better to select the serial number using the Validity cell range, or add a more descriptive column to your materials list for the purpose of selection. If the column for selection is not the first column of your table you would need something like the following instead of VLOOKUP. This searches column B and chooses value from column A. =IFNA(INDEX($'Material Stock'.A$3:A$1197,MATCH(A21,$'Material Stock'.B$3:B$1197,0)),"")  If you have experience with macros you might create a dialog to perform the selection but If you have a large list of material it might be better to start using a database. Edit: I think to allow either item number or Serial number to be entered and search for the other from the list would require a macro. Attached example. Untitled 21.ods I tried to base this on your description but it may need some modifications. The macro included is as follows and needs to be assigned to the sheet event "Content changed" on the workorder sheet. If column A is changed then the formula to search is entered in column B. If column B is changed the formula is entered in column A. REM ***** BASIC ***** Option Explicit Sub ContentChangedEvent(oEvent As Object) ' Assign this macro to the Sheet event "Content changed" or workorder sheet' Dim oRngAd As Object oRngAd = oEvent.RangeAddress ' Only if single cell was modified' If oRngAd.StartRow <> oRngAd.EndRow Then Exit Sub If oRngAd.StartColumn <> oRngAd.EndColumn Then Exit Sub ' If row of cell changed was greater than 20 and less than 26' If oRngAd.StartRow > 19 And oRngAd.StartRow < 25 Then If Len(oEvent.Formula) > 0 Then ' Ignore formula to avoid recursion of event' If Left(oEvent.Formula, 1) <> "=" Then Select Case oRngAd.StartColumn Case 0 ' Column A' oEvent.Spreadsheet.getCellByPosition(1, oRngAd.StartRow).Formula = _ "=IFNA(INDEX($'Material Stock'.B$3:B$1197;MATCH(A" & Format(oRngAd.StartRow + 1,"0") & ";$'Material Stock'.A$3:A$1197;0));"")" Case 1 ' Column B' oEvent.Spreadsheet.getCellByPosition(0, oRngAd.StartRow).Formula = _ "=IFNA(INDEX($'Material Stock'.A$3:A$1197;MATCH(B" & Format(oRngAd.StartRow + 1,"0") & ";$'Material Stock'.B$3:B\$1197;0));"")"
Case Else
'   Do Nothing'
End Select
End If
End If
End If
End Sub


text:
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Kirchhoff’s Circuit Law and Circuit Analysis

When we deal with the complex electric circuits (like T-circuits), we use some mathematical techniques to find the current and voltage drops in the circuit. It’s not as difficult as it may seem at the first sight. In this post, we will tell you how to analyze the circuit with the help of the Kirchhoff’s Circuit Law.

There are two simple rules about the voltage and the current in the closed electric circuit. The first one, called Kirchhoff’s Current Law, states that the sum of the currents flowing into a point (or node) in a circuit equals the sum of the currents flowing out. The second, Kirchhoff’s Voltage Law, states that the sum of the voltage drops around any closed circuit is zero. Let’s now figure out what a branch, a node and a loop mean. The branch is simply the component of the circuit such as the resistor or a battery. The node is the point of connection between two or more branches. And, finally, the loop is any closed path in the circuit.

Let’s consider, for example, T-circuit in the scheme below:

Let the first voltage source (or the battery) be $10\ V$ and the second one $15\ V$. And let the resistances be $R_1=15\ \Omega$, $R_2=22\ \Omega$ and $R_3=30\ \Omega$. Our task is to analyze the circuit and find the currents $I_1$, $I_2$, $I_3$ and the voltage drops on resistances $R_1$, $R_2$, $R_3$.

As we can see in the scheme above, this circuit has three branches, two nodes (designated as 1 and 2) and three loops (designated as A, B and C). So, firstly, applying Kirchhoff’s Current Law to the nodes 1 and 2, we get:

$I_3 = I_1 + I_2$

Then, we can apply Kirchhoff’s Voltage Law to the loops A, B and C:

$V_1 = I_1R_1 + I_3R_3$ (1)

$V_2 = I_2R_2 + I_3R_3$ (2)

$V_1 – V_2 = I_1R_1 – I_2R_2$ (3)

Let’s substitute $I_3$ in equations (1)-(2) to the sum of currents $I_1$ and $I_2$:

$V_1 = I_1(R_1+R_3) + I_2R_3$

$V_2 = I_2(R_2+R_3) + I_1R_3$

After substituting the numbers, we get the system of equations:

$45I_1 + 30I_2 = 10$ (4)

$30I_1 + 52I_2 = 15$ (5)

Let’s express $I_1$ from the equation (5) via $I_2$ and substitute it into the equation (4):

$I_1=(15-52I_2)/30$

$I_2=0.26\ A$

As we know $I_2$, we can find $I_1$:

$I_1=0.05\ A$

Finally, we can find $I_3$:

$I_3=I_1+I_2 = 0.31\ A$

Then, the voltage drops on resistances $R_1$, $R_2$, $R_3$ will be:

$I_1R_1 = 0.05\ A \times 15\ \Omega = 0.75\ V$

$I_2R_2 = 0.26\ A \times 22\ \Omega = 5.72\ V$

$I_3R_3 = 0.31\ A \times 30\ \Omega = 9.3\ V$

So, as you may see, it is easier than you expected. We are glad to help you with this and other problems in physics, just let us know about them!

Submit


text:
# What does a statistically significant correlation imply?

Sorry for the total lack of any semblance of statistical knowledge or understanding. That said:

If we have two variables, A and B, we can compute a Pearson's r for them. Knowing how many samples we have, we can use a lookup table to find a p value. So, assuming good data (taken using a reliable measurement instrument from a representative sample), what does this p value tell us? Specifically, is it the probability of none of the following being true:

Changes in A cause changes in B,

OR

Changes in B cause changes in A,

OR

There is some variable C, and changes in C cause changes in A and B?

Are there other possibilities? In short, is a statistically significant correlation "acceptably" likely to imply one of the above types of causation, assuming decent sampling but no imposition of treatment, or are there other complications?

• Correlation doesn't imply causation in any of the ways you describe. All that it says is that increases or decreases in A are related to either increases or decreases in B. – Michael Chernick Sep 1 '17 at 19:49
• Okay. Wikipedia lists seven possibilities: "A causes B (direct causation); B causes A (reverse causation); A and B are consequences of a common cause, but do not cause each other; A and B both causes C, which is (explicitly or implicitly) conditioned on; A causes B and B causes A (bidirectional or cyclic causation); A causes C which causes B (indirect causation); there is no connection between A and B and the correlation is a coincidence"... – Jean Luc Picard Sep 1 '17 at 19:53
• Of those, the first, second, and third are mentioned in my question. The fifth is a combination of the first and the second. The fourth seems like something a good sampling methodology should avoid. The sixth looks different, but I question the conceptual utility of any definition of causation where A causes C causes B does not imply that A causes B; it seems like we can find some intervening factor for virtually any case of causality. I thought the seventh was what our p value gave a probability of. With which of these assumptions do you disagree? – Jean Luc Picard Sep 1 '17 at 19:59
• Here is an example of an eighth possibility for your consideration: that correlation sometimes represents an arbitrary analytical decision. Give a paper map of a river to people and ask them to digitize locations along it. Each person creates a set of $(x,y)$ coordinates. In some of those sets the river might be oriented to run horizontally, in which case the correlation coefficient is close to zero; in some of them it might seem to run to the northeast, with a correlation coefficient close to $1$; in yet others to the northwest, with correlation close to $-1$. – whuber Sep 1 '17 at 21:38
• (Continued) There are other possibilities as well. The point is that correlation per se is a description of a set of $(x,y)$ data: nothing more. Thus, the value of $r$ by itself does not logically imply anything whatsoever about causation or even the existence of hypothetical variables that have any meaning or importance for a study. A p-value provides some indication of whether a particular probability model (the null hypothesis) may be appropriate or not. Probability models, in themselves, have no causal elements. – whuber Sep 1 '17 at 21:40

I'm simplifying a little bit, but basically: the p-value that you're probably referring to denotes the probability that, given a particular sample size, two random sets of numbers will have a correlation greater than or equal to the one you've observed.

Example: Say we roll a pair of dice 6 times. This generates 6 unique x,y points. If the order of the rolls doesn't matter, what are the chances that you'd end up with the following points:

1,1; 2,2; 3,3; 4,4; 5,5; 6,6?

Pretty low, right? This dataset has a correlation coefficent of 1 and an n of 6. If you look up the associated p-value, you'll find it's < 0.00001. There is less than 0.00001% chance that these numbers are simply random dice rolls.

Now for some actual dice rolls - here's numbers I generated randomly in excel:

1,6; 2,3; 3,4; 3,2; 5,2; 6,4; .

The correlation coefficient of this data set is = 0.1806. Again, the sample size is 6. The associated p-value for a two-tailed test is 0.3660225.

If these were data from an experiment, we'd say that the association between x and y is so weak that it could be adequately described by random chance. In fact, we'd say that there is a 36.6% chance that there is no true treatment effect, and that the (weak) observed correlation is simply due to random chance.

A significant correlation does not necessarily imply causation. To determine whether there is a causal relationship, you must use a randomized experiment. That is one reason why tobacco companies were so difficult to prosecute in the 1960's - no scientists were "randomly assigning" humans to be smokers or non-smokers, then waiting to see who got lung cancer. The prosecutor's primary evidence came from came from observational studies, with no randomized treatments. However, after decades of strong correlation between smoking and lung cancer, along with plausible medical explanations for how tobacco smoking harms the body, courts were finally convinced that smoker's lung cancer was caused by smoking and not something else. Statisticians are the same way - in the absence of experimental data, causation is only implied by strong, repeated, and prolonged correlations, usually with a plausible mechanism for how x might be acting on y.

• "A significant correlation does not necessarily imply causation." I assume that means any causation, not just any specific causation. Is there a simple explanation of why this is? If A correlates strongly with B in a well-designed sample, what possible scenarios exist beyond some combination of A causing B, B causing A, and a third factor C causing both A and B? – Jean Luc Picard Sep 1 '17 at 20:47
• The alternate scenario is random chance. With all of the things we can measure in this world, there is no limit to the random associations people have uncovered. This, of course, does not imply causation. See this humorous page for examples. – Anson Call Sep 1 '17 at 20:52
• Then the p value (communicates/represents/is used as a proxy for) something other than the probability that what we're observing is random chance? If so, what? – Jean Luc Picard Sep 1 '17 at 21:09
• The p-value, if computed correctly, is always the probability of finding data as different / unusual / extreme as the data being considered, assuming that the hypothesis is true that the data are not different / not correlated / not associated. That's p-value by definition... But unprobable things can be found. Especially when you look many times. The probability of finding something unprobable is high if you look often enough. – Sal Mangiafico Sep 2 '17 at 2:17

I'll add a couple of points that may be valuable in this conversation. 1) A significant / interestingly-large correlation (S/IL correlation) may be local to the sample you're studying. And, 2) As you do more correlation tests, the chance of finding a S/IL correlation increases just by random chance.

Here's an experiment you can perform. Find about 10 co-workers. Design a survey of about 20 questions. Their height. Their weight. What month they were born in. How dark their hair is on a scale of 1 to 5. How much they like dark chocolate vs. milk chocolate on a scale of 1 to 5. How much they smoke on a scale of 1 to 5. And so on. And then perform correlation tests for each pair of questions.

Illustrating point 2 above, you are likely to find S/IL correlations just by chance for at least some of these correlations. Because you are testing 19 + 18 + 17 + 16 ... (and so on) combinations of these questions.

And you will be tempted to tell yourself "just so" stories about these correlations. "Oh oh course people who like dark chocolate more are more likely to smoke more, because both are bitter." "Oh of course blonder people drink alcohol more, because as everyone says 'blonds have more fun.'"

But of course for at least some of correlations, these explanations will be bullshit. "The first principle is that you must not fool yourself — and you are the easiest person to fool." That applies to all of us. Especially when we are analyzing our hard-earned data.

Illustrating point 1 above, if you went out into the larger world with one of these S/IL correlations in hand, you would quickly find that it doesn't hold up. It just happened to be that among your 10 co-workers that smokers liked dark chocolate more or that blonds drink more alcohol. There's no causation here; there's just a random correlation among two variables for a finite population.


text:
# Portal:Mathematics

## The Mathematics Portal

Mathematics is the study of representing and reasoning about abstract objects (such as numbers, points, spaces, sets, structures, and games). Mathematics is used throughout the world as an essential tool in many fields, including natural science, engineering, medicine, and the social sciences. Used for calculation, it is considered the most important subject. Applied mathematics, the branch of mathematics concerned with application of mathematical knowledge to other fields, inspires and makes use of new mathematical discoveries and sometimes leads to the development of entirely new mathematical disciplines, such as statistics and game theory. Mathematicians also engage in pure mathematics, or mathematics for its own sake, without having any application in mind. There is no clear line separating pure and applied mathematics, and practical applications for what began as pure mathematics are often discovered. (Full article...)

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Credit: User:Avenue based on original by User:Schutz (data by Francis Anscombe)
Anscombe's quartet is a collection of four sets of bivariate data (paired xy observations) illustrating the importance of graphical displays of data when analyzing relationships among variables. The data sets were specially constructed in 1973 by English statistician Frank Anscombe to have the same (or nearly the same) values for many commonly computed descriptive statistics (values which summarize different aspects of the data) and yet to look very different when their scatter plots are compared. The four x variables share exactly the same mean (or "average value") of 9; the four y variables have approximately the same mean of 7.50, to 2 decimal places of precision. Similarly, the data sets share at least approximately the same standard deviations for x and y, and correlation between the two variables. When y is viewed as being dependent on x and a least-squares regression line is fit to each data set, almost the same slope and y-intercept are found in all cases, resulting in almost the same predicted values of y for any given x value, and approximately the same coefficient of determination or R² value (a measure of the fraction of variation in y that can be "explained" by x, or more intuitively "how well y can be predicted" from x). Many other commonly computed statistics are also almost the same for the four data sets, including the standard error of the regression equation and the t statistic and accompanying p-value for testing the significance of the slope. Clear differences between the data sets are apparent, however, when they are graphed using scatter plots. The plots even suggest particular reasons why y cannot be perfectly predicted from x using each regression line: (1) While the variables are roughly linearly related in the first data set, there is more variability in y than can be accounted for by x, as seen in the vertical spread of the points around the regression line; in this case, one or more additional independent variables may be needed to account for some of this "residual" variation in y. (2) The second scatter plot shows strong curvature, so a simple linear model is not even appropriate for the data; polynomial regression or some other model allowing for nonlinear relationships may be appropriate. (3) The third data set contains an outlier, which ruins the otherwise perfect linear relationship between the variables; this may indicate that an error was made in collecting or recording the data, or may reveal an aspect of the variation of y that has not been considered. (4) The fourth data set contains an influential point that is almost completely determining the slope of the regression line; the reliability of the line would be increased if more data were collected at the high x value, or at any other x values besides 8. Although some other common summary statistics such as quartiles could have revealed differences across the four data sets, the plots give additional information that would be difficult to glean from mere numerical summaries. The importance of visualizing data is magnified (and made more complicated) when dealing with higher-dimensional data sets. Multiple regression is a straightforward generalization of linear regression to the case of multiple independent variables, while "multivariate" regression methods such as the general linear model allow for multiple dependent variables. Other statistical procedures designed to reveal relationships in multivariate data (several of which are closely tied to useful graphical depictions of the data) include principal component analysis, factor analysis, multidimensional scaling, discriminant function analysis, cluster analysis, and many others.

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 Blaise PascalImage credit: User:Anarkman

Blaise Pascal (pronounced [blez pɑskɑl]), (June 19, 1623 – August 19, 1662) was a French mathematician, physicist, and religious philosopher. He was a child prodigy who was educated by his father. Pascal's earliest work was in the natural and applied sciences where he made important contributions to the construction of mechanical calculators, the study of fluids, and clarified the concepts of pressure and vacuum by generalizing the work of Evangelista Torricelli. Pascal also wrote powerfully in defense of the scientific method.

A mathematician of the first order, Pascal helped create two major new areas of research. He wrote a significant treatise on projective geometry at the age of sixteen and corresponded with Pierre de Fermat from 1654 on probability theory, strongly influencing the development of modern economics and social science.

Following a mystical experience in late 1654, he abandoned his scientific work and devoted himself to philosophy and theology. However, he had suffered from ill-health throughout his life and his new interests were ended by his early death two months after his 39th birthday. (Full article...)

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text:
# Convergence of $L^p$ of approximation

Let $f \in L^p(\mathbb R^n)$ be given. Consider a partition of rectangles $I_{ij}:=[x_i,x_{i+1}]\times [x_j,x_{j+1}]$ of $\mathbb R^2.$

Then, we may define the coefficients

$$\alpha_{ij}= \frac{1}{\left\lvert I_{ij} \right\rvert} \int_{I_{ij}} f(s) \ ds$$

and consider the function $$g(x):=\sum_{ij} \alpha_{ij}\chi_{I_{ij}}(x).$$

I ask: If the distance between $x_i$ and $x_{i+1}$ goes uniformly in $i$ to zero. Does this imply that $g$ converges in $L^p$ to $f$? The functions $\chi$ are just indicator functions.

• Since you have accepted @IosefPinelis's answer, it is probably polite to upvote it. – LSpice May 1 '18 at 19:19

The answer is yes, at least for $p\ge1$. This follows because compactly supported continuous functions are dense in $L^p$ and such functions are uniformly continuous.
Here are details. For any $f\in L^p(\R^n)$, let $\bar f$ be defined as your function $g$: \begin{equation*} \bar f:=\sum_{ij} \al_{f;ij}\chi_{I_{ij}}, \end{equation*} where \begin{equation*} \al_{f;ij}:=\frac1{|I_{ij}|} \int_{I_{ij}} f(s)\,ds. \end{equation*}
Take any real $\ep\in(0,1)$ and take any continuous function $f_\ep$ with a support $K_\ep\subseteq[-M_\ep,M_\ep]^n$ for some real $M_\ep>0$ such that \begin{equation*} \|f-f_\ep\|_p<\ep. \tag{1} \end{equation*} Let $$\ep_1:=\ep/(2M_\ep+2)^{n/p}.$$ By the uniform continuity of $f_\ep$, there is some real $\de_\ep>0$ such that \begin{equation*} |f_\ep-\bar{f_\ep}|=\sum_{ij} |f_\ep-\bar{f_\ep}|\,\chi_{I_{ij}}\le \sum_{ij} \ep_1\chi_{I_{ij}}=\ep_1 \end{equation*} as soon as \begin{equation*} \max_i(x_{i+1}-x_i)<\de_\ep. \tag{2} \end{equation*} Hence, \begin{multline*} \|f_\ep-\bar{f_\ep}\|_p^p =\int_{\R^n} |f_\ep-\bar{f_\ep}|^p =\int_{[-M_\ep-1,M_\ep+1]^n} |f_\ep-\bar{f_\ep}|^p \\ \le\ep_1^p(2M_\ep+2)^n=\ep^p. \tag{3} \end{multline*} Also, for $h:=f-f_\ep$, by Jensen's inequality and (1), \begin{multline*} \|\bar f-\bar{f_\ep}\|_p^p=\|\,\bar h\,\|_p^p=\int_{\R^n}|\,\bar h\,|^p =\sum_{ij}\int_{I_{ij}}|\,\bar h\,|^p =\sum_{ij}|\,\bar h\,|^p\,|I_{ij}| \\ \le\sum_{ij}\bar{|h|^p}\,|I_{ij}| =\sum_{ij}\int_{I_{ij}}|h|^p =\|h\|_p^p=\|f-f_\ep\|_p^p<\ep^p. \tag{4} \end{multline*} Thus, by (1), (3), (4), and Minkowski's inequality, \begin{equation*} \|f-\bar f\|_p\le3\ep \end{equation*} as soon as (2) holds.


text:
## Bifurcation of Internal Solitary Wave Modes from the Essential Spectrum

### Authors: P.G. Kevrekidis and C.K.R.T. Jones

ABSTRACT

The bifurcation of internal modes from the phonon band in models supporting solitary wave solutions has been one of the exciting new phenomena in the field. We will present a number of analytical and semi-analytical techniques for the detection, study and understanding of these modes. We will see how they appear, without threshold, due to the discretization of the continuum equations. This perturbation is viewed in terms of a singular continuum approximation and analyzed by both perturbation theory and the Evans' function method. It is shown that these methods give equivalent results. Moreover, they are corroborated by mixed analytical-numerical computations based on the discrete Evans' function recently developed by the first author. The extent to which these predictions survive to strong discretizations are discussed. The results will be presented in the context of both the sine-Gordon and the $\phi^4$ models.

Paper Available at: ftp://dimacs.rutgers.edu/pub/dimacs/TechnicalReports/TechReports/1999/99-38.ps.gz


text:
# [NTG-context] References broken when using more than one component?

Steffen Fritzsche steffen.fritzsche at uni-ulm.de
Wed Mar 21 13:54:23 CET 2012

Hi,

I've got a deep hierarchy of components, while everything builds fine I have a problem using references. Have a look at the following sample files:

reftest.tex:

\starttext

\chapter[one]{Chapter One}

\section[bla]{Blablabla}
See \in{Section}[three:foobar]

\chapter[two]{Chapter Two}

\section[blubb]{BlubbBlubbBlubb}

\subsection{BlubberBlubber}
See \in{Section}[one:bla]

\component refcomp

\stoptext

refcomp.tex:
\startcomponent refcomp
\chapter[three]{Chapter Three}
\component refcompsec
\stopcomponent

refcompsec.tex:
\section[foobar]{Foobar Section}
See \in{Section}[two:blubb] for more details!
\stopcomponent

If I move the second component "refcompsec.tex" into the "refcomp.tex" component the reference in chapter one is found. If I translate it as shown using two components the reference in the first chapter is lost.

Is this a bug or am I doing something wrong?

Cheers,
Steffen
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text:
Hey! I'm David, the author of the Real-World Cryptography book. I'm a crypto engineer at O(1) Labs on the Mina cryptocurrency, previously I was the security lead for Diem (formerly Libra) at Novi (Facebook), and a security consultant for the Cryptography Services of NCC Group. This is my blog about cryptography and security and other related topics that I find interesting.

more on the next page...

# Project Memento: An NFT art project posted Yesterday

Believe it or not, even though I was one of the participant in the standardization of ERC-721, the Ethereum standard for NFCs, I had never read the final standard, yet alone written an ERC-721 smart contract. At the time I was against the standard, at least in its form, due to its collision with the ERC-20 standard (you can't create a smart contract that is both a compliant ERC-20 token and a compliant ERC-721 NFT). My arguments ended up losing, ERC-721 was standardized with the same function names as ERC-20, and I heard that today another standard is surfacing (ERC 1155) to fix this flaw (among others).

Anyway, I recently decided to bury the hatchet and read the damn standard. And after a few weekends of hacking with my friends Eric Khun and Simon Patole, we wrote an ERC-721 compliant dapp as an art project. It's called project memento and it boasts a grid of tiles each containing a letter, a-la million dollar page. People can acquire letters, alter them, and form lines (or diagonals) of words together. I'm curious as to what will become of this, but this was fun!

You can check out project memento here

comment on this story

# The Real-World Cryptography book is done and shipping! posted 2 weeks ago

Today I received the first copy of my book Real-World Cryptography! You can't imagine how much sweat and love I put into this work in the last two years and a half. It's the book I've always wanted to read, and it's the book I thought the field was missing. It's the book I wanted to have as a student when I was learning about hash functions and ciphers, and it's the book I wish I could have referred to my fellow pentesters when they had questions about TLS or end-to-end encryption. It's the book I'd use as a developer looking for cryptographic libraries and best practice. It's the first book with a cryptocurrency chapter, and it's the book cryptographers will read to learn about password-authenticated key exchanges, sponge functions, the noise protocol framework, post-quantum cryptography, and zero-knowledge proofs. Real-world cryptography is what the field of applied cryptography really looks like today. It's all there.

Get the book here!

comment on this story

# Looking for a cryptography engineer job? Interested in zero-knowledge proofs? posted 3 weeks ago

Hey you!

If you've been following my blog recently, you must have seen that I joined O(1) Labs to work on the Mina cryptocurrency. Why? Simply because I felt like it was the most interesting project in the space, and now that I'm deep in OCaml and Rust code trying to understand PLONK and other state-of-the-art zero-knowledge proof systems work I can tell you that it is indeed the most interesting project in the space :)

My two zero-knowledge friends are too busy doing a PhD. They like self-inflicted pain. (Mathias and Michael, I'm looking at you.) So if you're free, ping me! And if you're not free, ping me anyway because we might be able to work something out.

Because people need a job description, check that one out: https://boards.greenhouse.io/o1labs/jobs/4023646004

But realistically, just contact me on twitter or here if you have any question about what the culture of the company is, what your day-to-day would look like, and if you can work remotely (yes).

comment on this story

# How does PLONK work? Part 11: Our final protocol! (Without the copy constraints) posted last month

In this eleventh video, I go back to the PLONK protocol and finally explain how it works with polynomial commitments. This version of the protocol is not finished, as it doesn't have zero-knowledgeness (the polynomial evaluations leak information about the polynomials) and the wiring (or copy constraint) has not been enforced (e.g. the output wire of this gate should be the left wire of this other gate). In the next video, I will explain how copy constraints can be enforced via the PLONK permutation argument.

Stay tuned for part 12... Check the full series here.

comment on this story

# How does PLONK work? Part 10: The Kate polynomial commitment scheme posted last month

In this tenth video, I explain how the Kate polynomial commitment scheme works. For more information about it, check this other blogpost I wrote. This polynomial commitment scheme will be useful to force the prover to commit to its polynomials before learning the random point they need to be evaluated at.

comment on this story

# How does PLONK work? Part 9: Our final protocol! (Without the copy constraints) posted last month

In this ninth video, I explain what polynomial commitment schemes are as well as their API. I also mention the Kate polynomial commitment scheme (KZG), based on pairings, and bootle/bulletproof types of polynomial commitments schemes, based on inner products.

comment on this story

# How does PLONK work? Part 8: A polynomial dance posted last month

In this eighth video, I explain how the prover and the verifier can perform a "polynomial dance" in order to construct the circuit polynomial $f$. The principle is simple: the prover doesn't want to leak information about the private inputs and the intermediary values in the circuit, and the verifier doesn't want to give the prover too much freedom in the way they construct the circuit polynomial $f$.

comment on this story

# How does PLONK work? Part 7: A sketch protocol with our polynomial posted last month

In this seventh video, I explain how we use our circuit polynomial $f$ in a protocol between a prover and a verifier to prove succinctly that $f$ vanishes on a number of specified points.

Stay tuned for part 9... Part 8 is here. Check the full series here.

comment on this story


text:
MathSciNet bibliographic data MR2139098 (2006b:17027) 17B37 (17B10) Ondrus, Matthew Whittaker modules for \$U\sb q({\germ{sl}}\sb 2)\$$U\sb q({\germ{sl}}\sb 2)$. J. Algebra 289 (2005), no. 1, 192–213. Article

For users without a MathSciNet license , Relay Station allows linking from MR numbers in online mathematical literature directly to electronic journals and original articles. Subscribers receive the added value of full MathSciNet reviews.


text:
# How to symbolically solve this transcendental equation

The equation below shows up when dealing with geodesic curves on elliptic paraboloids:

$$$$u-c^2=u(1+4c^2)\sin^2\left(v-2c \ln\left(k(2\sqrt{u-c^2}+\sqrt{4u+1})\right)\right) \tag{1}$$$$

Obtaining a solution for $$v(u)$$ is rather straightforward using Solve, like so:

Solve[u-c^2==u(1+4 c^2) Sin[v-2c Log[k (2 Sqrt[u-c^2]+Sqrt[4u+1])]]^2,v,Reals]

[output omitted]

It would be much more convenient to find the inverse function $$u(v)$$, so I tried solving for $$u$$, but MMA gave up with "... cannot be solve with the methods available to Solve". That's when I gave Reduce a shot:

Reduce[u-c^2==u(1+4c^2)Sin[v-2c Log[k\left(2 Sqrt[u-c^2]+Sqrt[4u+1]\right)]]^2,u,Reals]

Same message as before.

Is there a way to make this equation more tractable for Mathematica, to let it arrive at a symbolic solution for $$u(v)$$?

NB: $$c$$ and $$k$$ are constants, more precisely $$c,k \in \mathbb{R}$$; $$u \geq 0$$ ($$u$$ denotes the radius)

Edit: How can we obtain the paraboloid geodesic $$(1)$$ above?

First, define the paraboloid surface as follows:

$$$$(u,v)\mapsto\begin{pmatrix}\sqrt{u}\cos{v} \\ \sqrt{u}\sin{v} \\ u \end{pmatrix} \tag{2}$$$$

In order to trace out a complete paraboloid, let $$u\geq 0$$ and $$v \in [0,2\pi)$$. I would rather use $$r$$ and $$\varphi$$ to make their meaning a little more clear, but let's stick to the usual naming scheme w.r.t. differential geometry.

Next, calculate the general solution to the geodesics using the Euler-Lagrange equation:

$$$$\underbrace{\frac{\frac{\partial P}{\partial v}+2v\,'\frac{\partial Q}{\partial v}+v\,'^2\frac{\partial R}{\partial v}}{2\sqrt{P+2Qv\,'+Rv\,'^2}}}_{\displaystyle{=0}}-\frac{d}{du}\left(\frac{Q+Rv'}{\sqrt{P+2Qv\,'+Rv\,'^2}}\right)=0$$$$

$$P, Q, R$$ are the coefficients of the first fundamental form; I managed to verify that the solution indeed matches equation $$(1)$$ given in the literature, see here.

Rearrange $$(1)$$ so we have $$v(u)$$, and use that inside $$(2)$$ to describe geodesics that travel across the paraboloid surface.

I was able to come up with a Notebook which computes the integration constants $$c,k$$ to yield the specific geodesic passing through two points $$(u_1,v_1),(u_2,v_2)$$.

However, if these points have the same "height" on the paraboloid, that is $$u_1=u_2$$, Mathematica won't find a solution. I assume knowing the inverse function $$u(v)$$ would be helpful here.

• At best you can find an inverse series InverseSeries[ Series[c^2 + (1 + 4 c^2) u Sin[v - 2 c Log[k (2 Sqrt[-c^2 + u] + Sqrt[1 + 4 u])]]^2, {u, 0, 4}]] but it's extremely unlikely you'll ever find a closed form inverse for a transcendental problem like this. Only numerical solutions are possible, and that might include fixed point iteration too. Sep 5, 2020 at 22:16
• This could be a good question if you'd provide appropriate meaning of constants and variables with more or less detailed information how the equation can be derived from description of geodesics. I guess that the parametrisation in terms of u and v is not optimal, nevertheless providing appropriate restriction on the variables and parameteres one would get much more. If one can use $v(u)$ it is also possible to have $u(v)$ using InverseFunction with appropriate restrictions although I cannot promise that analytical formula can be obtained. Sep 5, 2020 at 23:52
• In any case it would be constructive to play with e.g. Manipulate[ContourPlot[u-c^2==u(1+4 c^2) Sin[v-2c Log[k (2 Sqrt[u-c^2]+Sqrt[4u+1])]]^2,{v,-4,4},{u,0,8},AspectRatio->Automatic,ContourStyle->Thick],{c,-3,3},{k,0.1,3}] to get an idea what one could expect. Sep 5, 2020 at 23:55
• @Artes: added some info to my post above. Sep 6, 2020 at 12:08
• I have never seen ContourPlot used in this way - how would I find an exact solution, if there is any? Sep 9, 2020 at 16:14

In situations like this I use InverseFunction. Although it does not generate an analytical expression, you still end up with something which returns an exact result rather than an approximate one:

lhsMinusRhs[c_, k_][u_, v_] := u-c^2 - u(1+4 c^2) Sin[v-2c Log[k (2 Sqrt[u-c^2]+Sqrt[4u+1])]]^2
assumptions[c_, k_][u_, v_] := u >= 0;
uFunction[c_, k_][v_] :=
InverseFunction[
Function[{u},
ConditionalExpression[
lhsMinusRhs[c, k][u, v],
assumptions[c, k][u, v]
] // Evaluate
]
][0] // Evaluate;

For example, uFunction[3, 4][5] will return an exact Root object corresponding to $$(c, k) = (3, 4)$$ and $$v = 5$$.

uFunction[3, 4][5]
(* Root[{..., 16.33...}] *)
lhsMinusRhs[3, 4][%, 5] // FullSimplify
(* 0 *)

Note that InverseFunction makes no guarantees on which branch of the inverse will be chosen. If I am after a specific branch, the approach I take is to make assumptions specific enough to pin down the unique value corresponding to that branch.

• It seems that InverseFunction might have a bug: (219309) Sep 9, 2020 at 6:38
• Your code for uFunction[3, 4][5] // N returns approximately 16.33 on my machine as well, so I tried to plot uFunction to see how it behaves for different values of $c,k$: Manipulate[Plot[uFunction[a, b][val] // N, {val, 0, 2 \[Pi]}], {{a, 3}, -5, 5, Appearance -> "Labeled"}, {{b, 4}, -5, 5, Appearance -> "Labeled"}], but I get an empty plot, no matter what... Sep 17, 2020 at 12:53
• (1/2) @user2286339 I'm not sure why this happens, but InverseFunction seems reluctant to evaluate when supplied with inexact input; compare the output of uFunction[3, 4][5] // N and uFunction[3, 4][5.] // N. An empty plot results because Plot numericises the sampling values of val. Sep 17, 2020 at 17:45
• (2/2) If the goal is to make a plot, I would not go down the analytical route. Instead I'd make a numerical function, e.g. uNumerical[u0_][c_, k_][v_] := FindRoot[lhsMinusRhs[c, k][#, v] &, {u0}] (where u0 is the initial guess), which is much faster to plot: Manipulate[Plot[uNumerical[15][a, b][val] // N, {val, 0, 2 \[Pi]}], {{a, 3}, -5, 5, Appearance -> "Labeled"}, {{b, 4}, -5, 5, Appearance -> "Labeled"}] Sep 17, 2020 at 17:45
• Originally, I wanted to rearrange eq. (1) in terms of $u$ which seems to have no analytical form, when @Artes hinted at using ContourPlot to find it, if there is one. How would I do that? His code snippet above (ContourPlot) suggests a 2\Pi periodicity, but that is all that I was able to gather from the plot. Sep 17, 2020 at 20:11


text:
# Question #2570f

$\frac{4 w \sqrt{5 w}}{3}$
80 does not have an exact square root, but it has factors which are perfect squares. $\sqrt{\frac{16 \times 5 \times {w}^{2} \times w}{9}} = \sqrt{\frac{{4}^{2} \times 5 \times {w}^{2} \times w}{3} ^ 2}$
= $\frac{4 w \sqrt{5 w}}{3}$


text:
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# Difficult Math Question

Author Message
Intern
Joined: 01 Oct 2008
Posts: 2
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### Show Tags

01 Oct 2008, 17:37
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Hey

Does anyone know how to solve this question? I'm having some trouble with it.

A Merchant purchased a jacket for $60.00 and then determined a selling price that equaled the purchase price of the jacket plus a markup that was 25% of the selling price. During a sale, the merchant discounted the selling price by 20% and sold the jacket. What was the merchant's gross profit on this sale? a)$0
b) $3 c)$4
d) $12 e)$15

I've narrowed the answer down to a choice between b and c since the other answer choices seems impossible, but I still can't seem to figure out how to solve the problem.

Any help would be appreciated!

Cheers
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Joined: 18 May 2008
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### Show Tags

01 Oct 2008, 17:57
Ans C
Let us denote Selling Price by S
By first statement , S=60+0.25*S
Thus S=80
But after discount of 20%,final selling Price is 64
Thus Gross profit is Final Selling Price-Purchase Price
=64-6=4

plinny wrote:
Hey

Does anyone know how to solve this question? I'm having some trouble with it.

A Merchant purchased a jacket for $60.00 and then determined a selling price that equaled the purchase price of the jacket plus a markup that was 25% of the selling price. During a sale, the merchant discounted the selling price by 20% and sold the jacket. What was the merchant's gross profit on this sale? a)$0
b) $3 c)$4
d) $12 e)$15

I've narrowed the answer down to a choice between b and c since the other answer choices seems impossible, but I still can't seem to figure out how to solve the problem.

Any help would be appreciated!

Cheers
Intern
Joined: 01 Oct 2008
Posts: 2
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02 Oct 2008, 15:21
Great! Thanks for all the help!
Re: Difficult Math Question   [#permalink] 02 Oct 2008, 15:21
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# Difficult Math Question

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text:
## Combining and comparing models using Model Confidence Set

By Gabriel Vasconcelos

In many cases, especially if you are dealing with forecasting models, it is natural to use a large set of models to forecast the same variable and then select the best model using some error measure. For example, you can break the sample into a training sample (in-sample) and a test sample (out-of-sample), estimate all models in the training sample and see how the perform in the test sample. You could compare the models using the root mean squared error (RMSE) or the mean absolute error (MAE).

If you want to make things a little more formal, then you can use some statistical test such as Diebold-Mariano to compare the models pairwise. This test will tell you if model A is better than model B. If you have $n$ models to compare you will need to perform $n(n-1)/2$ tests. If  $n$ is large you will have a lot of test results to look at and reporting the results will become a problem (imagine that you have 100 models, you will have 4950 p-values to report).

Fortunately, Hansen, Lunde and Nason proposed (here) a suitable solution in an article in 2011. Their solution is called Model Confidence Set (MCS). The MCS has an interpretation similar to a confidence interval, i. e. the $MCS_{1-\alpha}$ contains the best model with $100(1-\alpha)\%$ confidence. The number of models in the MCS increases as we decrease $\alpha$ just like the size of a confidence interval. You only need to perform the MCS procedure once to compare all models and you have to report only the p-value for each model, which is an output of the test.

## Application

Before going to the application, let us define a function that calculates the RMSE for several models at the same time, which we will be using a lot.

rmse=function(real,...){
forecasts=cbind(...)
sqrt(colMeans((real-forecasts)^2))
}

To the application!! We are going to generate some data with the following design:

• 300 observations ($t=1,\dots,300$) and 8 variables ($n=1,\dots,8$),
• All variables are generated from $y_{i,t}=\phi y_{i,t-1}+\lambda_if_t +\varepsilon_t$ with $\phi=0.6$, $\lambda$ is sampled from a uniform(-1,1) distribution and $\varepsilon_{i,t}$ is Normal with mean equals 0 and variance equals 4,
• $f_t$ is a common factor generated from $f_{t}=\rho f_{t-1}+\epsilon_t$ with $\rho=0.75$ and $\epsilon_t$ is Normal with mean equals 0 and variance equals 4,
• I generated two independent factors: four variables are generated from factor 1 and the remaining four variables are generated from factor 2.
• We assume that we do not know which factor generates each variable.
library(reshape2)
library(grid)
library(gridExtra)
library(ggplot2)
library(boot)
T = 300 # = number of observations = #
n = 8 # = Number of Variables = #
rho = 0.75 # = factors AR coefficient = #
phi = 0.6 # = Variables AR coefficient = #

set.seed(1) # = seed for replication = #

# = Generate two autoregressive factors = #
f = matrix(0, T, 2)
for(i in 2:T){
f[i, ] = rho * f[i-1, ] + rnorm(2)
}

# = Generate variables = #
Y = matrix(0, T, n)
for(i in 2:T){
# = Based on the first factor = #
Y[i, 1:4] = phi * Y[i-1, 1:4] + lambda[1:4] * f[i, 1] + rnorm(4, 0, 2)
# = Based on the second factor = #
Y[i, 5:8] = phi * Y[i-1, 5:8] + lambda[5:8] * f[i,2] + rnorm(4, 0, 2)
}

# = Organize variable lags = #
# = y is the first (dependent) variable = #
Y = as.data.frame(embed(Y, 2)[ , c(1, 9:16)])
# = y is the first (dependent) variable = #
colnames(Y) = c("y", paste("x", 1:n, sep = ""))

Now that we have our data, we are going to estimate all possible combinations of linear regressions using our 8 variables. The first variable will be the dependent variable, defined as $y$. The explanatory variables are the first lags of all eight variables (including the first), defined as $x_i, i=1,\dots,8$. This design results in 255 models to be estimated and compared.

# = First we will generate formulas for all the 255 equations = #
variables = colnames(Y)[-1]
combinations = lapply(1:n, function(x)combn(1:n, x))
formulas = lapply(combinations, function(y){
apply(y, 2, function(x){
paste("~", paste(variables[x], collapse = " + "), sep=" ")
})
})
formulas = unlist(formulas)

# = Now we break the sample into in-sample and out-of-sample = #
in_sample = Y[1:100, ]
out_of_sample = Y[-c(1:100), ]

# = Estimate the models = #
models = lapply(formulas, function(x) lm(paste("y", x), data = in_sample))
# = Compute forecasts = #
forecasts = Reduce(cbind, lapply(models, function(x) predict(x, out_of_sample)))
colnames(forecasts) = paste("model", 1:ncol(forecasts), sep = "")

# RMSES (Mean of all models, model with all variables)
rmse(out_of_sample$y,rowMeans(forecasts),forecasts[,ncol(forecasts)])  ## [1] 2.225175 2.322508  As you can see, the average forecast is more accurate than the model using all 8 variables. Now we are going to compute the MCS using the function below. If you want more details on the algorithm, just look at the original article here. mcs=function(Loss,R,l){ LbarB=tsboot(Loss,colMeans,R=R,sim="fixed",l=l)$t
Lbar=colMeans(Loss)
zeta.Bi=t(t(LbarB)-Lbar)

save.res=c()
for(j in 1:(ncol(Loss)-1)){
Lbardot=mean(Lbar)

t.stat=(Lbar-Lbardot)/sqrt(vard)
t.max=max(t.stat)
model.t.max=which(t.stat==t.max)

t.max.b=apply(t.stat.b,1,max)

p=length(which(t.max<t.max.b))/R

save.res=c(save.res,p)
names(save.res)[j]=names(model.t.max)

Lbar=Lbar[-model.t.max]
zeta.Bi=zeta.Bi[,-model.t.max]

}
save.res=c(save.res,1)
names(save.res)[j+1]=names(Lbar)
save.p=save.res
for(i in 2:(length(save.res)-1)){
save.p[i]=max(save.res[i-1],save.res[i])
}
aux=match(colnames(Loss),names(save.p))
save.p=save.p[aux]
save.res=save.res[aux]
return(list(test=save.p,individual=save.res))
}

The MCS is computed in a single line of code, but first we are going to break the out-of-sample into two sub-samples. We are going to compute the MCS in the first sub-sample and compare the models in the second sub-sample. The MCS function returns the p-value needed to include each model in the set. For example, if model1 has a p-value=0.1, it will be included in the MCS of 90\% confidence.

# = generate 2 subsamples = #
out1 = out_of_sample[1:100, ]
out2 = out_of_sample[-c(1:100), ]
f1 = forecasts[1:100, ]
f2 = forecasts[-c(1:100),]

# = Compute MCS = #
set.seed(123) # = Seed for replication = #
MCS=mcs( Loss=(f1-out1$y)^2, R=1000, l=3 )$test # = R block bootstrap samples with block length = 3  = #
# = If you want to see the p-values write (print(sort(MCS)) = #
# = Selected models in the 50% MCS  = #
selected=which(MCS>0.5)

# RMSE (Mean of all models, model with all variables, MCS models)
rmse(out2$y,rowMeans(f2),f2[,ncol(f2)],rowMeans(f2[,selected]))  ## [1] 2.142932 2.240110 2.013547  As you can see, if we combine only the models in the MCS the RMSE becomes smaller. This is not always the case, but it happens a lot. The combined results are better when the forecasts do not have a very high positive correlation. The MCS is not useful only to combine forecasts. Even if the combinations are not better than individual models, you still have a confidence set that includes the best model with the desired significance. If some model is excluded from the confidence set you have strong evidence that this model is worst than the ones in the set (depending on the confidence level, of course). Naturally, if you increase the confidence level, the number of models in the set also increases just like in confidence intervals. Finally, we can find some interesting relationships in the plots below. The first plot shows the forecasting errors as function of the confidence level. The optimal confidence level is around 55\%. If we increase it more the MCS starts to select a lot o bad models and the forecast becomes less accurate. The second plot shows the number of models as a function of the confidence level and the third plot shows the errors as a function of the number of models. alpha = seq(0,1,0.01) nmodels = sapply(alpha, function(x) length(which(MCS >= x))) errors = sapply(alpha, function(x) rmse(out2$y, rowMeans(as.matrix(f2[ , which(MCS >= x)]))))

df = data.frame(confidence = 1 - alpha, errors = errors, nmodels = nmodels)
dfmelt = melt(df, id.vars = "errors")

g1=ggplot(data=df) + geom_point(aes(x=confidence,y=errors))
g2=ggplot(data=df) + geom_point(aes(x=confidence,y=nmodels))
g3=ggplot(data=df) + geom_point(aes(x=nmodels,y=errors))
grid.arrange(g1,g2,g3,ncol=2)

## References

Hansen, Peter R., Asger Lunde, and James M. Nason. “The model confidence set.” Econometrica 79.2 (2011): 453-497.


text:
# Gödel and Formalism freeness Juliette Kennedy. Bill Tait.

## Presentation on theme: "Gödel and Formalism freeness Juliette Kennedy. Bill Tait."— Presentation transcript:

Gödel and Formalism freeness Juliette Kennedy

Bill Tait

A. Heyting

Carnap diary entry Dec 1929

Gödel “On Russell’s Mathematical Logic” 1944

Kreisel 1972

Gödel Princeton Bicentennial lecture 1946

Joint work with M. Magidor and J. Väänänen

Gödel’s two notions of definability Two canonical inner models: – Constructible sets Model of ZFC Model of GCH – Hereditarily ordinal definable sets Model of ZFC CH? – independent of ZFC

Constructibility Constructible sets (L):

Ordinal definability Hereditarily ordinal definable sets (HOD): – A set is ordinal definable if it is of the form {a : φ(a,α 1,…, α n )} where φ(x,y 1,…, y n ) is a first order formula of set theory. – A set is hereditarily ordinal definable if it and all elements of its transitive closure are ordinal definable.

Myhill-Scott result: Hereditarily ordinal definable sets (HOD) can be seen as the constructible hierarchy with second order logic (in place of first order logic): Chang considered a similar construction with the infinitary logic L ω 1 ω 1 in place of first order logic.

If V=L, then V=HOD=Chang’s model=L. If there are uncountably many measurable cardinals then AC fails in the Chang model. (Kunen.)

C( L *) L * any logic. We define C( L *): C( L *) = the union of all L´ α

Looking ahead: For a variety of logics C( L *)=L – Gödel’s L is robust, not limited to first order logic For a variety of logics C( L *)=HOD – Gödel’s HOD is robust, not limited to second order logic For some logics C( L *) is a potentially interesting new inner model.

Robustness of L Q 1 xφ(x)  {a : φ(a)} is uncountable C( L (Q 1 )) = L. In fact: C( L (Q α )) = L, where – Q α xφ(x)  |{a : φ(a)}| ≥ א α Other logics, e.g. weak second order logic, ``absolute” logics, etc.

Observations: avoiding L C( L ω 1 ω ) = L( R ) C( L ∞ω ) = V

Quantifiers Q 1 MM xyφ(x,y)  there is an uncountable X such that φ(a,b) for all a,b in X – Can express Suslinity. Q 0 cf xyφ(x,y)  {(a,b) : φ(a,b)} is a linear order of cofinality ω – Fully compact extension of first order logic.

Theorems C( L (Q 1 MM )) = L, assuming 0 # C( L (Q 0 cf )) ≠ L, assuming 0 # L µ ⊆ C( L (Q 0 cf )), if L µ exists

What Myhill-Scott really prove In second order logic L 2 one can quantify over arbitrary subsets of the domain. A more general logic L 2,F : in domain M can quantifier only over subsets of cardinality κ with F(κ) ≤ |M|. F any function, e.g. F(κ)=κ, κ +, 2 κ, ב κ, etc L 2 = L 2,F with F(κ)≡0 Note that if one wants to quantify over subsets of cardinality κ one just has to work in a domain of cardinality at least F(κ).

Theorem For all F: C( L 2,F )=HOD Third, fourth order, etc logics give HOD.

Bernays “It seems in no way appropriate that Cantor’s Absolute be identified with set theory formalized in standardized logic, which is considered from a more comprehensive model theory.” -Letter to Gödel, 1961. (Collected Works, vol. 4, Oxford)

Thank you!

Robustness of L (contd.) A logic L * is absolute if ``φ ∈ L *” is Σ 1 in φ and ``M ⊨ φ” is Δ 1 in M and φ in ZFC. – First order logic – Weak second order logic – L (Q 0 ): ``there exists infinitely many” – L ω 1 ω, L ∞ω : infinitary logic – L ω 1 G, L ∞G : game quantifier logic – Fragments of the above