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# If a six sided die is rolled three times, what is the probability of

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If a six sided die is rolled three times, what is the probability of  [#permalink]

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24 Jul 2019, 02:57
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If a six sided die is rolled three times, what is the probability of getting at least one even number and at least one odd number?

A. 1/8
B. 1/4
C. 1/2
D. 3/4
E. 7/8

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If a six sided die is rolled three times, what is the probability of  [#permalink]

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Updated on: 24 Jul 2019, 16:55
# of favourable cases = 3*3*3*6
# of total cases = 6*6*6
Probability = (3*3*3*6)/(6*6*6) = 3/4

IMO D

Posted from my mobile device

Originally posted by Kinshook on 24 Jul 2019, 03:01.
Last edited by Kinshook on 24 Jul 2019, 16:55, edited 1 time in total.
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If a six sided die is rolled three times, what is the probability of  [#permalink]

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Updated on: 24 Jul 2019, 04:23
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The probability of getting at least 1 even and 2 odd number: 3/6*3/6*3/6= 1/2*1/2*1/2=1/8
Also, number of ways of rolling a six sided die like: (EvenOddOdd, Odd,Even,Odd, OddOddEven)—3 ways —> 3*1/8

The probability of getting at least 1 odd and 2 even numbers:
3/6*3/6*3/6= 1/8
Also, number of ways of rolling a six sided die like: (EvenOddEven, Even,Even,Odd, OddEvenEven)—3 ways—> 3*1/8

3/8+3/8= 3/4

Posted from my mobile device

Originally posted by lacktutor on 24 Jul 2019, 03:05.
Last edited by lacktutor on 24 Jul 2019, 04:23, edited 1 time in total.
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Re: If a six sided die is rolled three times, what is the probability of  [#permalink]

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24 Jul 2019, 03:42
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Probability of getting at least one EVEN and at least one ODD $$=$$ Probability that the three rolls are NOT all ODD or NOT all EVEN

Probability of getting 3 ODD's $$= (\frac{3}{6})*(\frac{3}{6})*(\frac{3}{6}) = \frac{1}{8}$$

Probability of getting 3 EVEN's $$= (\frac{3}{6})*(\frac{3}{6})*(\frac{3}{6}) = \frac{1}{8}$$

Probability that the three rolls are all ODD or all EVEN $$= \frac{1}{8}+\frac{1}{8} = \frac{1}{4}$$

Probability that the three rolls are NOT all ODD or NOT all EVEN $$= 1-\frac{1}{4} = \frac{3}{4}$$

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Re: If a six sided die is rolled three times, what is the probability of  [#permalink]

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24 Jul 2019, 03:53
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lacktutor wrote:
The probability of getting at least 1 even and 2 odd number: 3/6*3/6*3/6= 1/2*1/2*1/2=1/8

The probability of getting at least 1 odd and 2 even numbers:
3/6*3/6*3/6= 1/8

1/8+1/8= 1/4

Posted from my mobile device

Dear lacktutor

What you appear to have failed to take into account is - the number of ways you can choose the one die that would be either even or odd, i.e. 3C1

3C1*3/6*3/6*3/6 (for 1 odd and 2 even) and 3C2*3/6*3/6*3/6 (for 2 odd and 1 even)

3*1/8 + 3*1/8 = 6/8 or 3/4, and hence, (D) is the correct answer choice

PLEASE HIT KUDOS IF YOU LIKE MY SOLUTION
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Re: If a six sided die is rolled three times, what is the probability of  [#permalink]

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24 Jul 2019, 04:09
2
Total outcomes = 6*6*6= 216
If all the numbers are odd 3*3*3 = 27
If all the numbers are even 3*3*3= 27
So sum of all even and all odd outcomes = 54
Probability of all odd or all even 54/216 = 1/4
Hence probability of at least one even and one add = 3/4
Hence D

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Re: If a six sided die is rolled three times, what is the probability of  [#permalink]

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24 Jul 2019, 04:27
RJ7X0DefiningMyX wrote:
lacktutor wrote:
The probability of getting at least 1 even and 2 odd number: 3/6*3/6*3/6= 1/2*1/2*1/2=1/8

The probability of getting at least 1 odd and 2 even numbers:
3/6*3/6*3/6= 1/8

1/8+1/8= 1/4

Posted from my mobile device

Dear lacktutor

What you appear to have failed to take into account is - the number of ways you can choose the one die that would be either even or odd, i.e. 3C1

3C1*3/6*3/6*3/6 (for 1 odd and 2 even) and 3C2*3/6*3/6*3/6 (for 2 odd and 1 even)

3*1/8 + 3*1/8 = 6/8 or 3/4, and hence, (D) is the correct answer choice

PLEASE HIT KUDOS IF YOU LIKE MY SOLUTION

Yes, you’re right. I didn’t add the number of combinations within Even and Odd.
Thanks
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Re: If a six sided die is rolled three times, what is the probability of  [#permalink]

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24 Jul 2019, 06:29
Bunuel wrote:
If a six sided die is rolled three times, what is the probability of getting at least one even number and at least one odd number?

A. 1/8
B. 1/4
C. 1/2
D. 3/4
E. 7/8

we are considering cases w/o replacement
so atleast even ; 3/6*3/6*3/6 ; 1/8
and odd ; 1/8
total cases ; 1/8+1/8 ; 1/4
1-1/4 ; 3/4
IMO D
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If a six sided die is rolled three times, what is the probability of  [#permalink]

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05 Jan 2020, 05:10
lacktutor wrote:
The probability of getting at least 1 even and 2 odd number: 3/6*3/6*3/6= 1/2*1/2*1/2=1/8
Also, number of ways of rolling a six sided die like: (EvenOddOdd, Odd,Even,Odd, OddOddEven)—3 ways —> 3*1/8

The probability of getting at least 1 odd and 2 even numbers:
3/6*3/6*3/6= 1/8
Also, number of ways of rolling a six sided die like: (EvenOddEven, Even,Even,Odd, OddEvenEven)—3 ways—> 3*1/8

3/8+3/8= 3/4

Posted from my mobile device

Why we are adding the probabilities at last to get the final resultant 3/4 when the question itself ask to calculate the probability of getting atleast one even number and atleast one odd number ?

*AND* is clearly mentioned : so we should perform multiplication, isn’t it ???

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If a six sided die is rolled three times, what is the probability of  [#permalink]

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05 Jan 2020, 05:14
firas92 wrote:
Probability of getting at least one EVEN and at least one ODD $$=$$ Probability that the three rolls are NOT all ODD or NOT all EVEN

Probability of getting 3 ODD's $$= (\frac{3}{6})*(\frac{3}{6})*(\frac{3}{6}) = \frac{1}{8}$$

Probability of getting 3 EVEN's $$= (\frac{3}{6})*(\frac{3}{6})*(\frac{3}{6}) = \frac{1}{8}$$

Probability that the three rolls are all ODD or all EVEN $$= \frac{1}{8}+\frac{1}{8} = \frac{1}{4}$$

Probability that the three rolls are NOT all ODD or NOT all EVEN $$= 1-\frac{1}{4} = \frac{3}{4}$$

Why can’t we approach like this ;

P(Atleast 1 odd ) = 1-P (All Even ) =1-(1/2)^3= 1-(1/8) = 7/8

P(Atleast 1 Even ) =1-P(All Odd ) = 1-(1/2)^3= 1-(1/8) = 7/8

So final probability = 7/8.7/8 = 49/64

Posted from my mobile device
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Re: If a six sided die is rolled three times, what is the probability of  [#permalink]

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05 Jan 2020, 22:23
LeenaSai wrote:
firas92 wrote:
Probability of getting at least one EVEN and at least one ODD $$=$$ Probability that the three rolls are NOT all ODD or NOT all EVEN

Probability of getting 3 ODD's $$= (\frac{3}{6})*(\frac{3}{6})*(\frac{3}{6}) = \frac{1}{8}$$

Probability of getting 3 EVEN's $$= (\frac{3}{6})*(\frac{3}{6})*(\frac{3}{6}) = \frac{1}{8}$$

Probability that the three rolls are all ODD or all EVEN $$= \frac{1}{8}+\frac{1}{8} = \frac{1}{4}$$

Probability that the three rolls are NOT all ODD or NOT all EVEN $$= 1-\frac{1}{4} = \frac{3}{4}$$

Why can’t we approach like this ;

P(Atleast 1 odd ) = 1-P (All Even ) =1-(1/2)^3= 1-(1/8) = 7/8

P(Atleast 1 Even ) =1-P(All Odd ) = 1-(1/2)^3= 1-(1/8) = 7/8

So final probability = 7/8.7/8 = 49/64

Posted from my mobile device

The probability of two independent events occurring together is given by the product of their probabilities. These two events (at least one odd and at least one even) are not independent. So this method is not correct.
From the total, if you remove the probability of all odd and all even (mutually exclusive events so you just subtract their probabilities one after the other) event, you get that the remaining cases will have at least one odd and one even outcome.
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Re: If a six sided die is rolled three times, what is the probability of  [#permalink]

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05 Jan 2020, 22:38
LeenaSai wrote:
firas92 wrote:
Probability of getting at least one EVEN and at least one ODD $$=$$ Probability that the three rolls are NOT all ODD or NOT all EVEN

Probability of getting 3 ODD's $$= (\frac{3}{6})*(\frac{3}{6})*(\frac{3}{6}) = \frac{1}{8}$$

Probability of getting 3 EVEN's $$= (\frac{3}{6})*(\frac{3}{6})*(\frac{3}{6}) = \frac{1}{8}$$

Probability that the three rolls are all ODD or all EVEN $$= \frac{1}{8}+\frac{1}{8} = \frac{1}{4}$$

Probability that the three rolls are NOT all ODD or NOT all EVEN $$= 1-\frac{1}{4} = \frac{3}{4}$$

Why can’t we approach like this ;

P(Atleast 1 odd ) = 1-P (All Even ) =1-(1/2)^3= 1-(1/8) = 7/8

P(Atleast 1 Even ) =1-P(All Odd ) = 1-(1/2)^3= 1-(1/8) = 7/8

So final probability = 7/8.7/8 = 49/64

Posted from my mobile device

The probability of two independent events occurring together is given by the product of their probabilities. These two events (at least one odd and at least one even) are not independent. So this method is not correct.
From the total, if you remove the probability of all odd and all even (mutually exclusive events so you just subtract their probabilities one after the other) event, you get that the remaining cases will have at least one odd and one even outcome.

Now it’s pretty clear VeritasKarishma ?
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Re: If a six sided die is rolled three times, what is the probability of  [#permalink]

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05 Jan 2020, 22:56
E O E/O . This anagram wheter EOO or EOE cn be permuted in 3!/2! ways = 3 ways.
Further,

getting probability of (E) (O) (E/O) = 1/2 * 1/2 * 6/6 = 1/4 and now we will permute this EOE or EEO in 3 ways so
multiply it with obtained probability = 1/4 * 3 = 3/4 ----> D
Re: If a six sided die is rolled three times, what is the probability of   [#permalink] 05 Jan 2020, 22:56
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text:
• 研究论文 •

中国云杉林的地理分布与气候因子间的关系

1. 1中国科学院植物研究所植被与环境变化国家重点实验室, 北京 100093
2中国科学院研究生院, 北京 100049
3徐州师范大学, 江苏徐州 221116
• 发布日期:2012-05-04
• 通讯作者: 王国宏
• 作者简介:* E-mail: ghwangaq@ibcas.ac.cn

Relationship between climatic factors and geographical distribution of spruce forests in China

LI He1,2, ZHANG Wei-Kang1,3, WANG Guo-Hong1,*()

1. 1State Key Laboratory of Vegetation and Environmental Change, Institute of Botany, Chinese Academy of Sciences, Beijing 100093, China
2Graduate University of Chinese Academy of Sciences, Beijing 100049, China
3Xuzhou Normal University, Xuzhou, Jiangsu 221116, China
• Published:2012-05-04
• Contact: WANG Guo-Hong

被引次数

36 | 11

Abstract:

Aims Our objective was to examine the relationship between climatic factors and geographical distribution of spruce forests in China.
Methods We sampled 613 points within the geographical range of Chinese spruce forests, of which 235 points were at the upper altitudinal limit and 228 at the lower altitudinal limit. The elevation for each point was determined using Google Earth while climatic data were from the Chinese meteorological interpolation database. Linear regression, comparison of coefficient of variation (CV) and principal component analysis (PCA) were conducted for data analysis.
Important findings Within the distribution range of Chinese spruce forests, mean values of mean annual air temperature (MAT), mean air temperature of the coldest month (MTCM), mean air temperature of the warmest month (MTWM), growing degree days on a 5 ℃ basis (GDD5) and on a 0 ℃ basis (GDD0), mean annual precipitation (MAP), soil moisture (SM) and aridity index (α) are 3.38 ℃, -9.75 ℃, 14.78 ℃, 1 227.83 ℃·d, 2 271.19 ℃·d, 712.23 mm, 80.02% and 0.50, respectively. Both the upper and lower limits of altitude were significantly correlated with each of the climatic factors. In terms of CV, MAT and MTCM are significantly higher than the other six climatic factors; however, no significant differences were detected among those six. In addition, GDD5 and GDD0 have higher loading on the first principal component, yet MAP and SM have higher loading on the second and third principal component. Major conclusions are that GDD0 and GDD5 are likely the key factors that influence the distribution of Chinese spruce forest, followed by MAP and SM.

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text:
# Split Poisson Process AND severity

I have a Poisson process whose statistics are interarrival times ($\bf X$), number of arrivals ($\bf N$), and arrival times ($\bf T$). Later, the process is split by a Bernoulli process that categorizes each event with a probability $p$ of hit, and $1-p$ of miss.

As a result I get two separate Poisson processes, one for hit ($\textbf{X}_h$,$\textbf{N}_h$,$\textbf{T}_h$) and the other for miss ($\textbf{X}_m$,$\textbf{N}_m$,$\textbf{T}_m$). But I'm not just happy with having both process categorized, I want to assign a probabilistic "severity" degree to hit events.

So suppose, (taking the example from this website) that the Poisson arrivals are radioactive emissions and each emitted particle is either (by Bernoulli) detected (type 1 with probability $p$) or missed (type 0 with probability $1-p$) by a counter:

My question: Is it possible to assign a quantity, like velocity, to the detected particle in the example, by sampling a proper density function?

UPDATE

Maybe what I'm asking is: given that a Poisson process ($\lambda$) is preservable under random selection when counted with probability $p$, and that the thinned process is still Poisson with parameter $\mu=\lambda p$, maybe $p$ can involve the probabilities of two events hapenning, sthg like $p = p_1 \times p_2$, i.e. partitioning $p$ under some criteria. Thus $\mu = \lambda(p_1p_2)$ Does it make sense?

UPDATE 2

Provided that the "thinning probability" $p$ must satify the binomial probability law $\binom {n} {k} p^k q^{n-k}$. Now, say that I take the example from Parzen's, Stoch Proc, 48: "Suppose that customers pass by a shop in accord with a Poisson'process at mean rate $\lambda$. If each customer has probability $p$ of entering the shop, customers enter the shop in accord with a Poisson process with mean rate $\mu=\lambda p$". Now, if I also know that entering customers will also buy cookies with prob $p_1$, how should I factor in this new piece of information to describe the new stochastic process? (PS. It doesn't matter if the new thinned process with isn't Poisson anymore.)"

Thanks!

• (1) Could you elaborate on what is meant by "probabilistic severity degree"? (2) Usually quantities like $X$, $N$, and $T$ do not "define" a Poisson process--they are statistics that can be derived by observations of it and are closely inter-related. Do you perhaps have in mind some kind of generalization of Poisson processes?
– whuber
Dec 1, 2014 at 22:20
• Hi whuber, I updated the post. please check and thanks for the clarification! Dec 1, 2014 at 22:34
• Thanks--that clears up a lot. But now the appearance of "velocity" is surprising. To what property of the process does that refer?
– whuber
Dec 1, 2014 at 22:49
• To no property of the process I guess, but it's related to the physical problem. I'm interested in knowing if there's a framework that contemplates such superposition of different pieces of information... Dec 1, 2014 at 22:52
• The difficulty we're facing here is that up until this comment you haven't provided any information at all to connect the statistical framing of the question (as a thinned Poisson process) to any physical characteristics of the system it is supposed to be describing. It is hard to see how your question could be answered, much less even understood, without such information. Perhaps you could be more specific about what you mean by "superposition" of "pieces of information" and how that might relate to the specific circumstances you contemplate?
– whuber
Dec 1, 2014 at 22:54


text:
## 8.2 Methods with trend

### Holt’s linear trend method

Holt (1957) extended simple exponential smoothing to allow the forecasting of data with a trend. This method involves a forecast equation and two smoothing equations (one for the level and one for the trend): \begin{align*} \text{Forecast equation}&& \hat{y}_{t+h|t} &= \ell_{t} + hb_{t} \\ \text{Level equation} && \ell_{t} &= \alpha y_{t} + (1 - \alpha)(\ell_{t-1} + b_{t-1})\\ \text{Trend equation} && b_{t} &= \beta^*(\ell_{t} - \ell_{t-1}) + (1 -\beta^*)b_{t-1}, \end{align*} where $$\ell_t$$ denotes an estimate of the level of the series at time $$t$$, $$b_t$$ denotes an estimate of the trend (slope) of the series at time $$t$$, $$\alpha$$ is the smoothing parameter for the level, $$0\le\alpha\le1$$, and $$\beta^*$$ is the smoothing parameter for the trend, $$0\le\beta^*\le1$$. (We denote this as $$\beta^*$$ instead of $$\beta$$ for reasons that will be explained in Section 8.5.)

As with simple exponential smoothing, the level equation here shows that $$\ell_t$$ is a weighted average of observation $$y_t$$ and the one-step-ahead training forecast for time $$t$$, here given by $$\ell_{t-1} + b_{t-1}$$. The trend equation shows that $$b_t$$ is a weighted average of the estimated trend at time $$t$$ based on $$\ell_{t} - \ell_{t-1}$$ and $$b_{t-1}$$, the previous estimate of the trend.

The forecast function is no longer flat but trending. The $$h$$-step-ahead forecast is equal to the last estimated level plus $$h$$ times the last estimated trend value. Hence the forecasts are a linear function of $$h$$.

### Example: Australian population

aus_economy <- global_economy %>% filter(Code == "AUS") %>%
mutate(Pop = Population/1e6)
fit <- aus_economy %>%
model(AAN = ETS(Pop ~ error("A") + trend("A") + season("N")))
fc <- fit %>% forecast(h=10)

In Table 8.2 we demonstrate the application of Holt’s method to the annual population of Australia. The smoothing parameters, $$\alpha$$ and $$\beta^*$$, and the initial values $$\ell_0$$ and $$b_0$$ are estimated by minimising the SSE for the one-step training errors as in Section 8.1.

Table 8.2: Forecasting Australian annual population using Holt’s linear trend method.
Year Time Observation Level Slope Forecast
$$t$$ $$y_t$$ $$\ell_t$$ $$\hat{y}_{t+1|t}$$
1959 0 10.05 0.2225
1960 1 10.28 10.28 0.2224 10.28
1961 2 10.48 10.48 0.2172 10.50
1962 3 10.74 10.74 0.2309 10.70
1963 4 10.95 10.95 0.2234 10.97
1964 5 11.17 11.17 0.2213 11.17
1965 6 11.39 11.39 0.2212 11.39
1966 7 11.65 11.65 0.2349 11.61
2014 55 23.50 23.50 0.3656 23.52
2015 56 23.85 23.85 0.3594 23.87
2016 57 24.21 24.21 0.3596 24.21
2017 58 24.60 24.60 0.3689 24.57
$$h$$ $$\hat{y}_{T+h|T}$$
2018 1 24.97
2019 2 25.34
2020 3 25.71
2021 4 26.07
2022 5 26.44
2023 6 26.81
2024 7 27.18
2025 8 27.55
2026 9 27.92
2027 10 28.29

The estimated smoothing coefficient for the level is $$\hat{\alpha} = 0.9999$$. The very high value shows that the level changes rapidly in order to capture the highly trended series. The estimated smoothing coefficient for the slope is $$\hat{\beta}^* = 0.3266$$. This is relatively large suggesting that the trend also changes often (even if the changes are slight).

### Damped trend methods

The forecasts generated by Holt’s linear method display a constant trend (increasing or decreasing) indefinitely into the future. Empirical evidence indicates that these methods tend to over-forecast, especially for longer forecast horizons. Motivated by this observation, Gardner & McKenzie (1985) introduced a parameter that “dampens” the trend to a flat line some time in the future. Methods that include a damped trend have proven to be very successful, and are arguably the most popular individual methods when forecasts are required automatically for many series.

In conjunction with the smoothing parameters $$\alpha$$ and $$\beta^*$$ (with values between 0 and 1 as in Holt’s method), this method also includes a damping parameter $$0<\phi<1$$: \begin{align*} \hat{y}_{t+h|t} &= \ell_{t} + (\phi+\phi^2 + \dots + \phi^{h})b_{t} \\ \ell_{t} &= \alpha y_{t} + (1 - \alpha)(\ell_{t-1} + \phi b_{t-1})\\ b_{t} &= \beta^*(\ell_{t} - \ell_{t-1}) + (1 -\beta^*)\phi b_{t-1}. \end{align*} If $$\phi=1$$, the method is identical to Holt’s linear method. For values between $$0$$ and $$1$$, $$\phi$$ dampens the trend so that it approaches a constant some time in the future. In fact, the forecasts converge to $$\ell_T+\phi b_T/(1-\phi)$$ as $$h\rightarrow\infty$$ for any value $$0<\phi<1$$. This means that short-run forecasts are trended while long-run forecasts are constant.

In practice, $$\phi$$ is rarely less than 0.8 as the damping has a very strong effect for smaller values. Values of $$\phi$$ close to 1 will mean that a damped model is not able to be distinguished from a non-damped model. For these reasons, we usually restrict $$\phi$$ to a minimum of 0.8 and a maximum of 0.98.

### Example: Australian Population (continued)

Figure 8.3 shows the forecasts for years 2018–2032 generated from Holt’s linear trend method and the damped trend method.

aus_economy %>%
model(
Holt's method = ETS(Pop ~ error("A") + trend("A") + season("N")),
Damped Holt's method = ETS(Pop ~ error("A") + trend("Ad", phi = 0.9) + season("N"))
) %>%
forecast(h=15) %>%
autoplot(aus_economy, level = NULL) +
ggtitle("Forecasts from Holt's method") + xlab("Year") +
ylab("Population of Australia (millions)") +
guides(colour=guide_legend(title="Forecast"))

We have set the damping parameter to a relatively low number $$(\phi=0.90)$$ to exaggerate the effect of damping for comparison. Usually, we would estimate $$\phi$$ along with the other parameters. We have also used a rather large forecast horizon ($$h=15$$) to highlight the difference between a damped trend and a linear trend.

### Example: Internet usage

In this example, we compare the forecasting performance of the three exponential smoothing methods that we have considered so far in forecasting the number of users connected to the internet via a server. The data is observed over 100 minutes and is shown in 8.4.

www_usage <- as_tsibble(WWWusage)
www_usage %>% autoplot(value) +
xlab("Minute") + ylab("Number of users")

We will use time series cross-validation to compare the one-step forecast accuracy of the three methods.

www_usage %>%
stretch_tsibble(.init = 10) %>%
model(
SES = ETS(value ~ error("A") + trend("N") + season("N")),
Holt = ETS(value ~ error("A") + trend("A") + season("N")),
Damped = ETS(value ~ error("A") + trend("Ad") + season("N"))
) %>%
forecast(h=1) %>%
accuracy(www_usage)
#> # A tibble: 3 x 9
#>   .model .type     ME  RMSE   MAE   MPE  MAPE  MASE  ACF1
#>   <chr>  <chr>  <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
#> 1 Damped Test  0.288   3.69  3.00 0.347  2.26 0.663 0.336
#> 2 Holt   Test  0.0610  3.87  3.17 0.244  2.38 0.701 0.296
#> 3 SES    Test  1.46    6.05  4.81 0.904  3.55 1.06  0.803

Damped Holt’s method is best whether you compare MAE or RMSE values. So we will proceed with using the damped Holt’s method and apply it to the whole data set to get forecasts for future years.

fit <- www_usage %>%
model(Damped = ETS(value ~ error("A") + trend("Ad") + season("N")))
# Estimated parameters:
tidy(fit)
#> # A tibble: 5 x 3
#>   .model term  estimate
#>   <chr>  <chr>    <dbl>
#> 1 Damped alpha   1.000
#> 2 Damped beta    0.997
#> 3 Damped phi     0.815
#> 4 Damped l      90.4
#> 5 Damped b      -0.0173

The smoothing parameter for the slope is estimated to be almost one, indicating that the trend changes to mostly reflect the slope between the last two minutes of internet usage. The value of $$\alpha$$ is very close to one, showing that the level reacts strongly to each new observation.

fit %>%
forecast(h = 10) %>%
autoplot(www_usage) +
xlab("Minute") + ylab("Number of users")

The resulting forecasts look sensible with decreasing trend, and relatively wide prediction intervals reflecting the variation in the historical data. The prediction intervals are calculated using the methods described in Section 8.5.

In this example, the process of selecting a method was relatively easy as both MSE and MAE comparisons suggested the same method (damped Holt’s). However, sometimes different accuracy measures will suggest different forecasting methods, and then a decision is required as to which forecasting method we prefer to use. As forecasting tasks can vary by many dimensions (length of forecast horizon, size of test set, forecast error measures, frequency of data, etc.), it is unlikely that one method will be better than all others for all forecasting scenarios. What we require from a forecasting method are consistently sensible forecasts, and these should be frequently evaluated against the task at hand.

### Bibliography

Gardner, E. S., & McKenzie, E. (1985). Forecasting trends in time series. Management Science, 31(10), 1237–1246. https://doi.org/10.1287/mnsc.31.10.1237

Holt, C. E. (1957). Forecasting seasonals and trends by exponentially weighted averages (O.N.R. Memorandum No. 52). Carnegie Institute of Technology, Pittsburgh USA. https://doi.org/10.1016/j.ijforecast.2003.09.015


text:
Chiral symmetry, Confinement and Nuclear Matter properties

# Chiral symmetry, Confinement and Nuclear Matter properties

G. Chanfray IPN Lyon, Université de Lyon/Univ. Lyon 1, IN2P3/CNRS    M. Ericson IPN Lyon, Université de Lyon/Univ. Lyon 1, IN2P3/CNRS Theory Division,CERN    E. Massot IPN Lyon, Université de Lyon/Univ. Lyon 1, IN2P3/CNRS
###### Abstract

We discuss the possible influence of fundamental QCD properties such as spontaneous chiral symmetry breaking and nucleon substructure on nuclear matter properties. We propose a chiral version of the relativistic model in which the attractive background scalar field is associated with the chiral invariant field governing the radial fluctuations of the quark condensate. Nuclear matter stability is ensured once the scalar response of the nucleon depending on the quark confinement mechanism is properly incorporated. The needed parameters are estimated from lattice results and a satisfactory description of bulk properties follows, the only really free parameter being the coupling constant. Pion loops can be also incorporated to obtain in a consistent way the finite density chiral susceptibilities. A good description of the asymmetry energy is obtained once the full rho meson exchange and Fock terms are included.

## 1 Introduction

A fundamental question of present day theoretical nuclear physics is the relationship between non perturbative QCD and the very rich structure of the nuclear many-body problem. A first trial beyond the standard non relativistic treatment based on effective forces (Skyrme or Gogny forces) is the relativistic mean field approach initiated by Walecka SW86 (). The next step should be to incorporate in this framework the effect associated with the most prominent properties of low energy QCD, i.e., the spontaneous breaking of chiral symmetry and the effect of hadron substructure and color confinement. In a bottom-to-top attitude this certainly constitutes a necessary basis to study matter under extreme density and temperature conditions.

## 2 Chiral symmetry and confinement

Basics on chiral symmetry. The chiral symmetry of the QCD Lagrangian is certainly a crucial key for the understanding of many phenomena in low energy hadron physics. This symmetry originates from the fact that the QCD Lagrangian is almost invariant under the separate flavor transformations of left-handed and right-handed light quark fields and . The explicit violation of chiral symmetry is governed by the quark mass 10 MeV which is much smaller than typical hadron masses of order GeV, indicating that the symmetry is excellent. It is however well established that the QCD vacuum does not possess the symmetry of the Lagrangian i.e., chiral symmetry is spontaneously broken (SCSB) as it is evidenced by a set of remarkable properties. The first one is the building-up of a chiral quark condensate : which mixes, in the broken vacuum, left-handed and right-handed quarks (). Another order parameter at the hadronic scale is the pion decay constant MeV which is related to the quark condensate by the Gell-Mann-Oakes-Renner (GOR) relation : valid to leading order in the current quark mass. It leads to a large negative value -(240 MeV) indicating a strong dynamical breaking of chiral symmetry. The second feature is the appearance of soft Goldstone bosons to be identified with the almost massless pions. Finally the chiral asymmetry of the vacuum associated with the condensation of quark-antiquark pairs is directly visible at the level of the the hadronic spectrum : there is no degeneracy between possible chiral partners such as or .

An illustration: the Nambu-Jona-Lasinio (NJL) model. Although chiral symmetry breaking and restoration can be studied on the lattice, the underlying physical mechanisms at the QCD level are not yet fully understood. It is thus useful to study this problem with models such as the very popular NJL model. In such a model the very complicated highly non perturbative multi-gluons exchanges between quarks are simulated by a very simple chiral invariant contact interaction with strength parameter . In its simplest form the Lagrangian is :

 LNJL=i¯ψγμ∂μψ−m¯ψψ+G12[(¯ψψ)2+(i¯ψγ5→τψ)2]. (1)

where is an isodoublet of quark fields. The model also incorporates the fact that only low momentum quarks strongly interact in QCD. This is achieved by including a cutoff of the order of GeV in the momentum expansion of the quark fields. It is very easy to show that if the strength is sufficiently large, the ground state wave function is of BCS type , , and made of interacting quark-antiquark pairs (created by the and operators acting on the bare perturbative vacuum ), hence building up a quark condensate. As well-known such a BCS ground state is the vacuum of quasi-particles, the constituant quarks, with mass solution of a gap equation :

 MQ=m−2G1⟨⟨¯qq⟩⟩=m+4NcG1∫p<Λdp(2π)3MQEp,Ep=√M2Q+p2. (2)

It is thus clear that the existence of such a constituant mass of the order of MeV is intimately related to the quark condensate which plays the role of an order parameter associated with the broken vacuum.

The mesons can be obtained using standard RPA exactly as in nuclear physics, as collective quark-antiquark excitations. This concerns in first place the pion (generated in the channel) which appears as a Goldstone boson and its chiral partner the scalar-isoscalar sigma meson ( channel) which comes out with a mass . This model insisting on chiral symmetry, once taken in a more sophisticated form, yields a rather good phenomenology, especially in the meson sector, but has an important weakness since it does not incorporate color confinement R97 ().

Interplay between chiral symmetry and confinement. When hadronic matter is heated and compressed, initially confined quarks and gluons start to percolate between the hadrons to be finally liberated. This picture is supported by lattice simulations showing that strongly interacting matter exhibits a sudden change in thermodynamic quantities (constituting a true phase transition or a rapid cross-over) within a narrow temperature window around MeV. This transition is accompanied by a sharp dropping of the chiral condensate indicating chiral symmetry restoration. However why these two transitions seem to occur simultaneously is still a major open question. This not fully understood duality between chiral symmetry and confinement is also present at the level of the nucleon structure and consequently in nuclear physics. Some of the pictures or models for nucleon structure put the emphasis on color confinement. This is particularly true for the MIT bag model where quarks with current quark mass, MeV, move freely in a cavity (bubble of perturbative vaccuum) generated by the confinement dynamics. A totally different view is provided by the constituant quark picture where the nucleon is made of three ”big” quarks (i.e., of the NJL type) getting their mass from the chiral condensate of the non perturbative QCD vacuum. A plausible hybrid picture mixing both aspects seems to be supported by lattice calculations B05 (). The nucleon looks like a shaped color string with three constituant quarks attached at the end-points of the . The introduction of a coupling between quarks and mesons allows to build the pion cloud which plays an extremely important role to account for the mass and the chiral properties of the nucleon (such as the sigma term or chiral susceptibilities defined below). Finally possible configurations where a string develops between one quark and one diquark in the color antitriplet channel, might also be present in the nucleon.

## 3 Chiral effective theory: nucleon structure and nuclear physics

How to build an effective chiral theory? NJL model as an example. For applications in hadronic or nuclear physics, one usually tries to formulate low energy QCD directly in terms of physical degrees of freedom, namely the hadrons. Mathematically one has to integrate out quarks and gluons in favor of hadrons. This is of course a formidable task that we will illustrate with the example of the NJL model. The appropriate technics is the path integral formalism where the partition function (vacuum persistence amplitude) is expressed as a functionnal integral over the quark fields: . The idea is to make a change of variable in the functionnal integral ( is a coupling constant to be fixed) and to integrate out the quarks in the Dirac sea. One thus obtains a very complicated highly non local effective Lagrangian (the fermion determinant) depending on the mesonic fields, . An approximative form can be obtained using a derivative or a loop expansion. To one loop, one gets: . The effective potential has the typical mexican hat shape associated with a broken (chiral) symmetry (see fig. 1). The degenerate vacuum corresponds to the chiral circle, i.e., the bottom of the mexican hat (fig.1). The pseudoscalar field (pion) represents the phase fluctuation of the condensate and it is massless since it does not cost energy to move on the bottom. The massive scalar mode (sigma) has a mass related to the curvature of the potential and represents the amplitude fluctuation of the condensate. The vacuum point with isospin zero is the point where and where has an expectation value identified with the pion decay constant .

Effective chiral theory and the background scalar field in nuclear matter. More generally we take the point of view that the effective theory has to be formulated first in term of the fields associated with the fluctuations of the condensate parametrized in a matrix form . The sigma and the pion are promoted to the rank of effective degrees of freedom. An alternative and very convenient formulation of the resulting sigma model is obtained by going from cartesian to polar coordinates (right panel of fig.1), i.e., going from a linear to a non linear representation, according to :

 W=σ+i→τ⋅→π=SU=(fπ+s)exp(i→τ⋅→ϕfπ). (3)

The new pion field corresponds to an orthoradial soft mode which is automatically massless (in the absence of explicit CSB) since it is associated with rotations on the chiral circle without cost of energy. The new sigma meson field , which is a chiral invariant, describes a radial mode associated with the fluctuations of the “chiral radius” . It can be associated with the ordinary sigma meson which gets a very large width from its strong decay into two pions. Since it has derivative couplings, it decouples from low energy pions whose dynamics is described by chiral perturbation theory. With increasing density, its fluctuations are associated with the shrinking of the chiral circle and it governs the evolution of the nucleon mass. In a recent paper CEG02 () we have proposed to identify this chiral invariant field with the sigma meson of nuclear physics and relativistic theories of the Walecka type, or, said differently, with the background attractive scalar field at the origin of the nuclear binding. This also gives a plausible solution to the long-termed problem of the chiral status of Walecka theories.

Tests of the effective theory with a chiral invariant scalar field. Once the appropriate couplings of the chiral fields to the baryons are introduced one can build an effective lagrangian to describe nuclear matter. Vector mesons ( and ) must be also included to get the needed short range repulsion and asymmetry properties. At the Hartree level, the pion and the rho do not contribute for symmetric nuclear matter whose energy density written as a function of the order parameter is :

 E0V=ε0=∫4d3p(2π)3Θ(pF−p)E∗p(¯s)+V(¯s)+g2ω2m2ωρ2. (4)

is the energy of an effective nucleon with the effective mass . is the scalar coupling constant of the model; in the pure linear sigma model it is . The effective potential when reexpressed in term of the new representation has the form :

 V(s)=12m2σ(s2+12s3fπ+...).

is obtained by minimization of the energy density and is given at low density by : . Its negative value is at the origin of the binding but the presence of the term (tadpole) has very important consequences. This tadpole is at the origin of the chiral dropping of the sigma mass (a effect at ), and generates an attractive three-body force which makes nuclear matter collapse and destroys the Walecka saturation mechanism. Hence the chiral theory does not pass the nuclear matter stability test.

This failure, which is in fact a long-standing problem KM74 (); BT01 (), is maybe not so surprising since the theory, as it is, also fails to describe some nucleon structure aspects as discussed below. The nucleon mass, as well as other intrinsic properties of the nucleon (sigma term, chiral susceptibilities), are QCD quantities which are in principle obtainable from lattice simulations. The problem is that lattice calculations of this kind are not feasible for quark masses smaller than MeV, or equivalently pion mass smaller than MeV, using the GOR relation. Hence one needs a technics to extrapolate the lattice data to the physical region. The difficulty of the extrapolation is linked to the non analytical behaviour of the nucleon mass as a function of (or equivalently ) which comes from the pion cloud contribution. The idea of Thomas et al TGLY04 () was to separate the pion cloud self-energy, , from the rest of the nucleon mass and to calculate it in a chiral model with one adjustable cutoff parameter . They expanded the remaining part in terms of as follows :

 MN(m2π)=a0+a2m2π+a4m4π+Σπ(mπ,Λ). (5)

The best fit value of the parameter shows little sensitivity to the shape of the form factor, with a value while (see ref. TGLY04 ()). The small value of reflects the fact that the non pionic contribution to the nucleon mass is almost linear in (i.e., in ). Taking successive derivatives of with respect to (i.e., to ), it is possible to obtain some fundamental chiral properties of the nucleon, namely the pion-nucleon sigma term and the scalar susceptibility of the nucleon. The non pionic pieces of these quantities are given by :

 σnon−pionN=mq∂Mnon−pionN∂mq≃m2π∂M∂m2π=a2m2π+2a4m4π≃29MeV. (6)
 χnon−pionNS=∂(σnon−pionN/2mq)∂mq≃2⟨¯qq⟩2vacf4π∂  ∂m2π(σnon−pionNm2π)=⟨¯qq⟩2vacf4π4a4. (7)

In the above equations the first equalities correspond to the definitions, the second equalities make use of the GOR relation and the last ones come from the lattice QCD analysis. With typical cutoff used in this analysis, GeV, which yields MeV, the total value of the sigma term is MeV, a quite satisfactory result in view of the most recent analysis. It is interesting to compare what comes out from the lattice approach with our chiral effective model. At this stage the only non pionic contribution to the nucleon mass comes from the scalar field, or more microscopically the nucleon mass entirely comes from the chiral condensate since the nucleon is just made of three constituant quarks with mass MeV. Hence the results for the non pionic sigma term and scalar susceptibility are identical to those of the linear sigma model :

 σ(σ)N=fπgSm2πm2σ,χ(σ)NS=−2⟨¯qq⟩2vacf3π3gSm4σ (8)

The identification of with of our model fixes the sigma mass to a value MeV, close to the one MeV that we have used in a previous article CE05 (). As it is the ratio which is thus determined this value of is associated with the coupling constant of the linear sigma model . Lowering reduces . Similarly, the comparison of with the lattice expression provides a model value for . The numerical result is while the value found in the expansion is only .

Nucleon structure effects and confinement mechanism. The net conclusion of the above discussion is that the model as such fails to pass the QCD test. In fact this is to be expected and even gratifying because it also fails the nuclear physics test. We will see that these two important failures may have a common origin. Indeed an important effect is missing, namely the scalar response of the nucleon, , to the scalar nuclear field, which is the basis of the quark-meson coupling model, (QMC), introduced in ref. G88 (). The physical reason is very easy to understand: the nucleons are quite large composite systems of quarks and gluons and they should respond to the nuclear environment, i.e., to the background nuclear scalar fields. This response originates from the quark wave function modification in the nuclear field and will oviously depend on the confinement mechanism. This confinement effect is expected to generate a positive scalar response , i.e., it opposes an increase of the scalar field, a feature confirmed by the lattice analysis (see below). This polarization of the nucleon is accounted for by the phenomenological introduction of the scalar nucleon response, , in the nucleon mass evolution as follows :

 MN(s)=MN+gSs+12κNSs2. (9)

This constitutes the only change in the expression of the energy density (eq. 4) but this has numerous consequences. The effective scalar coupling constant drops with increasing density but the sigma mass gets stabilized :

 g∗S(¯s)=∂M∗N∂¯s=MNfπ+κNS¯s,m∗2σ=∂2ε∂¯s2≃m2σ−(3gSfπ−κNS)ρS. (10)

The non-pionic contribution to the nucleon susceptibility is modified, as well :

 χ(σ)NS=−2⟨¯qq⟩2vacf2π(1m∗2σ−1m2σ)1ρ=−2⟨¯qq⟩2vacf2π1m4σ(3gSfπ−κNS). (11)

We see that the effect of confinement () is to compensate the pure scalar term. Again comparing with the lattice expression one gets a model value for the parameter :

 a4=−a222M(3−2C). (12)

where is the dimensionless parameter . Numerically gives , implying a large cancellation. As discussed below we will see that such a significant scalar response will generate other repulsive forces which restore the saturation mechanism. At this point it is important to come again to the underlying physical picture implying that the nucleon mass originates both from the coupling to condensate and from confinement. In the original formulation of the quark coupling model, nuclear matter is represented as a collection of (MIT) bags seen as bubbles of perturbative vacuum in which quarks are confined. Thus in such a picture the mesons do not move and do not couple to quarks as in the true non perturbative QCD vacuum. Consequently the bag picture is at best an effective realisation of confinement which must not to be taken too literally. Indeed, QCD lattice simulations strongly suggest that a more realistic picture is closer to a shaped color string (confinement aspect) attached to quarks. Outside this relatively thin string one has the ordinary non perturbative QCD vacuum possesing a chiral condensate from which the quarks get their constituant mass.

## 4 Results for nuclear matter and discussion

Stability and chiral properties of nuclear matter. In our first work devoted to the stability of nuclear matter CE05 () we limited ourselves to the Hartree approximation (eq. 4). The free sigma mass and the parameter characterizing the nucleonic scalar response were taken as adjustable parameters. The fact that the best fit came with values very close to what we deduced afterwards from lattice is certainly satisfactory. One problem was the too large incompressibility but is is nevertheless possible to improve our model description by adding an extra term in the expression of the effective nucleon mass in such a way that the nucleonic response vanishes at full chiral restoration (). In this case the set of parameters : , MeV and , leads to correct saturation properties, with an incompressibility value MeV. Another remarkable finding, according to the previous discussion, was the stability of the sigma mass which dropped by only MeV at . Ignoring the repusive tadpole associated with the nucleonic response would have given an almost vanishing sigma mass at this density.

The next step CE07 () has been to introduce pion loops which are necessary if we wish to address the question of the chiral properties of nuclear matter (namely the in-medium quark condensate and chiral susceptibilities) in a way which is consistent with nuclear matter stability since pion loops affect the energy. They do not enter at the Hartree mean field level but contribute through the Fock term and through the correlation energy. This energy can be calculated within a RPA ring approximation according to :

 Eloop=E−E0 ≡ Vεloop=3V∫+∞−∞idω2π)∫dq(2π)3∫10dλλ (13) ([VL(ω,q)ΠL(ω,q;λ)]+2[VT(ω,q)ΠT(ω,q;λ)]) .

is the full (RPA) longitudinal spin-isospin polarization propagator in the pionic channel. This calculation actually includes iterated pion exchange and also the part of the NN potential from the two-pion exchange with excitation. is nothing but the (non static) pion exchange potential corrected by short-range components via the Landau-Migdal parameters, . We have taken their values from a systematic survey of the data on spin-isospin physics by Ichimura et al. ISW06 () : . Beside pion exchange we have also introduced the transverse channel (), dominated by exchange together with the short-range component. All the other needed ingredients are taken from the pion-nucleon and pion-nucleus phenomenolgy. The scalar coupling constant is the one of the sigma model and for the sigma mass we have followed the lattice indications, allowing a small readjustment around the lattice value, which is . We have found a better fit with which corresponds to . The omega mass is known and the coupling constant is the only really free parameter. For the nucleon scalar response we have followed the indications of the lattice data but not strictly in view of the uncertainties attached to the higher derivatives. The value which fits the saturation properties is found to be , not far from the lattice value . With these inputs we have obtained a satisfactory description of the nuclear binding. The binding energy per particle is shown in fig. 2, with its different components. Taking the successive derivatives of the equation of state (the grand potential ) with respect to , it is possible to obtain the pseudo-scalar susceptibility (directly related to the quark condensate) and the scalar susceptibilities of nuclear matter CEG03 (). Since they are associated with chiral partners (pion and sigma), they should be identical at chiral restoration. The numerical calculation CE07 () on the right panel of fig. 2 indeed displays a convergence effect.

Relativistic Hartree-Fock (RHF). The description of asymmetric nuclear matter certainly necessitates the full rho exchange already at the Hartree level through its time component. We thus include it using the following interacting lagrangian :

 Lρ=−gρρaμΨ¯γμτaΨ−gρκρ2MN∂νρaμΨ¯σμντaΨ+12m2ρρaμρμa−14GμνaGaμν (14)

One specific comment is in order for the tensor coupling of vector mesons. The pure vector dominance picture (VDM) implies the identification of with the anomalous part of the isovector magnetic moment of the nucleon, i.e., . However pion-nucleon scattering data suggest (strong rho scenario). At the Hartree level we still have our previous picture. Symmetric and asymmetric nuclear matter are seen as an assembly of nucleons, i.e., of Y shaped color strings with massive constituant quarks at the end getting their mass from the chiral condensate. This nucleons move in self-consistent scalar () and vector background fields (). However to get a quantitative description, all the Fock terms, including not only pion and rho exchange but also sigma and omega exchange, are needed. These exchange terms are mediated by the propagation of the fluctuations of the previous meson fields. In particular it is possible to show that the scalar fluctuation, i.e., the fluctuation of the chiral condensate, propagates with an in-medium modified sigma mass . The results of preliminary RHF calculation MC07 () with rho coupling given by pure VDM (, ) are shown on fig. 3 and reproduce the saturation properties. is not very far from the quark model value and is larger than the lattice value. The resulting incompressibility MeV and asymmetry energy parameter MeV are obviously satisfactory. It is also possible to reproduce saturation properties in the strong rho scenario () but the asymmetry energy becomes too large : MeV. These results are certainly very encouraging but the ultimate test of this chiral theory will be the reproduction of finite nuclei (in particular the spin-orbit splitting) and the consequences for neutron star properties.

## References

• (1) B.D. Serot, J.D. Walecka, Adv. Nucl. Phys. 16(1986) 1; Int. J. Mod. Phys. E16 (1997) 15.
• (2) G. Ripka, ”Quarks bound by chiral fields”, Clarendon Press-Oxford, 1997.
• (3) F. Bissey et al, Nucl. Phys. B (Proc. Suppl.) 141 (2005) 22, hep-lat/0501004.
• (4) G. Chanfray, M. Ericson and P.A.M. Guichon, Phys. Rev C63 (2001)055202.
• (5) A.K. Kerman and L.D. Miller in “Second High Energy Heavy Ion Summer Study”, LBL-3675, 1974.
• (6) W. Bentz and A.W. Thomas, Nucl. Phys. A696 (2001) 138.
• (7) A. W. Thomas, P. A. M. Guichon, D. B. Leinweber and R. D. Young Progr. Theor. Phys. Suppl. 156 (2004) 124; nucl-th/0411014.
• (8) G. Chanfray and M. Ericson, EPJA 25 (2005) 151.
• (9) P.A.M. Guichon, Phys. Lett. B200 (1988) 235.
• (10) G. Chanfray and M. Ericson, Phys. Rev. C75 (2007) 015206. nucl-th/0611042.
• (11) M. Ichimura, H. Sakai and T. Wakasa, Prog. Part. Nucl. Phys., 56 (2006) 446.
• (12) G. Chanfray, M. Ericson and P.A.M. Guichon, Phys. Rev C61 (2003) 035209.
• (13) E. Massot and G. Chanfray, work in progress.
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# An unnamed professor is walking to class when an office chair (travelling in the same direction) collides with them. The professor fails int

Question

An unnamed professor is walking to class when an office chair (travelling in the same direction) collides with them. The professor fails into the chair and remains in the chair after collision. The professor, with a mass of 113kg, is travelling at 1.56 m/s. The chair has a velocity of 6.5 m/s, and a mass of 10kg. After the collision, and ignoring any effects of professor falling, what would be the speed of the chair professor system

in progress 0
1 year 2021-09-01T11:54:43+00:00 1 Answers 15 views 0

## Answers ( )

1. Answer: 1.96 m/s

Explanation:

Given

Mass of Professor $$m_1=113\ kg$$

Velocity of professor $$u_1=1.56\ m/s$$

mass of chair $$m_2=10\ kg$$

velocity of chair $$u_2=6.5\ m/s$$

Suppose  after the collision, v is the common velocity

Conserving momentum

$$\Rightarrow m_1u_1+m_2u_2=(m_1+m_2)v\\\\\Rightarrow v=\dfrac{113\times 1.56+10\times 6.5}{113+10}=\dfrac{241.28}{123}\\\\\Rightarrow v=1.96\ m/s$$


text:
# Given f(x) = 2 - 3x how do you evaluate f(a)?

Just replace $x$ with $a$
If $f \left(x\right)$ = $2 - 3 x$
So, $f \left(a\right) = 2 - 3 a$


text:
+0

0
41
2
+107

1 a. David's airplane trip took 3.8 hours. For one-half of that time, the airplane flew at a speed
of 680 miles per hour, and for the rest of the time, it flew at a speed of 740 miles per hour.
What distance did David travel?

2 a. Abigail travels in an airplane a distance of 1860 km. For one-third of the distance, the
airplane flies at a speed of 1032 km/h, and for the rest of the distance, it flies at a speed of
696 km/h. How long does the trip take?

Sep 16, 2020

#1
+9017
+2

1.

Remember:       distance   =   rate * time

Since he was going at two different rates for each part of the trip, lets first calculate the distance he flew during the first part of the trip and then we can calculate the distance he flew during the second part of the trip. The total distance traveled will be the sum of these two distances.

distance1   =   rate1 * time1   =   680 mi/hr  *  (3.8 / 2) hr   =   680 mi/hr   *   1.9 hr   =   1292 mi

distance2   =   rate2 * time2   =   740 mi/hr  *  (3.8 / 2) hr   =   740 mi/hr   *   1.9 hr   =   1406 mi

total distance   =   distance1  +  distance2   =   1292 mi  +  1406 mi   =   ?

2.

By solving the first formula for time,  we get:       time   =   distance / rate

Again let's split this into two parts for each part of the trip.

For the 1st part..... the distance is one third of the total distance, and the rate is 1032 km/hr.

For the 2nd part.... the distance is two thirds of the total distance, and the rate is 696 km/hr.

$$\text{time}_1\ =\ \dfrac{\text{distance}_1}{\text{rate}_1}\ =\ \dfrac{\frac13\cdot1860\ \text{km}}{1032\ \frac{\text{km}}{\text{hr}} } \ =\ \dfrac{620\ \text{km}}{1032\ \frac{\text{km}}{\text{hr}} }\ =\ \dfrac{620}{1032}\ \text{hr}$$

$$\text{time}_2\ =\ \dfrac{\text{distance}_2}{\text{rate}_2}\ =\ \dfrac{\frac23\cdot1860\ \text{km}}{696\ \frac{\text{km}}{\text{hr}} }\ =\ \dfrac{1240\ \text{km}}{696\ \frac{\text{km}}{\text{hr}} }\ =\ \dfrac{1240}{696}\ \text{hr}$$

$$\text{total time}\ =\ \text{time}_1+\text{time}_2 \ =\ \dfrac{620}{1032}\ \text{hr}\ +\ \dfrac{1240}{696}\ \text{hr}\ =\ \Big( \dfrac{620}{1032}\ +\ \dfrac{1240}{696}\Big)\ \text{hr}\ \approx\ 2.38\ \text{hr}$$ _

Sep 16, 2020
#2
+107
+1

Thank oy so much hectictar!!!!

Sep 16, 2020


text:
# Are there Guaranteed Proof-of-work Systems?

Proof-of-Work systems such as hashcash, it seems to me, do not actually prove that a certain amount of work has been performed. For example, in the case of hashcan, could not the first random hash choice just get lucky? Thus almost no actual "work" was performed?

My question is on the existence (and practicality) of methods that guarantee, within a probability, that a given minimum amount of "work" has been performed in each and every case, without compromising the average or maximum amounts.

On hashcash then, the probability of failing to prove work is the chance of "getting lucky". So, with a 20 bit zero hash target, that's around a few in a million (the first few random attempts out of a 2^20 space). Therefore a few in a million hashcans won't really be doing any work. But if i wanted to reduce this proof-of-work fail to a lower probability, simply increasing the zero bit space does not work, since it also drastically increases the average work.

To compare this to, say, the workload of compressing a file (eg zip). That's a much more controllable amount of work, roughly proportional to the input length.

Say we manufacture an unpredictable text generator, fed into zip, fed into a hash function. The proof of work is the generator seed and the final hash value. the generator output length controls the amount of work and saying you could "guess" the final hash is virtually impossible (compared to say 1:2^20).

But this example is not practical because it is symmetric, requiring the same work to be performed for verification.

so, are there any guaranteed (within probability) asymmetric proof-of-work systems?

• No cryptographic solution can protect you against an adversary making a lucky guess, you can merely make it "highly improbable". For example: I could get lucky guessing a private key of an Bitcoin address which has a positive balance, but my chances are roughly $1$ to $2^{256}-1$ (ignoring the birthday problem et al) Apr 26, 2018 at 10:51

## 2 Answers

You could use a challenge consisting of multiple subtasks.

For example, provide a unique random string of characters $R$ and ask for $n$ distinct strings that have $R$ as a prefix and whose SHA256 sum ends in $z$ zeros.

In a single hash calculation, the probability of a lucky guess is $2^{-4z}$, so the probability of multiple lucky guesses in one challenge is going to be very small if $z = 10$.

The number of SHA256 calculations needed to pass the challenge is of the order of $n \times 2^{4z}$, but only $n$ hash calculations are needed to check the results (plus a little extra to check that no two strings are the same).

• Isn't the probability of generating a specific n*B-bit string, equal to that of generating n specific B-bit strings? With that in mind, I don't think this changes anything other than requiring more calculations to verify the answers. Feb 24, 2018 at 17:53
• @squeamish. thanks for your answer. I think this is the right direction. Im currently working on a version of poncho's answer here. Feb 24, 2018 at 20:22

It's impossible, because if the solution is efficiently verifiable, it's always possible (even though it happens with negligible probability) to guess it and verify that it works efficiently


text:
The set $S_+(n, k)$ of rank-$k$ symmetric positive semidefinite matrices $X$ of size $n\times n$ is parametrized as $X = YY^T$ where $Y \in \mathbb{R}^{n\times k}_*$ is a full column rank matrix of size $n\times k$. Notice that for all orthogonal matrices $O$ of size $k$, the transformation $Y \rightarrow YO$ leaves $X$ unchanged. As a result, we may consider all matrices of the form $YO$ as being equivalent for our purpose. Quotienting this equivalence relation out results in a quotient space, $\mathbb{R}^{n \times k}_* / {\rm O}(k)$, with ${\rm O}(k)$ the group of orthogonal matrices. A point on the quotient space is (conceptually) an equivalence class $[Y] = \{YO : O \in {\rm O}(k)\}$. Numerically, we represent that equivalence class with (any) matrix $Y$ in $[Y]$.

A function $\phi : S_+(n,k) \to \mathbb{R}: X \mapsto \phi(X)$ induces a function $f: \mathbb{R}^{n \times k}_* \rightarrow \mathbb{R}: Y \mapsto f(Y)$ defined as $f(Y) = \phi(YY^T)$. Of course, $f$ has the appropriate invariance $f(Y) = f(YO)$ for all orthogonal $O$. Thus, $f$ is defined on the total space $\mathbb{R}^{n\times k}_*$ and descends as a well-defined function on the quotient space $\mathbb{R}^{n \times k}_* / {\rm O}(k)$. Endowing the total space with the usual Riemannian structure of a Euclidean space (the inner product $\langle H_1, H_2 \rangle = \operatorname{trace}(H_1^T H_2)$), a Riemannian structure for the quotient space follows.

See the paper [JBAS10] referenced below for details on this geometry, and [AMS08] for general background.

Factory call: M = symfixedrankYYfactory(n, k).

 Name Formula Numerical representation Set $S_+(n,k) = \{X \in \mathbb{R}^{n\times n} : X=X^T\succeq 0, {\rm rank}(X) = k\} \\ = \{ YY^T: Y \in \mathbb{R}^{n\times k}, {\rm rank}(Y) = k\}$ $X$ is represented by means of $Y$ as a matrix Y of size $n\times k$. Vertical space at $Y$ $\{ Y\Omega : \Omega \in \mathbb{R}^{k\times k}, \Omega + \Omega^T = 0 \}$ A vertical vector $U$ at $Y$ is represented as a matrix U of size $n\times k$: it is tangent to the equivalence classes. Horizontal space at $Y$ $\{ U \in \mathbb{R}^{n\times k}: U^T Y = Y^T U \}$ A horizontal vector $U$ at $Y$ is represented as a matrix U of size $n\times k$: it is orthogonal to all vertical vectors. Ambient space $\mathbb{R}^{n\times k}$ Points and vectors in the ambient space are, naturally, represented as matrices of size $n\times k$.

The following table shows some of the nontrivial available functions in the structure M. The tutorial page gives more details about the functionality implemented by each function.

 Name Field usage Formula Dimension M.dim()  $\operatorname{dim}\mathcal{M} = kn - k(k-1)/2$ Metric M.inner(Y, U, V)  $\langle U, V\rangle_Y = \operatorname{trace}(U^T V)$ Norm M.norm(Y, U)  $\|U\|_Y = \sqrt{\langle U, U \rangle_Y}$ Horizontal space projector M.proj(Y, H)  $P_Y(H) = H - Y\Omega$, where H represents a vector in the ambient space. $\Omega$ is a skew-symmetric matrix obtained as the unique solution to the Sylvester equation $\Omega (Y^T Y) + (Y^T Y)\Omega = Y^T H - H^T Y$ Euclidean to Riemannian gradient M.egrad2rgrad(Y, egrad) $\operatorname{grad} f(Y) = \nabla f(Y)$, where egrad represents the Euclidean gradient $\nabla f(Y)$, which is a vector in the ambient space. There is indeed nothing to do, since the Euclidean gradient of $f$ is already a horizontal vector thanks to the fact that $f(Y) = f(YO)$ for all orthogonal $O$. Euclidean to Riemannian Hessian M.ehess2rhess(Y, egrad, ehess, U) $\operatorname{Hess} f(Y)[U] = P_Y(\nabla^2 f(Y)[U])$, where egrad represents the Euclidean gradient $\nabla f(Y)$ (which happens to not be necessary in this case) and ehess represents the Euclidean Hessian $\nabla^2 f(Y)[U]$, both being vectors in the ambient space. Retraction M.retr(Y, U, t)  $\operatorname{Retr}_Y(tU) = Y + tU$ Random point M.rand()  Returns a point at random as follows: generate $Y$ with i.i.d. normal entries; return $Y$. With high probability, $Y$ will indeed be full-rank. Random vector M.randvec(Y)  Returns a horizontal vector at $Y$ with uniformly random direction, obtained as follows: generate $H$ with i.i.d. normal entries; return $P_Y(H)/\|P_Y(H)\|_Y$. Vector transport M.transp(Y, Z, U)  $\operatorname{Transp}_{Z\leftarrow Y}(U) = P_Z(U)$, where $U$ is a horizontal vector at $Y$ that is transported to the horizontal space at $Z$.

This is an example of low-rank matrix approximation. Let $A\in\mathbb{R}^{n\times n}$ be a symmetric matrix. We minimize the following cost function:

$$\phi(X) =\| X - A \|_F ^2,$$

such that $X \in S_+(n, k)$. $X$ is factorized as $YY^T$ with $Y \in \mathbb{R}^{n \times k}_*$. The function $f:\mathbb{R}^{n \times k}_* \rightarrow \mathbb{R}$ is defined as $f(Y) = \| YY^T - A \|_F ^2$. Compute the Euclidean gradient and Hessian of $f$:

$$\nabla f(Y) = 4(YY^T - A)Y,$$

$$\nabla^2 f(Y)[U] = 4(YU^T + UY^T)Y + 4(YY^T - A)U.$$

The Riemannian gradient and Hessian are obtained by applying the M.egrad2rgrad and M.ehess2rhess operators:

$$\operatorname{grad} f(x) = 4(YY^T - A)Y,$$

$$\operatorname{Hess} f(Y)[U] = \nabla^2 f(Y)[U] - Y\Omega$$ where $\Omega$ is obtained by solving the Sylvester equation, $$\Omega (Y^T Y) + (Y^T Y )\Omega = Y^T \nabla^2 f(Y)[U] - \nabla^2 f(Y)[U] ^T Y.$$

The Sylvester equation can be solved efficiently using lyap of MATLAB. Notice that there is no need to compute these explicitly: it suffices to write code for the Euclidean quantities and to apply the conversion tools on them to obtain the Riemannian quantities, as in the following code:

% Generate the problem data.
n = 1000;
k = 5;
Y_org = randn(n,k);
A = Y_org*Y_org';

% Create the problem structure.
manifold = symfixedrankYYfactory(n,k);
problem.M = manifold;

% Define the problem cost function and its gradient.
problem.cost = @(Y) norm( Y*Y' - A, 'fro')^2;egrad = @(Y) 4*(Y*Y' - A)*Y;ehess = @(Y, U) 4*(Y*U' + U*Y')*Y + 4*(Y*Y' - A)*U;
problem.grad = @(Y) manifold.egrad2rgrad(Y, egrad(Y));problem.hess = @(Y, U) manifold.ehess2rhess(Y, egrad(Y), ehess(Y, U), U);% Numerically check the differentials.checkgradient(problem); pause;checkhessian(problem); pause;% Execute the optimizationY = trustregions(problem);
[JBAS10] M. Journée, F. Bach, P.-A. Absil and R. Sepulchre, Low-rank optimization on the cone of positive semidefinite matrices, SIOPT, 2010.
[AMS08] P.-A. Absil, R. Mahony and R. Sepulchre, Optimization Algorithms on Matrix Manifolds, Princeton University Press, 2008.

We are still working on building a collection of examples for Manopt. The relevant ones will be referenced here when the time comes.


text:
# New year running contest?

#### TheEcologist

##### Global Moderator
Anybody want to set up a running goal before the end of the year?
For instance run 100 miles before January first?

I'm thinking of setting a goal for myself, but it helps in the motivation department if other people join in... as I'm pretty competitive and will not be the one that doesn't complete the goal.

#### TheEcologist

##### Global Moderator
Sweet. You can post your progress here. I just competed 10 km, last week and I'm on track to make 100 km before Jan 1st.

Code:
# 100 km challenge
#current day (since start of challenge 21-10-2011; 71 days total)
cday=5
#distance run (in km)
dist=100-c(seq(0,7.5,2.5),10)
day=c(0,1,2,3,5)
plot(dist~day, xlab='day left',ylab='distance to cover (km)',
xlim=c(0,71),ylim=c(0,100))
abline(lm(dist~day),lty='dashed')
abline(100,-1.4);abline(v=cday,lty="dotted")
legend(40,80,lty=c(2,1,3),legend=
c("Current projection","Minimum required","last update"))

#### TheEcologist

##### Global Moderator
Well that depends on how far I get before 1 jan! 100 miles would be a nice goal though. I need to update this too.


text:
# Writing and Comparing Numbers in Scientific Notation - Grade Eight

Write numbers in scientific notation and compare amounts. This two-page document contains approximately 35 problems and each page may be assigned independently. Be aware that this seems to be a scanned copy of an original learning exercise, so the font is somewhat fuzzy.


text:
# Start a new discussion

## Not signed in

Want to take part in these discussions? Sign in if you have an account, or apply for one below

## Discussion Tag Cloud

Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.

• CommentRowNumber1.
• CommentAuthorzskoda
• CommentTimeOct 5th 2011

New stub ergodic theory wanted at measure theory.

• CommentRowNumber2.
• CommentAuthorTodd_Trimble
• CommentTimeOct 5th 2011

Speaking of this, I had wanted to get my hands on a proof that the shift transformation on the space of irrationals between 0 and 1, which sends a continued fraction $[a_1, a_2, a_3, \ldots]$ to $[a_2, a_3, \ldots]$, is ergodic with respect to the measure $\frac1{\log 2}\frac{d x}{1 + x}$. Any ideas where to find such a proof? (Might be nice for an article ergodic theory as well, since this measure space provides archetypal examples of chaotic phenomena, or so I sense.)

• CommentRowNumber3.
• CommentAuthorzskoda
• CommentTimeOct 5th 2011
• (edited Oct 5th 2011)

I think that there is an encyclopaedia of mathematics and applications: dynamical systems (red cover, edit: or this was about operator algebras and dynamical systems? edit: no, it is very abstract one, I just peeked into it) which has large part dedicated to the shift-type dynamical systems. I had it in hands years ago, so I can not say about the particular result, but I would not be surprised to find it there.

Edit: I added a link to ergodic theory to one course page having some canonical references in ergodic theory.

• CommentRowNumber4.
• CommentAuthorUrs
• CommentTimeApr 25th 2012

added to ergodic theory this reference which discusses applications of cohomology theory to the subject. This reminded me of David Corfield’s search for cohomological notions in “random” contexts, though maybe it’s not relevant, I am not sure.


text:
## Elementary Algebra

The solution set is {x|x $\gt$ 7} , or (7,$\infty$) in interval notation.
We use the properties of inequalities to solve this problem. 4x - 6 $\gt$ 3x + 1 4x - 3x $\gt$ 1 + 6 x $\gt$ 7 The solution set is {x|x $\gt$ 7} , or (7,$\infty$) in interval notation.


text:
# Appendix Divider Page Roman Numbering

I want to continue the numbering of my Appendix in Roman. But if I switch after \begin{appendices} to Roman, the divider page is not included. If I switch the numbering before the appendix, my last chapter has Roman numbering, too. Can I switch to Roman beginning with the Divider Page?

\documentclass[a4paper,12pt,oneside]{report}
\usepackage[toc,page]{appendix}
\newcounter{savepage}
\begin{document}
\pagenumbering{Roman}
\pagestyle{plain}

\tableofcontents
\cleardoublepage

\chapter{Introduction}
\setcounter{savepage}{\arabic{page}}
\pagenumbering{arabic}
\chapter{Conclusion and Outlook}\label{ch_conclusion}

\begin{appendices}
\pagenumbering{Roman}
\setcounter{page}{\thesavepage}
\makeatletter
\begingroup
\let\protect\l@chapter\protect\l@section
\let\protect\l@section\protect\l@subsection
}
\makeatother
\end{appendices}
\end{document}

Thanks for your help!

• Welcome to TeX.SX! Please make your code compilable (if possible), or at least complete it with \documentclass{...}, the required \usepackage's, \begin{document}, and \end{document}. That may seem tedious to you, but think of the extra work it represents for TeX.SX users willing to give you a hand. Help them help you: remove that one hurdle between you and a solution to your problem. – Stefan Pinnow Oct 6 '19 at 10:38

Use \cleardoublepage followed by the change of the pagenumbering before \begin{appendices}:

\documentclass[a4paper,12pt,oneside]{report}
\usepackage[toc,page]{appendix}
\newcounter{savepage}
\begin{document}
\pagenumbering{Roman}
\pagestyle{plain}
\tableofcontents
%\cleardoublepage
\chapter{Introduction}

\cleardoublepage
\setcounter{savepage}{\value{page}}
\pagenumbering{arabic}
\chapter{Conclusion and Outlook}\label{ch_conclusion}

\cleardoublepage
\pagenumbering{Roman}
\setcounter{page}{\value{savepage}}
\begin{appendices}
\makeatletter
\begingroup
\let\protect\l@chapter\protect\l@section
\let\protect\l@section\protect\l@subsection
}
\makeatother
\chapter{Appendix chapter}
\section{Appendix section}
\end{appendices}
\end{document}

• Thank you very much! – shermes Oct 6 '19 at 14:01

You need to eject the page you need to know the number of, which is done by setting the value of savepage after \cleardoublepage.

I also suggest not to use \arabic{page} or \thesavepage, but \value{page} and \value{savepage}.

Also it's better to use abstract command and to avoid lengthy code in the document body. It's easy to patch appendices to do the necessary job. Thus the code in the body is much cleaner.

\documentclass[a4paper,12pt,oneside]{report}
\usepackage[toc,page]{appendix}
\usepackage{etoolbox}

\usepackage{lipsum} % for context

\newcounter{savepage}
% when appendices starts we need to reset the page number
\pretocmd{\appendices}
{\cleardoublepage\pagenumbering{Roman}\setcounter{page}{\value{savepage}}}
{}{}
% before the appendices we need to setup the toc
\apptocmd{\appendices}
{%
\begingroup
\let\string\l@chapter\string\l@section
\let\string\l@section\string\l@subsection
}%
}
{}{}
% when the appendices end we need to end the group

% front matter with Roman numbers
\newcommand{\frontmatter}{%
\cleardoublepage\pagenumbering{Roman}\pagestyle{plain}%
}
% main matter with arabic numbers (but save the page number up to here)
\newcommand{\mainmatter}{%
\cleardoublepage
\setcounter{savepage}{\value{page}}%
\pagenumbering{arabic}%
\pagestyle{headings}% or whatever you like
}

\begin{document}

\frontmatter

\tableofcontents

\chapter*{Introduction}

\lipsum

\mainmatter

\chapter{First}

\lipsum

\chapter{Conclusion and Outlook}\label{ch_conclusion}

\begin{appendices}

\chapter{First appendix}

\lipsum

\section{First section}

\lipsum

\chapter{Second appendix}

\lipsum

\end{appendices}

\end{document}


text:
Raymond Devillers ; Ronny Tredup - Some Basic Techniques allowing Petri Net Synthesis: Complexity and Algorithmic Issues

fi:8813 - Fundamenta Informaticae, October 21, 2022, Volume 187, Issues 2-4: Petri Nets 2021
Some Basic Techniques allowing Petri Net Synthesis: Complexity and Algorithmic Issues

Authors: Raymond Devillers ; Ronny Tredup

In Petri net synthesis we ask whether a given transition system $A$ can be implemented by a Petri net $N$. Depending on the level of accuracy, there are three ways how $N$ can implement $A$: an embedding, the least accurate implementation, preserves only the diversity of states of $A$; a language simulation already preserves exactly the language of $A$; a realization, the most accurate implementation, realizes the behavior of $A$ exactly. However, whatever the sought implementation, a corresponding net does not always exist. In this case, it was suggested to modify the input behavior -- of course as little as possible. Since transition systems consist of states, events and edges, these components appear as a natural choice for modifications. In this paper we show that the task of converting an unimplementable transition system into an implementable one by removing as few states or events or edges as possible is NP-complete -- regardless of what type of implementation we are aiming for; we also show that the corresponding parameterized problems are $W[2]$-hard, where the number of removed components is considered as the parameter; finally, we show there is no $c$-approximation algorithm (with a polynomial running time) for neither of these problems, for every constant $c\geq 1$.

Volume: Volume 187, Issues 2-4: Petri Nets 2021
Published on: October 21, 2022
Accepted on: July 4, 2022
Submitted on: December 8, 2021
Keywords: Computer Science - Formal Languages and Automata Theory


text:
1. May 20, 2013

### whynot314

I am curious if I am setting this problem up right.

Three forces act on the bracket. Determine the magnitdue and direction theta of F2 so that the resultant force is directed along the positive u axis and has a magnitude of 50lb.

[URL=http://s1341.photobucket.com/user/nebula-314/media/20130520_221430_zps840432f0.jpg.html][PLAIN]http://i1341.photobucket.com/albums/o745/nebula-314/20130520_221430_zps840432f0.jpg[/URL][/PLAIN]

[URL=http://s1341.photobucket.com/user/nebula-314/media/20130520_222748_zps3f68f545.jpg.html][PLAIN]http://i1341.photobucket.com/albums/o745/nebula-314/20130520_222748_zps3f68f545.jpg[/URL][/PLAIN]

This is how i set it up. If this set up is correct I am having trouble evluating the tan(25+theta) part. If there is an easier way please help.

2. May 20, 2013

### whynot314

sorry for the verticle photo.

3. May 20, 2013

### SammyS

Staff Emeritus
You could use the angle sum identity for tangent: $\displaystyle \ \tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}\ .$

But I think it would be better to solve for tan(25° + θ), then take the arctan of both sides to get a numerical result for 25° + θ . (Make sure that's in degrees.) Subtract 25° from that.

Last edited by a moderator: May 6, 2017
4. May 20, 2013

### whynot314

ahhhhh ok thanks, ill try this in the morning.


text:
Math Descriptive Statistics Article

# Spearman's Correlation

Spearman's correlation is a measure of rank correlation between two numerical variables. It's often denoted as $\rho$ or $r_{s}$.

For example, a Spearman's correlation test can help better identify the relationship between carats in a diamond ring and its price. Does more carats equate to a higher price?

The Spearman's rank correlation is a non-parametric test so there are no requirements of the distributions of the variables to be tested.

On the other hand, in a Pearson's correlation test, there is assumed to be a linear relationship between two variables. However, the Spearman's rank correlation test can identify non-linear relationships between two variables.

### Import Modules¶

In [2]:
import numpy as np
from scipy import stats
import pandas as pd
import seaborn as sns
import matplotlib.pyplot as plt
% matplotlib inline

In [3]:
sns.set_context("talk")
sns.set_style("darkgrid")

### Data Requirements¶

The assumption for the Spearman's rank correlation test is that there is a monotonic relationship between the two variables being tested. A monotonic relationship exists when one of two following situations exists:

1) As one variable increases, the other variable increases.

2) As one variable increases, the other variable decreases.

Spearman's correlation determines the strength (how large or small) and direction (positive or negative) of the monotonic relationship between two variables.

### Visual Examples of Monotonic and Non-Monotonic Relationships¶

Below, as the x-values increase, the y-values almost always increase too so there is a monotonic relationship.

Note there is an observation in which the x-value at $-1$ has a smaller y-value of $-1$ than the previous x-value of $-2$. So there is not a perfect positive monotonic relationship. Generally, as long as the far majority of observations follow one of the two conditions outlined above, there's a monotonic relationship between two variables.

In [4]:
x = [-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5]
y = [-3.2, -2.5, -2.1, -0.8, -1, 1.1, 2.1, 3.8, 6.5, 9.1, 13.8]
plt.figure(figsize=(8, 6))
plt.scatter(x, y, color='red')
plt.title("Monotonic Relationship Between Two Variables", y=1.015);

Below is a monotonic relationship because as the x-values increase, the y-values decrease.

In [5]:
x = [-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5]
y = [5, 4.1, 3.2, 1.8, 1.3, 0.1, -1.1, -2.1, -3.05, -3.7, -4.9]
plt.figure(figsize=(8, 6))
plt.scatter(x, y, color='blue')
plt.title("Another Monotonic Relationship Between Two Variables", y=1.015);

Here is an example of a non-monotonic relationship because as the x-values increase, the y-values first decrease and then later increase.

In [6]:
x = [-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5]
y = [5, 4.1, 3.2, 1.8, 1.3, 0.1, 1.1, 2.1, 3.05, 3.7, 4.9]
plt.figure(figsize=(8, 6))
plt.scatter(x, y, color='green')
plt.title("Non-Monotonic Relationship Between Two Variables", y=1.015);

### Explanation of Coefficient Values¶

If the Pearson's rank correlation value $r_{s}$ is positive, as values from one variable increase, values from the other variable increase.

If the rank correlation value is negative, as values from one variable decrease, the values from the other variable increase.

The table below provides some examples of $r_{s}$ and explanations.

Spearman's rank correlation Interpretation
0 There's no correlation between the two variables
-0.25 There's a small negative correlation between the two variables
-0.75 There's a large negative correlation between the two variables
-1.0 There's a perfect negative correlation between the two variables
0.25 There's a small positive correlation between the two variables
0.75 There's a large positive correlation between the two variables
1.0 There's a perfect positive correlation between the two variables

### Example: Diamonds Dataset - Carat and Price¶

#### Get Diamonds Dataset¶

Import diamonds dataset from Seaborn library and assign to DataFrame df_diamonds.

Each row of df_diamonds contains details about a specific diamond purchased. We'll just utilize the price and carat columns below.

In [7]:
df_diamonds = sns.load_dataset('diamonds')

Preview the first few rows of df_diamonds.

In [8]:
df_diamonds.head()

Out[8]:
carat cut color clarity depth table price x y z
0 0.23 Ideal E SI2 61.5 55.0 326 3.95 3.98 2.43
1 0.21 Premium E SI1 59.8 61.0 326 3.89 3.84 2.31
2 0.23 Good E VS1 56.9 65.0 327 4.05 4.07 2.31
3 0.29 Premium I VS2 62.4 58.0 334 4.20 4.23 2.63
4 0.31 Good J SI2 63.3 58.0 335 4.34 4.35 2.75

#### Explanation of carats¶

carat is a unit of measurement in diamonds to characterize its mass. Typically, bigger/heavier diamonds are known to be more expensive.

Here's a helpful image from womengettingmaried.com.

#### View Distribution of Carats versus Diamond Prices¶

The plot below has carats on the x-axis and price on the y-axis.

In [9]:
df_diamonds.plot.scatter(x='carat', y='price', s=22, alpha=0.25, c='plum', figsize=(11, 8))
plt.ylabel("price [$]", labelpad=15) plt.title("Carats versus Price of Diamonds", y=1.015);  At first glance, it seems there's a positive correlation between the two variables. Generally, as carats increase, price increases. However, this relationship doesn't like like an exact linear correlation. Let's first examine the linear relationship between the two variables by calculating the Pearson's correlation coefficient. In [10]: round(stats.pearsonr(df_diamonds['carat'], df_diamonds['price'])[0], 2)  Out[10]: 0.92 There is a strong positive linear relationship between the two variables. Next, let's examine the Spearman's rank correlation value. In [11]: round(stats.spearmanr(df_diamonds['carat'], df_diamonds['price'])[0], 2)  Out[11]: 0.96 The Spearman's rank correlation value of$0.96$is higher than the Pearson's correlation value of$0.92$. This high value of$0.96$and also because it's greater than$0.92\$ means that the relationship between the two variables can be characterized as something that's not linear. This is evident in the visualization. For every unit increase in carats, there's an increasingly greater unit increase in the price. The relationship between the two variables could be characterized closer to exponential than linear.