text:
Existence results for first order Hamilton Jacobi equations
Annales de l'I.H.P. Analyse non linéaire, Tome 1 (1984) no. 5, pp. 325-340.
@article{AIHPC_1984__1_5_325_0,
author = {Barles, Guy},
title = {Existence results for first order {Hamilton} {Jacobi} equations},
journal = {Annales de l'I.H.P. Analyse non lin\'eaire},
pages = {325--340},
publisher = {Gauthier-Villars},
volume = {1},
number = {5},
year = {1984},
zbl = {0574.70019},
mrnumber = {779871},
language = {en},
url = {http://archive.numdam.org/item/AIHPC_1984__1_5_325_0/}
}
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PY  - 1984
DA  - 1984///
SP  - 325
EP  - 340
VL  - 1
IS  - 5
PB  - Gauthier-Villars
UR  - http://archive.numdam.org/item/AIHPC_1984__1_5_325_0/
UR  - https://zbmath.org/?q=an%3A0574.70019
UR  - https://www.ams.org/mathscinet-getitem?mr=779871
LA  - en
ID  - AIHPC_1984__1_5_325_0
ER  -
Barles, G. Existence results for first order Hamilton Jacobi equations. Annales de l'I.H.P. Analyse non linéaire, Tome 1 (1984) no. 5, pp. 325-340. http://archive.numdam.org/item/AIHPC_1984__1_5_325_0/

[1] S.H. Benton, The Hamilton Jacobi Equation. A global Approach, Academic Press, New York, 1977. | MR 442431 | Zbl 0418.49001

[2] E.D. Conway and E. Hopf, Hamilton's theory and generalized solutions of Hamilton Jacobi equations. J. Math. Mech., t. 12, 1964, p. 939-986. | MR 182761 | Zbl 0178.11002

[3] M.G. Crandall, L.C. Evans, P.L. Lions, Some properties of viscosity solutions of Hamilton Jacobi equations. To appear in Trans. Amer. Math. Soc., 1983. | MR 732102 | Zbl 0543.35011

[4] M.G. Crandall and P.L. Lions, Viscosity solutions of Hamilton Jacobi equations. To appear in Trans. Amer. Math. Soc., 1983. | MR 690039 | Zbl 0599.35024

[5] A. Douglis, The continuous dependance of generalized solutions of non linear partial differential equations upon initial data. Comm. Pure Applied Math., t. 14, 1961, p. 267-284. | MR 139848 | Zbl 0117.31102

[6] W.H. Fleming, The Cauchy problem for a non linear partial differential. J. Diff. equations, t. 5, 1969, p. 515-530. | MR 235269 | Zbl 0172.13901

[7] W.H. Fleming, Nonlinear partial differential equations-probabilistie and game theoretic method. In Problems in Nonlinear Analysis, CIME Ed. Cremonese, Roma, 1971. | MR 282048 | Zbl 0225.35020

[8] W.H. Fleming, The Cauchy problem for degenerate parabolic equations. J. Math. Mech., t. 13, 1964, p. 987-1008. | MR 179473 | Zbl 0192.19602

[9] A. Friedman, The Cauchy problem for first order partial differential equations. Indiana Univ. Mat. J., t. 23, 1973, p. 27-40. | MR 326136 | Zbl 0243.35014

[10] S.N. Kruzkov, Generalized solutions of Hamilton Jacobi equations of Eikonal type. I. USSR Sbornik, t. 27, 1975, p. 406-446. | Zbl 0369.35012

[11] S.N. Kruzkov, Generalized solutions of nonlinear first order equations and certain quasilinear parabolic equations Vestnik Moscow unio Ser. I. Math. Mech., t. 6, 1964, p. 67-74 (In Russian).

[12] S.N. Kruskov, Generalized solutions of first order nonlinear equations in several independant variables. I. Mat. Sb., t. 70, 112, 1966, p. 394-415, II. Mat. Sb. (N. S.), t. 72, 114, 1967, p. 93-116 (in Russian). | Zbl 0164.40802

[13] P.L. Lions, Generalized solutions of Hamilton Jacobi equations. Pitman London, 1982, | MR 667669 | Zbl 0497.35001

[14] P.L. Lions, Existence results for first order Hamilton Jacobi equations. To appear in Ric. Mat. | MR 740198 | Zbl 0552.70012

[15] P.E. Souganidis, Existence of viscosity solutions of Hamilton Jacobi equations. | Zbl 0506.35020

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# Generalizations of the Mean Value Theorem

One of my hobbies here is running, and I recently ran 3km on the track, which led me to the following thought:

Was there a mile (~1600m) which I ran at least as fast as my average pace?

This can be thought of as a generalization of the mean value theorem, which says that a differentiable curve $f:[a,b] \to \mathbb{R}$ has at least one point where the derivative is equal to the “average” slope.  However, this asks whether there exist some (connected) interval where the derivative is less than the average slope.

This statement can be seen to be false, and we might as well use the running example.  Suppose the 3km was run at an average pace of 5:00min/mile, and in the race, for (counter-)example, I rode a bike to the 1,599m mark (which is still considered cheating so far as the IAAF is concerned), and got there after 1:00.  I stay there for 5:01, then ride my bike to the finishline (taking another minute).  I’ve just completed a 7:01 3k (WR, by the way) but there was never a 5minute period where I covered a mile.  With faster and faster bikes (or cars!) my “moving” time could be reduced until it takes me 5:00+$\epsilon$ to run the 3km.  Of course, this is the best I can do without covering a mile in under 5minutes, since as soon as cover 3km in under 5 minutes, I must have (by continuity) covered a mile in under 5 minutes.

Now, since I hate to be a downer, there’s a positive result to share, which can be thought of as a generalization of the above discussion, but does not require any differentiability, or even continuity (unless there are philosophical objections to the discontinuous crossing of a mile).  I’ll state the theorem for the unit interval, then translate that to the running example:

Suppose  $f:[0,1] \to \mathbb{R}$ has average speed $M$ (i.e., $f(1)-f(0) = M$) and $n$ is fixed.  Then for any $p \in (0,1]$ with $p \leq 1/n$, there is an interval with length $p$ where the average speed is at least $M/n$.

To see why this is true, just split up the interval into $n$ equal subintervals, and observe that the “average speed” on at least one of them must be at least $M/n$, or else the total change would not be great enough.

Hence, in the running example, if I ran a 9:00 3k, then I must have covered at least one of the kilometers in 3minutes or less, even if I didn’t cover any mile in under 5minutes!


text:
# Accessing data on local machine from remote kernel

I have a remote kernel on a server that I would like to use to process some data. The connection works, i.e, I'm able to use the local front end and the remote kernel. However, I cannot figure out how should I access from the remote kernel the data that are on my local machine.

I thought I could simply use Import with the path to my computer and to the file that I want to import, but I didn't succeed. Something like:

Import["computerName@ip_address:/Users/myUserName/Desktop/data.txt","Table"]

-
Import the data in the frontend and do DistributeDefinitions? –  Ajasja Jan 18 at 16:33
@Ajasja I looked at the documentation for DistributeDefinitions, but I don't get it how I should use it. For instance, f[n_] := PrimeQ[2^n - 1] and then DistributeDefinitions[f]; should transport the definition to all kernels. But when I try f[2] from the FE connected to the remote kernel the output is f[2] instead of True. –  VLC Jan 18 at 17:13
Ahh, sorry I thought they were slave kernels (for parallel computations), but you have the FE connected directly to a remote kernel. Then you have to share a disk or a drive or copy your data on some webserver... –  Ajasja Jan 18 at 17:17

After some fiddling I found a solution to directly access my local files from the remote kernel via SSH. If localuser is the user name on the machine that runs the front end on which are also the data you want to import from the remote kernel and 111.111.111.11 is the IP address of the same machine, you just need to type:

Import["!ssh localuser@111.111.111.11 cat /Users/localuser/Desktop/data.txt", "Table"]

To make this work smoothly you first need to set up public key authentication over SSH. Let remoteuser be the user name on the machine that runs the remote kernel and let 222.222.222.22 be the IP address of the remote machine.

On your local machine launch the terminal:

1. Generate a public key:
• ssh-keygen -t rsa
2. Ok, when asks to save it in:
• .ssh/id_rsa
3. Press return when asks for passphrase.
4. Press return againg when asks for passphrase.
5. Copy the public key on the remote machine:
• scp id_rsa.pub remoteuser@222.222.222.22:/home/remoteuser
6. Connect to remote machine:
• ssh remoteuser@222.222.222.22
7. Append the public key (previously copied from local machine to remote machine) to the list of authorized keys:
• cat id_rsa.pub >> .ssh/authorized_keys
8. Generate a public key on the remote machine:
• ssh-keygen -t rsa
9. Ok, when asks to save it in: -.ssh/id_rsa
10. Press return when asks for passphrase.
11. Press return againg when asks for passphrase.
12. Copy the public key on the local machine:
• scp id_rsa.pub localuser@111.111.111.11:/Users/localuser
13. Exit from remote server: CTRL+D
14. Append the public key (previously copied from remote machine into local home) to the list of authorized keys:
• cat id_rsa.pub >> .ssh/authorized_keys
-
Nice hack :) +1 –  sebhofer Jan 21 at 14:19
Is there a similar solution for the Export? –  ybeltukov Sep 9 at 23:16
@ybeltukov I don't know if it works, but did you try to follow all the steps above and then simply Export["!ssh localuser@111.111.111.11 cat /Users/localuser/Desktop/newfile.txt", data, "TSV"], where data is the dataset you want to export? –  VLC Sep 16 at 12:13


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# Number of electron for complete Discharge [closed]

What is the answer for question 17 and what is mean by "number of electrons for complete discharge?"

Which of the following needs the largest number of electrons for complete discharge during an electrolysis process?
a) 3 mol of $$\ce{Al^{3+}}$$
b) 4 mol of $$\ce{Al^{3+}}$$ ions
c) 5 mol of $$\ce{OH^-}$$ ions
d) 7 mol of $$\ce{O^{2-}}$$ ions

## closed as off-topic by airhuff, Tyberius, Ivan Neretin, orthocresol♦Oct 4 '18 at 11:00

This question appears to be off-topic. The users who voted to close gave this specific reason:

If this question can be reworded to fit the rules in the help center, please edit the question.

## 1 Answer

$$\text{a)} 3\times+3 = 9$$ $$\text{b)} 4\times+3 = 12$$ $$\text{c)} 5\times-1= -5$$ $$\text{d)} 7\times-2 = -14$$ This seems like an ambiguous question.

• b) "needs" the most electrons supplied to become neutral
• d) "gives up" more electrons than b) to become neutral.

I'd have to take "needs" literally and go with b) as the answer.


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# Math Help - Synthetic division 2

1. ## Synthetic division 2

sorry, this is a hard work sheet. I have another little synthetic division.

you have to divide $3x^5-3x^4+3x^2-3x+10$ by $x+2$. when i did this i got $(3x^5+3x^4+6x^3+16x^2+29x)$+ $\frac {68}{3x^5-3x^4+3x^2-3x+10}$

so when i do $((3x^5+3x^4+6x^3+16x^2+29x)$+ $\frac {68}{3x^5-3x^4+3x^2-3x+10}$ times $(x+2)$ the graph is different from $3x^5-3x^4+3x^2-3x+10$, but only by a little bit. can anyone see where i messed up?

Thank you so much and i appreciate all the help!

2. Originally Posted by OnMyWayToBeAMathProffesor
sorry, this is a hard work sheet. I have another little synthetic division.

you have to divide $3x^5-3x^4+3x^2-3x+10$ by $x+2$. ...
Hi,

I don't know what you have actually done but your result is not correct.

$(3x^5-3x^4+3x^2-3x+10) \div (x+2) = 3x^4-9x^3+18x^2-33x+63- \frac{116}{x+2}$

3. ## thanks

thank you very much, i will figure it out.


text:
# A pizza

A pizza place offers 14 different toppings. How many different three topping pizzas can you order?

Correct result:

n =  364

#### Solution:

$C_{{ 3}}(14)=\dbinom{ 14}{ 3}=\dfrac{ 14! }{ 3!(14-3)!}=\dfrac{ 14 \cdot 13 \cdot 12 } { 3 \cdot 2 \cdot 1 }=364 \ \\ \ \\ n={ { 14 } \choose 3 }=364$

We would be pleased if you find an error in the word problem, spelling mistakes, or inaccuracies and send it to us. Thank you!

Tips to related online calculators
Would you like to compute count of combinations?

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Page NaN of 1946

In a survey on NFL viewership, $589$ subjects were asked if they watch NFL football on a regular basis during the season. In this survey, $43\%$ responded favorably, with a margin of error of $4\%$.

The confidence interval used in this situation is a

A

$95\%$ confidence one-proportion $Z$-interval.

B

$99\%$ confidence one-proportion $Z$-interval.

C

$95\%$ confidence one-sample $T$-interval.

D

$90\%$ confidence proportion $Z$-interval.

E

Cannot be determined with the given information.

Accuracy 0%
Select an assignment template


text:
Positive integers expressed in the form : $\frac{a^b+c}{a+c}$

Note : This problem has no specific source .

Is it true that :

Every positive integer $n$ greater than $1$ can be expressed in the form :

$n=\frac{a^b+c}{a+c}$ , where $a,b>1$ , and $c \in \mathbb{Z} \backslash \{0\}$

I am able to prove following :

$\forall a , \exists ~b,c$ such that : $a+c \mid a^b+c$ , where $b>1$

Proof :

$n=\frac{a^b+c}{a+c}=\frac{a^b-a}{a+c}+1$ , therefore it is sufficient to prove :

$a+c \mid a(a^{b-1}-1)$

If $a+c \mid a$ then $b$ can be any positive integer .

If $a+c \not \mid a$ then we have to prove : $a+c \mid a^{b-1}-1$

or , in other words :

$a^{b-1} \equiv 1 \pmod {a+c}$

Now , for every $a$ there is a number $c$ such that : $\gcd(a,a+c)=1$

According to Euler Theorem :

$a^{\varphi(a+c)} \equiv 1 \pmod {a+c}$

Hence , we can write :

$b-1=\varphi(a+c) \Rightarrow b=1+\varphi(a+c)$

Since $\varphi(a+c)\geq 1$ it follows $b>1$

Q.E.D.

But , this is only necessary condition .

-

Let $a = n$, $b = 3$, $c = n^2$. This does it.

-

Hint $\$ If $\rm\:b,c\:$ are independent of $\rm\:n\:$ and $\rm\:a\:$ is a polynomial in $\rm\:n\:$ of degree $\rm\:d\:$ then comparing degrees in $\rm\:a^b+c = n\:(a+c)\:$ $\Rightarrow$ $\rm\:db = d+1,\:$ so $\rm\:d(b-1) = 1,\:$ hence $\rm\:d=1,\:b=2.\:$ Plugging in undetermined coefficients $\rm\:a\: =\: a_1\: n + a_0 \:$ yields $\rm\:a_1 = 1,\:a_0 = -1,\:$ hence

$$\rm\ \frac{a^b+c}{a+c}\ =\ \frac{(n-1)^2 - 1}{(n-1)-1}\ =\ \frac{n^2 - 2n}{n-2}\ =\ n$$

-

Do you know Fermat's Little Theorem? It is a useful tool here...

-


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# Question #e9bc7

Nov 13, 2016

$\text{_2^4"He" + ""_2^4"He" -> ""_4^8"Be}$

#### Explanation:

The thing to remember about nuclear equations is that they must conserve mass and charge.

In other words, the overall mass number and the overall atomic number must be equal on both sides of the equation.

In your case, you're dealing with a fusion reaction in which two helium-4 nuclides, $\text{^4"He}$, also called alpha particles, fuse to form a new element, beryllium, $\text{Be}$. Start by grabbing a periodic table and looking for the atomic number of the two elements, i.e. of helium and of beryllium.

You will find

$\text{For He: } Z = 2$

$\text{For Be: } Z = 4$

Now, in isotope notation, the atomic number is added to the bottom-left and the mass number is added to the top-left of the chemical symbol.

You will thus have

$\text{_2^4"He} \to$ isotope notation for helium-4

$\text{_4^A"Be} \to$ isotope notation for the resulting isotope

Your goal now is to find the value of $A$, the mass number of the beryllium isotope. You can now write

$\text{_color(blue)(2)^color(orange)(4)"He" + ""_color(blue)(2)^color(orange)(4)"He" -> ""_color(blue)(4)^color(orange)(A)"Be}$

Notice that you have

$\textcolor{b l u e}{2} + \textcolor{b l u e}{2} = \textcolor{b l u e}{4} \to$ charge is conserved

You also need to have

$\textcolor{\mathmr{and} a n \ge}{4} + \textcolor{\mathmr{and} a n \ge}{4} = \textcolor{\mathmr{and} a n \ge}{A} \to$ mass must be conserved as well

The value of $A$ will thus be

$4 + 4 = A \implies A = 8$

Therefore, the resulting isotope is beryllium-8 and the balanced nuclear equation that describes this fusion reaction looks like this

$\text{_2^4"He" + ""_2^4"He" -> ""_4^8"Be}$

As a side note, notice what happens when the beryllium-8 isotope fuses with another alpha particle

$\text{_ 4^8"Be" + ""_ 2^4"He" -> ""_ (color(white)(1)6)^12"C}$

The resulting nuclide is carbon-12, the most abundant isotope of carbon.


text:
# Problems Section: Problems 1371 - 1380

Q1371 Consider shuffles of a standard $52$-card pack.  (See the article in this issue for terminology and basic information.)  Let $\mathop{rev}\nolimits$ be the shuffle which reverses the pack -- that is, the first card is swapped with the last, the second with the second last, and so on.  As in the article, $\mathop{out}\nolimits$ denotes the outshuffle.

(a) Write a formula for $\mathop{rev}\nolimits(k)$ in terms of $k$; also, write $\mathop{rev}\nolimits$ as a product of cycles.
(b) Without any calculation, write down the cycle type of the composite shuffle $\mathop{rev}\circ\mathop{out}\circ\mathop{rev}$.
(c) Show that if we shuffle a pack of cards with both an outshuffle and a reverse shuffle, it makes no difference which one we do first.

Q1372 Show that a composition of $n-1$ cycles
$$(\,1\ 2\,)\circ(\,1\ 3\,)\circ(\,1\ 4\,)\circ\cdots\circ(\,1\ n\,)$$
can be written as a single cycle.  Is
$$(\,1\ n\,)\circ\cdots\circ(\,1\ 4\,)\circ(\,1\ 3\,)\circ(\,1\ 2\,)$$
the same cycle?  For any numbers $a_1,a_2,\ldots,a_m$, describe the shuffle
$$(\,a_1\ a_2\ \cdots\ a_m\,)\circ(a_m\ \cdots\ a_2\ a_1\,)\ .$$


text:
# Question about the autocorrelation function of the fluctuating force in the Langevin model for Brownian motion

According to the Langevin model, we have, for the motion of Brownian particles, $$\frac{dv}{dt} = -M\gamma v + \zeta(t)$$ with $$\zeta(t)$$ the random force acting on the particle due to fluctuations.

Then I was told that the autocorrelation time $$\tau_{c}$$ of this fluctuating force is typically of the order of the time interval between two collisions of the ﬂuid particles on the Brownian particle.

This statement puzzles me. I know these two terms $$-M\gamma v$$ and $$\zeta(t)$$ come from the collision (scattering) between the Brownian particles and the fluid particles But I wonder does the statement mean that the autocorrelation function of the fluctuating force $$\langle\zeta(t)\zeta(t+\Delta t)\rangle$$ have the mathematical relation that $$\langle\zeta(t)\zeta(t+\Delta t)\rangle \sim e^{-\gamma t}$$ and $$\tau_{c} \sim 1/\gamma$$?

If there is such a kind of relation, why? And if not, how to show

that the autocorrelation time $$\tau_{c}$$ of this fluctuating force is typically of the order of the time interval between two collisions of the ﬂuid particles on the Brownian particle.

It is possible to write down a generalized Langevin equation, in which the random force term has a memory: $$\frac{dv(t)}{dt} = -\int_0^t dt' \Gamma(t-t') v(t') + \zeta(t)$$ where I have set $$M=1$$ for simplicity. In this case we have $$\Gamma(t) = \frac{\langle \zeta(0)\zeta(t)\rangle}{\langle v^2\rangle} .$$ So, if you want to have an exponentially decaying correlation function for $$\zeta(t)$$, $$\Gamma(t) = \frac{\gamma}{\tau_c} \exp(-t/\tau_c)$$ you must also have an integral of the frictional term involving a memory kernel based on that correlation function. In the limit that $$\Gamma(t)=\gamma\delta(t)$$ (which comes from the above formula by allowing $$\tau_c\rightarrow 0$$) you recover the Langevin equation $$\frac{dv}{dt} = -\gamma v(t) + \zeta(t) .$$ You can find some discussion of this in various books on statistical mechanics, but also in this online account.


text:
## Methods of modern mathematical physics. I: Functional analysis. Rev. and enl. ed.(English)Zbl 0459.46001

New York etc.: Academic Press, A Subsidiary of Harcourt Brace Jovanovich, Publishers, XV, 400 p. \$ 24.00 (1980).

### MSC:

 46A03 General theory of locally convex spaces 46-01 Introductory exposition (textbooks, tutorial papers, etc.) pertaining to functional analysis 46B03 Isomorphic theory (including renorming) of Banach spaces 46B10 Duality and reflexivity in normed linear and Banach spaces 46C05 Hilbert and pre-Hilbert spaces: geometry and topology (including spaces with semidefinite inner product) 47B06 Riesz operators; eigenvalue distributions; approximation numbers, $$s$$-numbers, Kolmogorov numbers, entropy numbers, etc. of operators 47B25 Linear symmetric and selfadjoint operators (unbounded)

Zbl 0242.46001


text:
RZXGate¶

class RZXGate(theta)[source]

A parameteric 2-qubit $$Z \otimes X$$ interaction (rotation about ZX).

This gate is maximally entangling at $$\theta = \pi/2$$.

The cross-resonance gate (CR) for superconducting qubits implements a ZX interaction (however other terms are also present in an experiment).

Circuit Symbol:

     ┌─────────┐
q_0: ┤0        ├
│  Rzx(θ) │
q_1: ┤1        ├
└─────────┘

Matrix Representation:

\begin{align}\begin{aligned}\newcommand{\th}{\frac{\theta}{2}}\\\begin{split}R_{ZX}(\theta)\ q_0, q_1 = exp(-i \frac{\theta}{2} X{\otimes}Z) = \begin{pmatrix} \cos(\th) & 0 & -i\sin(\th) & 0 \\ 0 & \cos(\th) & 0 & i\sin(\th) \\ -i\sin(\th) & 0 & \cos(\th) & 0 \\ 0 & i\sin(\th) & 0 & \cos(\th) \end{pmatrix}\end{split}\end{aligned}\end{align}

Note

In Qiskit’s convention, higher qubit indices are more significant (little endian convention). In the above example we apply the gate on (q_0, q_1) which results in the $$X \otimes Z$$ tensor order. Instead, if we apply it on (q_1, q_0), the matrix will be $$Z \otimes X$$:

     ┌─────────┐
q_0: ┤1        ├
│  Rzx(θ) │
q_1: ┤0        ├
└─────────┘

\begin{align}\begin{aligned}\newcommand{\th}{\frac{\theta}{2}}\\\begin{split}R_{ZX}(\theta)\ q_1, q_0 = exp(-i \frac{\theta}{2} Z{\otimes}X) = \begin{pmatrix} \cos(\th) & -i\sin(\th) & 0 & 0 \\ -i\sin(\th) & \cos(\th) & 0 & 0 \\ 0 & 0 & \cos(\th) & i\sin(\th) \\ 0 & 0 & i\sin(\th) & \cos(\th) \end{pmatrix}\end{split}\end{aligned}\end{align}

This is a direct sum of RX rotations, so this gate is equivalent to a uniformly controlled (multiplexed) RX gate:

$\begin{split}R_{ZX}(\theta)\ q_1, q_0 = \begin{pmatrix} RX(\theta) & 0 \\ 0 & RX(-\theta) \end{pmatrix}\end{split}$

Examples:

$R_{ZX}(\theta = 0) = I$
$R_{ZX}(\theta = 2\pi) = -I$
$R_{ZX}(\theta = \pi) = -i Z \otimes X$
$\begin{split}RZX(\theta = \frac{\pi}{2}) = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 0 & -i & 0 \\ 0 & 1 & 0 & i \\ -i & 0 & 1 & 0 \\ 0 & i & 0 & 1 \end{pmatrix}\end{split}$

Create new RZX gate.

Attributes

 RZXGate.decompositions Get the decompositions of the instruction from the SessionEquivalenceLibrary. RZXGate.definition Return definition in terms of other basic gates. RZXGate.label Return gate label RZXGate.params return instruction params.

Methods

 RZXGate.add_decomposition(decomposition) Add a decomposition of the instruction to the SessionEquivalenceLibrary. Assemble a QasmQobjInstruction RZXGate.broadcast_arguments(qargs, cargs) Validation and handling of the arguments and its relationship. RZXGate.c_if(classical, val) Add classical condition on register classical and value val. RZXGate.control([num_ctrl_qubits, label, …]) Return controlled version of gate. RZXGate.copy([name]) Copy of the instruction. Return inverse RZX gate (i.e. Return True .IFF. DEPRECATED: use instruction.reverse_ops(). RZXGate.power(exponent) Creates a unitary gate as gate^exponent. Return a default OpenQASM string for the instruction. Creates an instruction with gate repeated n amount of times. For a composite instruction, reverse the order of sub-instructions. Return a numpy.array for the RZX gate.


text:
# Coefficients of homology

I am wondering why people use different coefficients when defining homology of simplicial complex, like homology over $R$, $Z$, $Z/2$, etc? Is one better then the other and why?

Moreover, which one(s) is(are) most useful in practical computation other than abstract theory?

Thanks a lot.

• – Watson Jun 5 '16 at 16:00
• In some sense, if you don't go to deeply in the theory, knowing with $\mathbb{Z}$ coefficients is everything. This is the content of the universal coefficient theorem. It often occurs that one can compute homology with other coefficients more easily (e.g. with Z/2 coeffients poincare duality works directly for non orientable manifolds). Sometimes the way (co) homology is defined leads one directly to different coefficients. A great example is the de Rham cohomology. – Thomas Rot Jun 5 '16 at 20:54

There is no best coefficient in general, but there is indeed a best coefficient for specific case.

Let me show you some examples.

Suppose you want to compute the Euler Characteristic of a space $X$. Suppose that for some reason, counting cell is not a viable option, let's say you don't have an explicit cell structure of your space. You are left with trying to compute the rank of the homology groups. Now you can verify that $\text{rank}H_*(X;\mathbb{Z})=\dim H_*(X,\mathbb{Q})$. The latter group is indeed a vector space, so it has no torsion and it is easier to compute (every short exact sequence splits for example). Even more is true, computing Euler Characteristic is invariant from the chosen field coefficients. In conclusion, it is natural to trying to compute $H_*(X,\mathbb{Q})$ in this case.

Suppose you have a non-orientable closed manifold $M$: When you want to study closed manifolds (from an homology viewpoint), Poincaré Duality is a powerful tool. A powerful tool with some hypothesis to verify, namely orientability. If you don't have orientability you are left with two option, trying to compute $H_*(M;\mathbb{Z}_{\rho})$ (homology with local coefficient - something pretty technical which was let's say created for dealing with this kind of situation) or more easily $H_*(M;\mathbb{Z}_2)$, why? well because every manifold is $\mathbb{Z}_2$ orientable, and therefore you have P-D with these coefficients.

Clearly, homology with $\mathbb{Z}$ coefficients contains all the homological information, since we have UCT which means that if you know homology with $\mathbb{Z}$ for a space, then you know the homology for that space with coefficient any abelian group.

What's the problem: You don't always know how to compute homology with $\mathbb{Z}$-coefficient. By its very property that it bring a lot of informations, computations in general are cumbersome: you have torsion to deal with for instance which is a pain in the ass. So we came up with some weaker version of it, easier to compute but still interesting. This is how research is done usually, you find some great invariant and then realise that it's impossible to compute (very common) and then he must find something weaker but computable and still interesting (very hard).

A last example to back up what I've said. When you will deal with characteristic classes, in particular Stiefel-Whitney ones, you will realise what I mean by sometimes computing cohomology with integer coefficients is hard. But if you turn to $\mathbb{Z}_2$ coefficients, the cohomology ring turns out to be a free graded ring and the generators are the universal classes.

• I can't really see any way of making the assertion "$\operatorname{rank} H_*(M;\mathbb{Z}) = \dim H_*(M;k)$" true if the characteristic of $k$ is positive... – Najib Idrissi Jun 6 '16 at 8:41
• maybe I rushed a little, but if I recall correctly, $\text{rank}H_*(M;\mathbb{Z})=\dim H_*(M;\mathbb{Z})\otimes \mathbb{Q}$ and then one verifies at once that euler char doesn't depend on the field you are computing it. But you are right, I need to modify it a little, since what I wrote it's formally wrong – Riccardo Jun 6 '16 at 8:44
• Yes, by the UCT $\operatorname{rank} H_k(M;\mathbb{Z}) = \dim H_k(M;\mathbb{Q})$, and yes, the Euler characteristic does not depends on the field. But for example $H_2(\mathbb{RP}^2;\mathbb{Z}) = 0$ has rank zero whereas $H_2(\mathbb{RP}^2;\mathbb{F}_2) = \mathbb{F}_2$ has dimension one. ("Formally wrong" is just another way of saying "wrong"...) – Najib Idrissi Jun 6 '16 at 8:46
• yes, and thank you for pointing out this:) – Riccardo Jun 6 '16 at 8:52


text:
# Test 2(Mock Exam)

These mock exams are provided to help you prepare for Term/Final tests. The best way to use these practice tests is to try the problems as if you are taking the test. Please don't look at the solution until you have attempted the question(s). Only reading through the answers or studying them, will typically not be helpful in preparing since it is too easy to convince yourself that you understand it.

Exercise $$\PageIndex{1}$$

Solve the following Initial Value Problems:

1. $$3y^2y^{\prime}-xy^3=x, y(0)=0.$$
2. $$\displaystyle xy' -y = x , y(1) = -1$$
1. $$y=(e^{\frac{ x^2}{2}}-1)^{1/3}$$
2. $$y=x^2-2x.$$
Solution

1. $$3y^2y^{\prime}-xy^3=x \implies 3y^2y^{\prime}=xy^3+x=x(y^3+1).$$

\begin{align*}\displaystyle \int \frac{3y^2 dy}{y^3+1} &= \int xdx\\ ln|y^3+1| &=\displaystyle\frac{ x^2}{2}+C\\ y^3+1 &=Ae^{\frac{ x^2}{2}}\\ y &=(Ae^{\frac{ x^2}{2}}-1)^{1/3}\\ \end{align*}

Now, $$x=0 , y=0 \implies A=1.$$

Hence $$y=(e^{\frac{ x^2}{2}}-1)^{1/3}.$$

2.

$$\displaystyle xy' -y = x , y(1) = -1 \implies \displaystyle y' - \frac{1}{x}y = 1 , y(1) = -1.$$

Hence the integrating factor $$= e^{\int \frac{-1}{x} dx}=e^{-ln(x)}=e^{ln(x^{-1})}=x^{-1}.$$

\begin{align*} \displaystyle x^{-1} y' -x^{-1}\frac{1}{x}y & = 1 \\\displaystyle \frac{d}{dx}(x^{-1}y) &= 1 \\\displaystyle x^{-1}y &= \int dx \\\displaystyle x^{-1}y & = x+C \\\displaystyle y & = x^2+Cx \\\end{align*}

Since $$y(1)=-1,-1=(1)^2+C.$$ Hence $$C=-2.$$Thus the solution is $$y=x^2-2x.$$

Exercise $$\PageIndex{2}$$

Use an appropriate test to determine if the following series converges or diverges. Justify your answer.

1. $$\sum_{k=2}^{\infty} \displaystyle \frac{k}{7^k}$$
2. $$\sum_{k=1}^{\infty} \displaystyle \frac{k+2}{(2k+1)}$$
3. $$\sum_{k=2}^{\infty} \displaystyle \frac{k^k}{(ln(k))^k}$$

Converges, diverges, diverges.

Solution

1. By Ratio Test,

$$\rho=\lim_{k\to \infty} \displaystyle \frac{ (\frac{k+1}{7^{k+1}})}{(\frac{k}{7^k})}= \lim_{k\to \infty} \displaystyle \frac{k+1}{k} \frac{1}{7}= \displaystyle \frac{1}{7} <1.$$ Hence the series converges.

2. By divergence test $$\lim_{k\to \infty} \frac{k+2}{(2k+1)} =\frac{1}{2}\ne 0$$, the series diverges.

3. \begin{align*} \rho &=\lim_{k\to \infty} \displaystyle \sqrt[k]{\frac{ k^k}{(ln(k))^k}} \\&= \lim_{k\to \infty} \displaystyle \frac{ k}{ln(k)} \\&= \lim_{k\to \infty} \displaystyle \frac{1}{ \frac{1}{k} }\mbox{( by L'H\^{o}pital's rule)}\\&= \infty.\end{align*}

By the Root test,  the series diverges.

Exercise $$\PageIndex{3}$$

Determine whether each of the following series converges or diverges, and if it converges, find its sum.  Justify your answer.

1. $$\sum_{k=1}^{\infty} \displaystyle \frac{1}{(k+1)(k+2)}$$
2. $$1+ \displaystyle \frac{\pi}{e}+ \displaystyle \frac{\pi^2}{e^2}+ \displaystyle \frac{\pi^3}{e^3}+\cdots$$
1.  Convergent telescoping series with sum is $$\frac{1}{2}$$.
2. Divergent geometric series.
Solution

1. By using partial fraction decomposition, we get  $$\displaystyle \frac{1}{(k+1)(k+2)}= \displaystyle \frac{1}{k+1}-\frac{1}{k+2}.$$

Thus $$\sum_{k=1}^{\infty} \displaystyle \frac{1}{(k+1)(k+2)} = \sum_{k=1}^{\infty}( \displaystyle \frac{1}{k+1}- \displaystyle \frac{1}{k+2}).$$ The second part of each term cancels with the first part of the  succeeding term, hence $$S_n= \displaystyle \frac{1}{2}- \displaystyle \frac{1}{n+2}.$$ Therefore, $$\lim_{n\to \infty} S_n= \displaystyle \frac{1}{2}$$.

2. This is a geometric series with first term $$a=1$$ and the ratio $$r= \displaystyle \frac{\pi}{e} >1$$. Thus this series diverges.

Exercise $$\PageIndex{4}$$

A glass of juice with a temperature of $$40^{\circ} F$$ is placed in a room with a constant temperature of $$75^{\circ}F,$$ and 1 hour later its temperature is $$55^{\circ}F.$$ Using Newton’s Law of Cooling, show that $$t$$ hours after the juice is placed in the room its temperature is approximated by $$T=75-35e^{-.56t}.$$

Solution

The ambient temperature (surrounding temperature) is $$75°F$$, so $$T_s=75$$. The temperature of the glass of juice when it placed in a room  is $$40°F$$, which is the initial temperature (i.e., initial value), so $$T_0=40$$. Therefore Equation becomes

$\dfrac{dT}{dt}=k(T−75) \nonumber$

with $$T(0)=40.$$

Rewrite the differential equation by multiplying both sides by $$dt$$ and dividing both sides by $$T−75$$:

$\dfrac{dT}{T−75}=k\,dt. \nonumber$

Integrate both sides:

\begin{align*} ∫\dfrac{dT}{T−75} &=∫k\,dt \\ \ln|T−75| &=kt+C.\end{align*}

Solve for $$T$$ by first exponentiating both sides:

\begin{align*}e^{\ln|T−75|} &=e^{kt+C} \\ |T−75| &=C_1e^{kt}, & & \text{where } C_1 = e^C. \\ T−75 &=\pm C_1e^{kt} \\ T−75 &=Ce^{kt}, & & \text{where } C = \pm C_1\text{ or } C = 0.\\ T(t) &=75+Ce^{kt}. \end{align*}

Solve for $$C$$ by using the initial condition $$T(0)=40:$$

\begin{align*}T(t) &=75+Ce^{kt}\\ T(0) &=75+Ce^{k(0)} \\ 40 &=75+C \\ C &=-35.\end{align*}

Therefore the solution to the initial-value problem is

$T(t)=75-35e^{kt}.\nonumber$

To determine the value of $$k$$, we need to use the fact that after $$1$$ hour the temperature of the juice is $$52°F$$. Therefore $$T(1)=55.$$ Substituting this information into the solution to the initial-value problem, we have

$T(t)=75-35e^{kt}\nonumber$

$T(1)=55=75-35e^{k}\nonumber$

$-20=-35e^{k}\nonumber$

$e^{k}=\dfrac{4}{7}\nonumber$

$\ln e^{k}=\ln(\dfrac{4}{7})\nonumber$

$k=\ln(\dfrac{4}{7})\nonumber$

$k≈−0.56.\nonumber$

So now we have $$T(t)=75-35e^{−0.56t}.$$

Exercise $$\PageIndex{5}$$

For each of the following series determine whether the series converges absolutely, converges conditionally, or diverges.  Justify your answer.

1. $$\sum_{k=1}^\infty \displaystyle (-1)^{k+1}{3^k \over k!}$$
2. $$\sum_{k=1}^\infty \displaystyle (-1)^{k+1}{\sqrt{k+3} \over k}$$
3. $$\sum_{k=1}^\infty \displaystyle (-1)^{k+1} \left( \frac{k+1}{k} \right)^k$$

Converges absolutely, Converges conditionally, diverges

Solution

1. \begin{align*}\rho & = \lim_{k\to\infty}\left|a_{k+1}\over a_k\right|\\&=\lim_{k\to\infty} \dfrac{\left|(-1)^{k+2} {3^{k+1}\over (k+1)!} \right|}{\left|(-1)^{k+1} {3^k \over k!} \right|}\\&= 3 \lim_{k\to\infty} \dfrac{1}{(k+1)}\\&= 0<1. \end{align*}

Hence by the Ratio test, the series converges absolutely.

2. Converges conditionally by alternating series test, since $$\displaystyle \sqrt{k+3}/k$$ is decreasing. Does not converge absolutely by comparison with p-series, $$\displaystyle p=1/2$$

3. Diverges by divergence test since $$\displaystyle \lim_{k→∞}|a_k|=e$$.

Test 2(Mock Exam) is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by Pamini Thangarajah.


text:
# Math Help - Complex numbers

1. ## Complex numbers

The question:

Simplify $(\sqrt{3 + 4i}+\sqrt{3-4i})^2$ where we assume $\sqrt{z}$ has a non negative real part.

My attempt:

$(3+4i) + 2(9-16i^2) + (3-4i)$ (perfect square)

$3 + 4i + 18 + 32 + 3 - 4i$

= 56

That's not the answer though, it's supposed to be 16. I'm not sure what I'm going wrong. Any assistance would be great!

2. Originally Posted by Glitch
The question:

Simplify $(\sqrt{3 + 4i}+\sqrt{3-4i})^2$ where we assume $\sqrt{z}$ has a non negative real part.

My attempt:

$(3+4i) + 2(9-16i^2) + (3-4i)$ (perfect square)

$3 + 4i + 18 + 32 + 3 - 4i$

= 56

That's not the answer though, it's supposed to be 16. I'm not sure what I'm going wrong. Any assistance would be great!

Who knows what you did here: $(\sqrt{a}+\sqrt{b})^2=a+2\sqrt{ab}+b^2$ , and you did something different (I think though

that you forgot the square root in the middle term), so

$(\sqrt{3 + 4i}+\sqrt{3-4i})^2=3+4i+2\sqrt{9+16}+3-4i=3+10+3=16$ , since $9-16i^2=9+16$

Tonio

3. Originally Posted by Glitch
The question:

Simplify $(\sqrt{3 + 4i}+\sqrt{3-4i})^2$ where we assume $\sqrt{z}$ has a non negative real part.

My attempt:

$(3+4i) + 2(9-16i^2) + (3-4i)$ (perfect square)

$3 + 4i + 18 + 32 + 3 - 4i$
The middle term is wrong. You should still have a square root- it should be $2\sqrt{9- 16i^2}= 2(5)$, not 2(5)(5).

= 56

That's not the answer though, it's supposed to be 16. I'm not sure what I'm going wrong. Any assistance would be great!

4. Thanks! I can't believe I missed that!


text:
# How is target return optimized, in practice?

Suppose we have $$A$$ a $$T \times N$$ matrix of daily returns for an asset universe of $$N$$ items, $$b$$ a $$(T,)$$ vector of daily returns for a target asset, $$x$$ a $$(N,)$$ weighting vector. We want a portfolio that replicates daily returns $$b$$ as closely as possible, subject to the constraint that portfolio weights $$x$$ sum to 1 and $$x$$ is non-negative (no shorting). One might use this for tax loss harvesting or hedging.

If we first consider the unconstrained case:

$$\min_x \sum_t^{T} (A \cdot x - b)$$

This is nothing more than least-squares approximation / regression, whose objective can be re-written as:

$$|Ax-b|^2 = (Ax-b)^T(Ax-b) = x^T{A^T}Ax - 2x^TA^Tb + b^2$$

Taking the derivative wrt $$x$$ and setting to 0,

$$2A^TAx - 2A^Tb = 0 \\ x = (A^TA)^{-1}A^Tb$$

To handle equality constraints such as $$x$$ summing to 1, and inequality constraints like $$x$$ being non-negative, one can set up a constrained least squares problem, solve the KKT matrix, and still derive a closed form solution, albeit one that has cubic complexity due to necessity of inverting matrices.

However, the portfolio manager may have a few other constraints - maybe they prefer to keep $$x$$ relatively sparse, so they can just buy about 3-4 assets rather than allocating a small percentage across thousands of assets. There are more complex techniques like ridge regression, lasso regression, and so on, but these start to involve stepwise optimization (e.g. coordinate descent).

My question is one about how the portfolio optimization industry actually optimizes in practice: is it more similar to analytic convex optimization with simple inequality / equality constraints, and maybe a bit of hacks on top of the Least-square regression (for instance, rounding off small values of $$x$$), or is more step-wise optimization the norm? Given a sufficiently complex objective, would it make sense just to switch to a more black-box optimization algorithm, like a genetic algorithm or some kind of stochastic gradient descent optimizer?

• I have seen the latter become normal at a number of places - e.g. pick as many constraints as you want and chuck your objective function into scipy.optimize, of course while monitoring convergence and checking whether you are hitting constraints etc. Jun 4 at 16:45

Your practical problem here is the insistence on non-negative weights, that sum to 1. There is no closed-form equation that solves this problem.

However, it can be done via iterative methods. You weight each stock i by Ai that is unbounded, Its portfolio weight Wi is e(AI)/sum(e(Ai)). The weights sum to 1 and are all positive.

SumW = sum(e(Ai))

dWi/dAi = Wi x (1-Wi) / SumW

dUtility/dAi thus becomes above x dUtility/dWi


text:
Following up from my previous post, here’s the next problem:

The public key is $(n, e) = (221, 11)$. What is the RSA Decryption of $7$?

• What is RSA? wikipedia: RSA (Rivest–Shamir–Adleman) is one of the first public-key cryptosystems and is widely used for secure data transmission. In such a cryptosystem, the encryption key is public and distinct from the decryption key which is kept secret (private). In RSA, this asymmetry is based on the practical difficulty of factoring the product of two large prime numbers, the “factoring problem”.

• How does RSA work?
Let’s look at an example from wikipedia:

1. Choose two distinct prime numbers, such as
$p = 61$ and $q = 53$
2. Compute $n = pq$ giving
$n = 61 \times 53 = 3233$
3. Compute the Carmichael’s totient function of the product as
$\lambda (n)=\operatorname {lcm} (p-1,q-1)$ giving
$\lambda (3233)=\operatorname {lcm} (60,52)=780$
4. Choose any number $1 \lt e \lt 780$ that is coprime to $780$. Choosing a prime number for $e$ leaves us only to check that $e$ is not a divisor of $780$. Let $e = 17$.
5. Compute $d$, the modular multiplicative inverse of $e \bmod \lambda(n)$ yielding, $d = 413$
Worked example for the modular multiplicative inverse:
$d\times e \equiv 1{\bmod {\lambda }}(n)$
$413\times 17 \equiv 1{\bmod {7}}80$

The public key is $(n = 3233, e = 17)$. For a padded plaintext message m, the encryption function is $c(m)=m^{17} \bmod 3233$

The private key is $(n = 3233, d = 413)$. For an encrypted ciphertext c, the decryption function is $m(c)=c^{413} \bmod 3233$

For this problem we have $n$, $e$, and $c$. We have to find $d$ and calculate $m$ using the above decryption function.
$n = 221 = 13 \times 17$. Carmichael’s totient function, $\lambda (221)=\operatorname {lcm} (12,16)=48$. We already have $e = 11$.
$d$ is the modular multiplicative inverse of $e \bmod \lambda(n)$ i.e.,
$d \times e \equiv 1 \bmod \lambda (n)$ or $d \times 11 \equiv 1 \bmod 48$.

We can write a simple program to calculate d:

d = 0
while (11 * d) % 48 != 1: d+=1
print(d)

35

Now we can find $m$ as: $m = c ^ d \bmod n = 7 ^{35} \bmod 221 = 54$

Thus, 54 is our final answer.

• What is Carmichael’s totient function?
wikipedia: The Carmichael function associates to every positive integer $n$ a positive integer $\lambda (n)$, defined as the smallest positive integer $m$ such that
$$a^m \equiv 1 \bmod n$$
for every integer $a$ between $1$ and $n$ that is coprime to $n$.

wikipedia: The exponent of the multiplicative group of integers mod n (previous post), that is, the least common multiple of the orders in the cyclic groups, is given by the Carmichael function $\lambda (n)$. In other words, $\lambda (n)$ is the smallest number such that for each $a$ coprime to $n$, $a^{\lambda(n)} \equiv 1 \pmod n$ holds.

• How can we find the modular multiplicative inverse using pen and paper? By making use of the Extended Euclidean Algorithm. Here’s a youtube video that walks you through the procedure and here’s a Khan Academy article that explains it well. Adding these here for myself for future reference.

Now let’s try another one:

The public key is $(n, e) = (437, 13)$. What is the RSA Digital Signature of $17$?

• What is RSA Digital Signature?
This article explained it very well. From the article, here is the “textbook” definition of RSA signing compared with the Encryption and Decryption functions.

• $Enc(m; e) = R(m,e)$
• $Dec(c; d) = R(c,d)$
• $Sign(m; d) = R(m,d)$
• $Ver(m; s; e) = R(s,e) == m$

We are basically applying the sender’s private key on the message they are sending to generate a signature. That means, for this problem, we have to find $m^d \bmod n$.

We can follow the procedure from the previous problem now. $d$, or the private key, is the modular multiplicative inverse of $e \bmod \lambda (n)$. Here, $\lambda (n)$ is the Carmichael totient function, calculated as $\lambda (n) = lcm (p-1, q-1)$, where $p$ and $q$ are the prime factors of $N$.

Through trial and error, we find that the prime factors of $n = 437$ are $19$ and $23$. Of course this would be a very difficult problem if the prime numbers were sufficiently large; that’s the whole idea behind RSA encryption. Anyways, now that we have $p$ and $q$, we can find $\lambda (437) = lcm (18, 22) = 198$. Now, using the same algorithm I used for the previous problem,

d = 0;
while (d * 13) % 198 != 1: d+=1
print(d)

61

Therefore, the solution to this problem is $17^{61}\bmod 437 = 195$.

Until next time!


text:
# For her cell phone plan Heather pays $30 per month plus$0.05 per text. She wants to keep her bill under $60 per month. What inequality represents how many texts Heather can send each month while staying within her budget? ##### 1 Answer Mar 29, 2018 #### Answer: The monthly cost is $30 + 0.05 n$if we call the number of texts $n$. Heather wants to keep this less than $60$, so the inequality is: $30 + 0.05 n < 60$#### Explanation: Note that we can also subtract $30$from each side: $0.05 n < 30$We can also divide both sides by $0.05$: $n < 600\$

This is the most useful inequality, since it tells Heather she needs to send fewer than 600 texts per month, or about 20 per day.


text:
# Kind (type theory)

In the area of mathematical logic and computer science known as type theory, a kind is the type of a type constructor or, less commonly, the type of a higher-order type operator. A kind system is essentially a simply typed lambda calculus "one level up", endowed with a primitive type, denoted $*$ and called "type", which is the kind of any data type which does not need any type parameters.

A kind is sometimes confusingly described as the "type of a (data) type", but it is actually more of an arity specifier. Syntactically, it is natural to consider polymorphic types to be type constructors, thus non-polyomorphic types to be nullary type constructors. But all nullary constructors, thus all monomorphic types, have the same, simplest kind; namely $*$.

Since higher-order type operators are uncommon in programming languages, in most programming practice, kinds are used to distinguish between data types and the types of constructors which are used to implement parametric polymorphism. Kinds appear, either explicitly or implicitly, in languages whose type systems account for parametric polymorphism in a programatically accessible way, such as Haskell and Scala.[1]

## Examples

• $*$, pronounced "type", is the kind of all data types seen as nullary type constructors, and also called proper types in this context. This normally includes function types in functional programming languages.
• $* \rightarrow *$ is the kind of a unary type constructor, e.g. of a list type constructor.
• $* \rightarrow * \rightarrow *$ is the kind of a binary type constructor (via currying), e.g. of a pair type constructor, and also that of a function type constructor (not to be confused with the result of its application, which itself is a function type, thus of kind *)
• $(* \rightarrow *) \rightarrow *$ is the kind of a higher-order type operator from unary type constructors to proper types. See Pierce (2002), chapter 32 for an application.

## Kinds in Haskell

(Note: Haskell documentation uses the same arrow for both function types and kinds.)

Haskell's kind system[2] has just two rules:[vague]

• $*$, pronounced "type" is the kind of all data types.
• $k_1 \rightarrow k_2$ is the kind of a unary type constructor, which takes a type of kind $k_1$ and produces a type of kind $k_2$.

An inhabited type (as proper types are called in Haskell) is a type which has values. For instance, ignoring type classes which complicate the picture, 4 is a value of type Int, while [1, 2, 3] is a value of type [Int] (list of Ints). Therefore, Int and [Int] have kind $*$, but so does any function type, for instance Int -> Bool or even Int -> Int -> Bool.

A type constructor takes one or more type arguments, and produces a data type when enough arguments are supplied, i.e. it supports partial application thanks to currying.[3][4] This is how Haskell achieves parametric types. For instance, the type [] (list) is a type constructor - it takes a single argument to specify the type of the elements of the list. Hence, [Int] (list of Ints), [Float] (list of Floats) and even [[Int]] (list of lists of Ints) are valid applications of the [] type constructor. Therefore, [] is a type of kind $* \rightarrow *$. Because Int has kind $*$, applying it to [] results in [Int], of kind $*$. The 2-tuple constructor (,) has kind $* \rightarrow * \rightarrow *$, the 3-tuple constructor (,,) has kind $* \rightarrow * \rightarrow * \rightarrow *$ and so on.

### Kind inference

Standard Haskell does not allow polymorphic kinds. This is in contrast to parametric polymorphism on types, which is supported in Haskell. For instance, in the following example:

data Tree z  = Leaf | Fork (Tree z) (Tree z)

the kind of z could be anything, including $*$, but also $* \rightarrow *$ etc. Haskell by default will always infer kinds to be $*$, unless the type explicitly indicates otherwise (see below). Therefore the type checker will reject the following use of Tree:

type FunnyTree = Tree []     -- invalid

because the kind of [], $* \rightarrow *$ does not match the expected kind for z, which is always $*$.

Higher-order type operators are allowed however. For instance:

data App unt z = Z (unt z)

has kind $(* \rightarrow *) \rightarrow * \rightarrow *$, i.e. unt is expected to be a unary data constructor, which gets applied to its argument, which must be a type, and returns another type.

GHC has the extension PolyKinds, which, together with KindSignatures, allows polymorphic kinds. For example:

data Tree (z :: k) = Leaf | Fork (Tree z) (Tree z)
type FunnyTree = Tree []     -- OK


text:
# Tensor contraction

Suppose, there is a tensor of rank three, given by $C^{ij}_k$ and it is a product of two other tensors. The relation is as follows:

$C^{ij}_k$ = $A^{ij}$$B_k This C^{ij}_k has 27 elements in it. If I contract the indices j and k Then I get the following: C^i = A^{ij}$$B_j$

By equating $j$ and $k$ the tensor has been reduced to a rank 1 tensor and has only 3 elements. The process is now similar to ordinary matrix multiplication between a 3X3 and a 3X1 matrices.

I get that by equating two indices, we are selecting only a few of the elements of original $C^{ij}_k$ and summing them up, which eventually represents a vector.

My question is, what happens to the other elements? Do they become zero for some reason? If so, what property makes them zero? If they are not zero, why we are allowed to simply ignore them? what do they represent?

$$C^{1} = C^{11}B_{1} + C^{12}B_{2} + C^{13}B_{3}$$
each of the $C^{i}$ is a combination of several elements from the old matrix.
• What about the element $C^{11}B_2$? It is in none of the sums. – Samapan Bhadury Oct 20 '17 at 18:06
• @SamapanBhadury: it's not set to zero, $C^{i}$ just doesn't depend on it. – Jerry Schirmer Oct 20 '17 at 18:39


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# Does this generalisation of Latin squares have a name?

I am interested in looking at $n\times n$ tableaux (or matrices) in which (WLOG) each integer in $\{ 1, 2, \ldots, n \}$ occurs exactly $n$ times. This is a generalisation of a Latin (or even semi-Latin) square, which obviously has this property. It is a proper generalisation, as there are tableaux with this property that are not semi-Latin, such as $$\left[\begin{matrix} 1&2&2\\3&1&3\\3&1&2\end{matrix}\right].$$ I have counted the number of such tableaux, up to a suitable (for my applications) notion of isomorphism for $n = 2,3,4$, and there are lots of them. There are $5$ with $n = 2$; $305$ with $n = 3$; and $2630904$, with $n = 4$. As these generalise Latin squares, it seems like the sort of thing that people in the combinatorics community might have investigated. Have such tableaux been studied before? Do they have a name? (If these have a name, I might do better with Google.) Many thanks.

EDIT: For the $2\times 2$ case, there are $6 = {4\choose 2}$ squares, since we can choose any two of the four matrix positions in which to place a $1$, and then the other two spots must contain a $2$. These are: $$\left\{ \left[\begin{matrix}1&1\\2&2\end{matrix}\right], \left[\begin{matrix}1&2\\1&2\end{matrix}\right], \left[\begin{matrix}1&2\\2&1\end{matrix}\right], \left[\begin{matrix}2&1\\1&2\end{matrix}\right], \left[\begin{matrix}2&1\\2&1\end{matrix}\right], \left[\begin{matrix}2&2\\1&1\end{matrix}\right]\right\}.$$ Now, of these, only the pair $\left[\begin{matrix}1&2\\2&1\end{matrix}\right]$ and $\left[\begin{matrix}2&1\\1&2\end{matrix}\right]$ are equivalent.

The notion of equivalence or isomorphism used is this. Two such $n\times n$ matrices $(a_{i,j})$ and $(b_{i,j})$ as above, are regarded as essentially the same if there is a permutation $\sigma\in S_{n}$ for which $\sigma( a_{i,j} ) = b_{\sigma i, \sigma j}$, for all $i$ and $j$.

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What is said notion of isomorphism? – Alex Becker Dec 31 '11 at 2:05
@AlexBecker: Isomorphism is defined as follows. Think of these as $n\times n$ matrices. Then $(a_{i,j})$ and $(b_{i,j})$ are isomorphic if there is a permutation $\sigma\in S_{n}$ for which $\sigma a_{i,j} = b_{\sigma i,\sigma j}$. – James Dec 31 '11 at 4:52

These are called "equi-n-squares."

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Before your isomorphism, the tableau representation is not important and you can just view it as a list. As such, you can pick $n$ of the $n^2$ positions for $1$, $n$ of the remaining $n^2-n$ for $2$ and so on. It becomes $$\binom {n^2}n \binom{n^2-n}n \binom {n^2-2n}n \ldots \binom nn$$ I didn't find 2630904 in OEIS. Without the isomorphism it is $\frac{(n^2)!}{(n!)^n}$ which is A034841.

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I didn't find the sequence (counted up to equivalence) in the OEIS either. I didn't think of trying the sequence without eliminating equivalents, though. Nice. – James Dec 31 '11 at 9:13
@James: I'm not clear on the isomorphism range you are talking about. For $n=2$ it seems you could just define that the upper left is $1$, then the other $1$ can go in any location, so there should be $3$, not $5$. For this, you just divide by one more factor of $n!$. That one is oeis.org/A057599 – Ross Millikan Dec 31 '11 at 16:06
I've added some further explanation to the question body. (It seemed too long for a comment.) Thanks. – James Jan 1 '12 at 2:34

Courtiel and Vaughan, Gerechte designs with rectangular regions, JCD (2012) calls them a gerechte framework.

A gerechte framework is a partition of an $n \times n$ array into $n$ regions of $n$ cells each.

This definition comes from a gerechte design, which is a Latin square together with the entries partitioned into sets of size $n$ such that each part contains each symbol exactly once (like sudoku).

In 1956, W. U. Behrens [4] introduced a specialisation of Latin squares which he called “gerechte”. --- Bailey, Cameron, Connelly. AMM (2008). (pdf)

W. U. Behrens, Feldversuchsanordnungen mit verbessertem Ausgleich der Bodenunterschiede, Zeitschrift für Landwirtschaftliches Versuchs- und Untersuchungswesen 2 (1956), 176-193.

(Unfortunately, I don't have access to this paper, so my trail ends here.)

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Many thanks! I don't think I can get that (Behrens) paper either but, even if I could, I can't read German. (I guess it is German.) But I'll look at the other papers you've kindly provided. – James Jul 13 '14 at 6:12


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Partial optional randomized response technique with calibration weighting to adjust non-response in successive sampling

Kumari Priyanka1,a, Pidugu Trisandhyab , Ajay Kumarb

aDepartment of Mathematics, Shivaji College (University of Delhi), India
bDepartment of Mathematics, University of Delhi, India
Correspondence to: 1 Department of Mathematics, Shivaji College (University of Delhi), New Delhi 110 027, India. priyanka.ism@gmail.com
Received March 19, 2021; Revised May 4, 2021; Accepted June 8, 2021.
Abstract
The present article endeavours to develop partial optional randomized - response technique (PORT) to deal with sensitive issues in presence of non-response in successive sampling. Calibration techniques have been embedded with PORT to estimate sensitive population mean at current move in two move successive sampling in presence of non-response. Optimum calibration weights are computed at each move with the aid of constraints based on auxiliary information. Detailed properties of the proposed estimators have been discussed. Possible cases in which non-response may creep at two moves has been explored. The proposed technique has been compared with the modified existing technique. Simulation results indicate that the proposed technique is more ecient than existing, modified one. Suitable recommendations are forwarded.
Keywords : partial optional randomized-response technique, sensitive variable, successive moves, calibration estimators, non-response, population mean, auxiliary information
1. Introduction

In many social surveys, people often do not respond genuinely when socially sensitive questions are asked. To account for this behaviour, different techniques were introduced by many statisticians to reduce no reporting, under-reporting or over reporting.

The sensitive issues may be matters of medical/legal malpractices, addiction of drugs, criminal conviction, induced abortions, acid attacks, domestic violence, etc. Two widely practised indirect questioning techniques dealing with these issues are randomized response technique (RRT) and scrambled response technique (SRT) that protect the privacy of respondents and mask the sensitive response, thereby motivating the respondents to give accurate response.

Warner (1965) was the first to provide such a randomizing model and followed by sizeable literature was added by Horvitz et al. (1967), Greenberg et al. (1971), Christofides (2003), Kim and Elam (2007), Wu et al. (2008), Yan et al. (2009), Arnab (2011), Diana and Perri (2011), Arnab et al. (2012). The SRT was initiated by Pollock and Bek (1976). Further, Eichhorn and Hayre (1983), Saha (2007), Diana and Perri (2010, 2011), Perri and Diana (2013) and Hussain and Al-Zahrani (2016) added substantive literature in this area. In RRT or SRT the respondent need to provide randomized or scrambled response. However, depending on the thought of respondents it might be possible that some issue may be sensitive to one respondent but the same issue may not be sensitive to other. Hence, to address these problems optional randomized response technique (ORT) may be used. ORT can be classified into two categories: full optional randomized response technique (FORT) and partial optional randomized response technique (PORT). The ORT is more efficient than compulsory randomized response technique (CRT) because the probability of obtaining true responses in ORT is much higher than that in the CRT (Arnab, 2004). PORT was discussed by many authors such as Mangat and Singh(1994), Gupta et al. (2002, 2013), Pal (2008), Chaudhuri and Dihidar (2009), and Sanaullah et al. (2020).

It is common that the sensitive variable may be dynamic over time, in such a situation one point survey may not be sufficient. Continuous monitoring of sensitive variable may be required. The dynamics of such sensitive variables may be studied using successive sampling. Addressing the sensitive variable, Arnab and Singh (2013), Yu et al. (2015), Naeem and Shabbir (2016), Singh et al. (2017), Priyanka et al. (2018), Priyanka and Trisandhya (2019a, 2019b), Priyanka et al. (2019), Singh et al. (2018) contributed rich literature. These researchers used simple random sampling design in successive sampling and used either RRT or SRT to deal with sensitive issues. However, no attempt has been made to estimate sensitive population mean using PORT in successive sampling.

Generally, all sample surveys are affected by the problem of non-response. When the issues are sensitive, then they are more prone to occur and can severely affect the validity and generalizability of the results. Non-response are generally of two types namely unit non-response and item non-response. In unit non-response, sampled unit fails to respond completely, however in item non-response, the sampled unit responds to the survey but fails to respond to a particular question.

Hence, before proceeding with any method the kind of non-response creeping in the survey must be identified and suitable measure must be devices.

Therefore, in this paper an attempt has been made to estimate sensitive population mean at current move using PORT with calibration weighting to adjust unit non-response in two move successive sampling. A new model for PORT has been proposed and the existing model by Sanaullah et al. (2020) have been modified to work as PORT model for successive sampling. The paper is structured as follows, in Section 2, the proposed PORT models on two successive moves have been presented along with the calibration estimators in presence of unit non-response at both the moves for the coded response variable at current move. In Section 3, the asymptotic behaviour of the estimators developed in previous section have been studied. Section 4 is devoted to discuss the study under simple random sampling without replacement. In Section 5, the possible cases in which the non-response may creep has been discussed. Section 6 presents the corresponding estimators for sensitive population mean along with the expression for their variances. Section 7 is dedicated to a simulation study to compare the precision of the proposed estimators with the usual modified estimator and the comparison of the proposed PORT models. Finally, Section 8 is devoted to the discussion of result and conclusion.

2. Survey design and development of estimator

### 2.1. Formulation and notation

A finite population U = (U1,U2, . . .,UN) of N identifiable units is considered for sampling over two moves. The size of population is constant while the value of units do change over the moves. Let the sensitive study character be denoted by y1 the first move and y2 the second move. It is assumed that the information on non-sensitive auxiliary variable x, whose population mean is known and stable over moves, is readily available on both the moves. In two moves successive sampling, we intend to focus on the probability design to draw relevant samples on different moves under unit non-response for the estimation of population mean of sensitive variable. At the first move, a sample sn of size n is drawn with the design d1 having probability pd1 (sn). We assume that sr1 of size r1 be the response set from sn; sr1snU, having probability pd1 (sr1). A random sub-sample sm of m units from sr1 (the set of responding units at first move) are preserved with design d2 having the probability pd2 (sm) to be used at current move. Next, an unmatched sample su of size u is drawn afresh (without replacement) at second move with the design d3 having probability pd3 (su). Let the response set r2 of size sr2 is obtained from su; sr2su ⊂ (Usn) having the probability pd3 (sr2). Let S1 and S2 be two scrambling variables that will be used to code the response for the sensitive variables. It is assumed that two scrambling variables are independent of each other and their mean and variance are known. Further, let z1(z2) be the coded response variable corresponding to sensitive variables y1(y2) on two moves. The first order and second order positive inclusion probabilities for different samples on two moves are shown in Table 1 and sampling design basic weights are presented in

Let the known mean and variance of scrambling variables be assumed as E(S1) = 1, E(S2) = 2, $V(S1)=σs12,V(S2)=σs22$.

Based on the considered sampling design, we intend to apply partial optional randomization technique (PORT) on successive moves to handle sensitivity of study variable.

### 2.2. Partial optional randomization technique on successive moves

Motivated by recent work of Sanaullah et al. (2020), we intend to modify their generalized randomised response model to be applicable for successive sampling under PORT. The model is modified as,

• PORT-I: Modified model (Sanaullah et al., 2020)

The response obtained from the jth respondent on first and second move are respectively given as,

$z1j={y1j,with probability p,y1jS1j+aS2j,with probability (1-p),$$z2j={y2j,with probability p,y2jS1j+aS2j,with probability (1-p),$

where a ∈ (−∞, ∞) is a suitable constant chosen by the investigator (Himmelfarb and Edgell, 1980).

Taking expectation on both sides of equation (2.1) and equation (2.2) respectively, we get

$Z¯1=pY¯1+(1-p)[Y¯1S¯1+aS¯2],Z¯2=pY¯2+(1-p)[Y¯2S¯1+aS¯2].$

Therefore, the population mean of sensitive variable at current move is given as

$Y2=Z¯2-(1-p)aS¯2p+(1-p)S¯1,$

such that

$ρz1z2=p2(ρy1y2σy1σy2)+2p(1-p) (ρy1y2σy1σy2S¯1)+(1-p)2[(ρy1y2σy1σy2) (σs12+S¯12)+Y¯1Y¯2σs12]I1I2ρz1y1=(p+(1-p)S¯1)σy1I1,ρz2y2=(p+(1-p)S¯1)σy2I2,I1=p2σy12+(1-p)2[σy12(σs12+S¯12)+σs12Y¯12+a2σs22],I2=p2σy22+(1-p)2[σy22(σs12+S¯12)+σs12Y¯22+a2σs22].$
• PORT-II: Proposed model

Since the modified Sanaullah et al. (2020) model involves an unknown constant to be chosen by the investigator. This selection of constant may involve an added source of randomness in the model. Hence, we intend to propose a model for PORT on successive moves which is independent of arbitrary chosen constant. The response received from the jth respondent under the proposed PORT model on first and second move are respectively given as

$z1j={y1j,with probability p,y1jS1j+S2j-S¯2S¯1with probability (1-p),$$z2j={y2j,with probability p,y2jS1j+S2j-S¯2S¯1with probability (1-p).$

On taking expectation on both sides of equation (2.4), we observe that

$E(z1j)=pE[y1j]+(1-p)E[y1jS1j+S2j-S¯2S¯1],=pY¯1+(1-p)[Y¯1+S¯2S¯1-S¯2S¯1],⇒Z¯1=Y¯1.$

Similarly, taking expectations on both the sides of equation (2.5) we get the population mean of sensitive variable at current move as

$Y¯2=Z¯2,$

such that

$ρz1z2=p2(ρy1y2σy1σy2)+p(1-p)[2ρy1y2σy1σy2+Y¯1Y¯2]+(1-p)2[I3]I4I5I3=(ρy1y2σy1σy2+Y¯1Y¯) (σs12+S¯12)+σs22+S¯22+σs22/nσs12/n+S¯12+S¯2Y¯2-Y¯1Y¯2,ρz1y1=pσy12+(1-p)σy12σy1I4,ρz2y2=pσy22+(1-p)σy22σy2I5,I4=p2σy12+(1-p)2[(σs12+S¯12) (σy12+Y¯12)σs12/n+S¯12-Y¯12+σs22+S¯22σs12/n+S¯12-σs22/n+S¯22σs12/n+S¯12 ]+2p(1-p)σy12,I5=p2σy22+(1-p)2[(σs12+S¯12) (σy22+Y¯22)σs12/n+S¯12-Y¯22+σs22+S¯22σs12/n+S¯12-σs22/n+S¯22σs12/n+S¯12 ]+2p(1-p)σy22.$
Remark 1

The mean and variance of sensitive variable at current move in two move successive sampling are obtained in terms of mean of coded response variable. Hence, efficient estimators need to be investigated to estimate coded response variable so that the estimate of sensitive variable get improved and became more effective. Hence, in next section we investigate the suitable estimators for coded response variable in presence of non-response at both the moves.

Remark 2

Since the study character is sensitive in nature. Even though the investigator try so hard, there will always be scope for some non-response. Hence, in order to deal with non-response, calibration technique applied over successive moves may be a good alternative. The calibration technique becomes more effective if auxiliary information is available and in successive sampling the information from previous move may also be used as an auxiliary information at current move together with the availability of additional auxiliary variable. Hence, in the next section a weighted calibration estimator for coded response variable have been proposed to adjust the effect due to presence of non-response.

2.3. Calibration estimator for coded response variable

Devil and Särndal (1992) invoked calibration technique in survey sampling, which is proved to be an efficient technique to adjust non-response by Lundström and Särndal (1999). Therefore, calibration technique has been used to adjust non-response with the aid of an additional auxiliary variable to estimate coded response variable which will be further used to estimate population mean of sensitive variable.

### 2.3.1. Calibration estimator based on fresh sample

Let the basic design weight $βi*$ be replaced by the new weight wr2. Hence, for the estimation of sensitive population mean at current move, the proposed calibration estimator $T^cuNR$ based on fresh sample of size u is given as

$T^cuNR=1N∑i∈sr2wr2iz2i,$

In order to obtain the calibrated weight wr2i, we minimize the chi-square type function

$∑i∈sr2(wr2i-βi*)2quiβi*,$

subject to calibration constraint

$1N∑i∈sr2wr2ixi=X¯,$

with qui being known positive constant unrelated to $βi*$, xi and . This results in to calibrated weights given by

$wr2i=βi*+βi*quiN[(X¯-1N∑i∈sr2βi*xi)xi∑i∈sr2quiβi*xi2].$

After substituting the calibrated weights wr2i in equation (2.7) we obtain the final calibrated estimator $T^cuNR$ as,

$T^cuNR=[1N∑i∈sr2βi*z2i+b^u(X¯-1N∑i∈sr2βi*xi)],$

with

$b^u=(∑i∈sr2βi*quixi2)-1(∑i∈sr2βi*quixiz2i),$

where qui being known positive constant unrelated to $βi*$ and .

### 2.3.2. Calibration estimator based on matched sample

In successive sampling, to ameliorate the performance of the estimators on the current move, it is quotidian practice to use the information collected on the first move as auxiliary information in addition to availability of additional non-sensitive auxiliary variable. The calibration estimator in presence of non-response is proposed based on sample of size m at current (second) move with the new calibrated weights as

$T^cmNR=1N∑i∈smwmiz2i.$

To find the calibrated weight wmi, we minimize the chi-square function

$∑i∈sm(wmi-αi*)2qmiαi*,$

subject to calibration constraints

$1N∑i∈smwmiz1i=z¯1n*NR,$$1N∑i∈smwmixi=X¯,$

with qmi as known positive constant unrelated to $αi*$ and . Following similar procedure, we obtain the calibrated estimator for $z¯1n*NR$ based on sample sr1, the response set obtained from sn drawn at first move as

$z¯1n*NR=[1N∑i∈sr1α1iz1i+(X¯-1N∑i∈sr1iα1ixi)b^n],$

with

$b^n=(∑i∈sr1α1iqnixi2)-1(∑i∈sr1α1iqnixiz1i),$

where qni is known positive constant unrelated to α1i and .

Now, minimizing the chi-square function in equation (2.13) subject to constraints in equation (2.14) and (2.15) lead to the calibrated weights given by

$wmi=αi*+[1Nxiαi*qmi{(X¯-1N∑i∈smαi*xi) (∑i∈smαi*qmiz1i2)-(z¯1n*NR-1N∑i∈smαi*z1i) (∑i∈smαi*qmiz1ixi)}+1Nz1iqmiαi*{(z¯1n*NR-1N∑i∈smαi*z1i) (∑i∈smαi*qmixi2)-(X¯-1N∑i∈smαi*xi) (∑i∈smαi*qmiz1ixi)}]b^m,$

where

$b^m=[(∑i∈smαi*qmi(z1i)2) (1N2∑i∈smαi*qmixi2)-1N2(∑i∈smαi*qmiz1ixi)2]-1.$

Substituting the calibrated weight wmi in equation (2.12), the final proposed calibrated estimator $T^cmNR$ based on sample size m at current move becomes

$T^cmNR=[1N∑i∈smαi*z2i+b^m1(z¯1n*NR-1N∑i∈smαi*z1i)+b^m2(X¯-1N∑i∈smαi*xi)],$

with

$b^m1=(1N2∑i∈smαi*qmiz1iz2i) [(∑i∈smαi*qmixi2)-(∑i∈smαi*qmiz1ixi)]b^m,b^m2=(1N2∑i∈smαi*qmixiz2i) [(∑i∈smαi*qmi(z1i)2)-(∑i∈smαi*qmiz1ixi)]b^m,$

qmi being known positive constant unrelated to $αi*$ and $z¯1n*NR$.

### 2.3.3. Combined calibrated estimator

The final calibrated estimator in presence of non-response at both moves is considered as convex linear combination of the two calibrated estimators $T^cuNR$ and $T^cmNR$ respectively and is given as,

$T^cNR=φT^cuNR+(1-φ)T^cmNR,$

where $T^cuNR$ and $T^cmNR$ are given in equation (2.11) and (2.18) respectively and ϕ ∈ [0, 1] is a scalar quantity to be chosen suitably.

3. Asymptotic variance

This section is dedicated to elaboration of asymptotic properties of proposed calibration estimator $T^cNR$ . Since, the estimator $T^cNR$ depends on the estimators $T^cuNR$ and $T^cmNR$ given in equation (2.11) and (2.18) respectively, so we first discuss the asymptotic properties of $T^cuNR$ and $T^cmNR$. In addition, the results suggested by Randles (1982), may be used to discuss the asymptotic variance of estimators.

### Proposition 1

The asymptotic behaviour of the calibrated estimator $T^cuNR$is same as that of

$T^cu∣bNR=1N∑i∈sr2βi*z2i+(X¯-1N∑i∈sr2βi*xi)b,$

with

$b=(∑i∈sr2quixi2)-1(∑i∈sr2quixiz2i).$
Proof

Assuming,

$T^cuNR(γ)=1N∑i∈sr2βi*z2i+(X¯-1N∑i∈sr2βi*xi)γ,$

for any variable γ, equation (3.3) shows $T^cuNR(γ)$ is the calibration estimator $T^cuNR$ when u in equation(2.11) is replaced by γ. Therefore, the limiting mean of $T^cuNR(γ)$ when the actual parameter value is b, given in equation (3.2), can be written as

$μ(γ)=limu→+∞Eb[T^cuNR(γ)]=Z˜2,$

where 2 is the limiting value of 2 as N → ∞. Using Randles (1982), the estimator $T^cuNR$ has the same asymptotic behaviour as that of the estimator in equation (3.1).

### Proposition 2

The variance of the estimator $T^cu∣bNR$in equation (3.1) is given by

$V(T^cu∣bNR)=[1N2∑i∈U∑j∈UΔijz2i(πi)cz2j(πj)c]+E1[1N2∑i∈(sn)c∑j∈(sn)cΔij∣(sn)c(π1i)c(π1j)ceiπi∣sncejπj∣snc],$

where ei = z2ixib, Δi j = πi jπiπj, Δi j|(sn)c = πi j|(sn)cπi|(sn)cπj|(sn)c and E1is the expectation under the design d1.

Proof

Since the estimator $T^cu∣bNR$ is unbiased, so its variance is given by

$V(T^cu∣bNR)=V1(E3[T^cu∣bNR])+E1[V3(T^cu∣bNR)],$

where E1 and V1 are the expectation and variance under the design d1 respectively, and E3 and V3 represent the conditional expectation and conditional variance under design d3 respectively

$V1(E3[T^cu∣bNR])=V1(1N∑i∈sr2βi*z2i)=1N2∑i∈U∑j∈UΔijz2i(πi)cz2j(πj)c.$

Now,

$E1[V3(T^cu∣bNR)]=E1[V3{(1N∑i∈sr2βi*z2i)+(X¯-1N∑i∈sr2βi*xi)b}]=E1[V3(1N∑i∈sr2βi*z2i-1N∑i∈sr2βi*xib)]=E1[V3(1N∑i∈sr2βi*ei)]=E1[1N2∑i∈(sn)c∑j∈(sn)cΔij∣(sn)c(π1i)c(π1j)ceiπi∣sncejπj∣(sn)c].$

Using equation (3.7) and (3.8) in equation (3.6), we get the expression for variance as in equation (3.5).

Remark 3

From Proposition 1 and Proposition 2, the estimator $T^cuNR$ and $T^cmNR$ are asymptotically unbiased and their asymptotic variances are given as

$V(T^cuNR)=1N2[∑i∈U∑j∈UΔijz2i(πi)cz2j(πj)c]+1N2E1[∑i∈(sn)c∑j∈(sn)cΔij∣(sn)c(π1i)c(π1j)ceiπi∣(sn)cejπj∣(sn)c].$

Similarly,

$V(T^cmNR)=1N2[∑i∈U∑j∈UΔij′z2i(πi)cz2j(πj)c]+1N2E2[∑i∈sr1∑j∈sr1Δij∣sr1π1iπ1jeiπi∣sr1ejπj∣sr1],$

where $Δij′=π1ij-π1iπ1j$ and E2 is the expectation under the design d2.

Proposition 3

The asymptotic variance of proposed calibration estimator is obtained as

$V(T^cNR)=φ2V(T^cuNR)+(1-φ)2V(T^cmNR),$

where $V(T^cuNR)$and $V(T^cmNR)$are given in equation (3.9) and (3.10) respectively.

### Proof

The asymptotic variance of calibration estimator $T^cNR$ is given by

$V(T^cNR)=E[T^cNR-Z¯2]2,=E[φT^cuNR+(1-φ)T^cmNR-Z¯2]2,=φ2V(T^cuNR)+(1-φ)2V(T^cmNR)+2φ(1-φ)cov(T^cuNR,T^cmNR).$

The values of $T^cuNR$ and $T^cmNR$ have been computed in equation (3.9) and (3.10) respectively and as the estimators $V(T^cuNR)$ and $V(T^cmNR)$ are based on two non-overlapping samples of sizes u and m respectively. So, $cov(T^cuNR,T^cmNR)=0$. By using these values in equation (3.12), we have the expression for the asymptotic variance of the calibrated estimator in presence of non-response as in equation (3.11).

Remark 4

From the equation (3.11), it can be concluded that, $V(T^cNR)$ is a function of unknown constant ϕ. Therefore, it is optimized with respect to ϕ and subsequently the optimum value of ϕ is obtained as

$φopt.=V(T^cmNR)V(T^cuNR)+V(T^cmNR).$

Substituting the value of ϕ(opt.) from equation (3.13) in (3.11), we get the optimum variance of the proposed estimator $T^cNR$

as

$V(T^cNR)opt.=V(T^cuNR)×V(T^cmNR)V(T^cuNR)+V(T^cmNR).$
4. Study under simple random sampling without replacement sampling design

In this section, calibration estimator in presence of non-response has been considered for simple random sampling without replacement (SRSWOR) design on both the moves. For that the relevant suppositions are given as

$π1i=r1N, π1ij=r1(r1-1)N(N-1).$

Because the sample sn is drawn from U with SRSWOR of size n, it implies that the complement, $snc=U-sn$ is a simple random sample without replacement of size Nn, therefore we have

$π1ic=N-nN, π1ijc=(N-n)(N-n-1)N(N-1).$

Also, we suppose that the matched sample sm is drawn from sr1, with SRSWOR of size m so

$πi∣sr1=mr1, πij∣sr1=m(m-1)r1(r1-1).$

Finally, the unmatched sample su is drawn from $snc$ with SRSWOR of size u. Thus, we have

$πi∣snc=r2N-r2, πij∣snc=r2(r2-1)(N-n)(N-n-1)$

Now, based on sample of size r2 on current move, the proposed calibration estimator $T^cuNR$ under SRSWOR sampling design becomes

$T^cuNR(s)=[z¯2u+b^u(s)(X¯-x¯u)],$

with

$b^u(s)=(∑i∈sr2quixi2)-1(∑i∈sr2quixiz2i).$

Similarly, based on sample of size m on current move, the proposed calibration estimator $T^cmNR$ under SRSWOR scheme is obtained as

$T^cmNR(s)=[z¯2m+b^m1(s)(z¯1n*NR(s)-z¯1m)+b^m2(s)(X¯-x¯m)].$

with

$b^m1(s)=(∑i∈smqmiz1iz2i) [(∑i∈smqmixi2)-(∑i∈smqmiz1ixi)]b^m(s),b^m2(s)=(∑i∈smqmixiz2i) [(∑i∈smqmi(z1i)2)-(∑i∈smqmiz1ixi)]b^m(s),b^m(s)=[(∑i∈smqmi(z1i)2) (∑i∈smqmixi2)-(∑i∈smqmiz1ixi)2]-1,z¯1n*NR(s)=[z¯1n+b^n(s)(X¯-x¯n)],b^n(s)=(∑i∈sr1qnixi2)-1(∑i∈sr1qnixiz1i).$

Now, the estimator $T^cNR(s)$ becomes

$T^cNR(s)=φsT^cuNR(s)+(1-φs)T^cmNR(s),$

where $T^cuNR(s)$ and $T^cmNR(s)$ are given in equation (4.1) and (4.2) respectively and ϕsε[0, 1] is a scalar quantity to be chosen suitably.

### Remark 5

Further if we assume, qui = qni = qmi = 1 in $T^cNR$, then the calibration estimator $T^cNR$ be denoted as $T^c*NR$.

### Remark 6

The proposed calibration estimator have been compared with general successive sampling estimator in presence of non-response at both the moves, so the general successive sampling estimator have been modified for estimation of coded response variable and is given as

$T^gNR=ΨgT^guNR+(1-Ψg)T^gmNR,where Ψg∈[0,1],$

with $T^guNR=z¯2r2,T^gmNR=z¯2m+βz2z1(z¯1r1-z¯1m)βz2z1=sz2z1sz12$.

5. Possible cases

There might be a possibility that non-response may occur only at current move or only at previous move or there may be no non-response at any move. Therefore, in order to retain similar estimators in all possible situations, the calibration technique have been retained and possible modifications has been done in the constraints as per the situation and calibration estimators in different possible situations have been obtained, which are described in following cases.

### 5.1. Case-I: when there is non-response only at current move

In this situation the proposed estimator $T^cNR$ of the coded response variable 2 changes to

$T^c1NR=Ψ1T^cuNR+(1-Ψ1)T^cm;Ψ1∈[0,1],$

where the estimator $T^cm$ can be obtained by replacing r1 by n in equation (2.16) and other equations that depends on it and $T^cuNR$ is defined in equation (2.11).

### 5.2. Case-II: when there is non-response only at first (previous) move

In the presence of non-response only at first (previous) move, the estimator $T^cNR$ of the coded response variable 2 changes to

$T^c2NR=Ψ2T^cu+(1-Ψ2)T^cmNR;Ψ2∈[0,1],$

where the estimator cu can be obtained by replacing r2 by u in equation (2.11) and other equations that depends on it and $T^cmNR$ is defined in equation (2.18).

### 5.3. Case-III: when there is no non-response at any move

In the presence of no non-response at any move, the estimator $T^cNR$ of the coded response variable 2 changes to

$T^c3=Ψ3T^cu+(1-Ψ3)T^cm;Ψ3∈[0,1],$

where the estimator cu can be obtained by replacing r2 by u in equation (2.11) and cm can be obtained by replacing r1 by n in equation (2.16) and other equations that depends on it and $T^cmNR$ is defined in equation (2.18).

6. Estimators for sensitive population mean under PORT

Replacing the population mean of coded response variable 2 in equation (2.3) and (2.6) by its estimators $T^c*NR,T^g*NR,T^c1*NR,T^c2*NR$ and c3, the respective estimators for sensitive population mean at current move becomes $Y¯^2c,Y¯^2g,Y¯^2c1,Y¯^2c2$ and $Y¯^2c3$ under PORT-I model and $Y¯^2c*,Y¯^2g*,Y¯^2c1*,Y¯^2c2*$ and $Y¯^2c3*$ under PORT-II model respectively, which are presented in

### Remark 7

For the considered model in equation (2.3) and (2.6), the two scrambling variables 1 and 2 used to perturb the true response through the PORT models may follow any distribution in two move successive sampling. Hence, following Pollock and Bek (1976) and Eichhorn and Hayre (1983), we consider scrambling variable S1 to follow normal distribution with mean 0 and variance 1. However, the scrambling variable S2 has been assumed to follow normal distribution with mean 1 and variance 1.

7. Simulation study

In this section, a simulation study has been carried out to reveal the behaviour of the proposed estimators. For this purpose, a natural population has been considered from statistical abstracts of United States. The considered population comprise of N = 51 states, which is described as: y1 : Rate of abortion in 2007 y2 : Rate of abortion in 2008 x : Rate of abortion in 2005. As discussed in Remark 6, the scrambling variables S1 ~ N(0, 1) and S2 ~ N(1, 1). The data for S1 and S2 have been generated by MATLAB software.

To judge the performance of both the PORT models under the proposed calibration estimators in presence of non-response to estimate the sensitive population mean in two move successive sampling, we have studied the behaviour of the estimators by considering different choices for rate of non-response at both moves. For simulation, 10, 000 independent replications of considered sampling design in two move successive sampling via MATLAB have been considered. All the samples are obtained under simple random sampling without replacement. An environment through simulation process has been created for non-response by assuming non-response rates as 10%, 20%, and 30% at both the moves.

The entire simulation has been replicated for different values of n, m and u which are considered as different sets given as

$Set-I:n=20; m=12;u=8, Set-II:n=20; m=10;u=10.$

Calibration estimators have been compared with general successive sampling estimator under both the considered PORT models in terms of percent relative efficiency (PRE), which are given as

$PRE1=MSE(Y¯^2g)MSE(Y¯^2c)×100,PRE2=MSE(Y¯^2g*)MSE(Y¯^2c*)×100,$

where $MSE(Y¯^2g)=1/10,000∑i=110,000[Y¯^2gi-Y¯2]2$ . Similarly, $MSE(Y¯^2g*ht),MSE(Y¯^2c)$ and $MSE(Y¯^2c*)$ can be computed. The simulation results have been represented in Tables 45 respectively. Further, in order to identify the better PORT model, the percent relative efficiency (PRE) of calibration estimator under PORT-I with respect to calibration estimator under PORT-II have been computed as

$PRE3=MSE(Y¯^2c)MSE(Y¯^2c*)×100,$

where $MSE(Y¯^2c)=1/10,000∑i=110,000[Y¯^2ci-Y¯2]2$ and similarly, $MSE(Y¯^2c*)$ can be computed. The results are presented in

8. Discussion of results and conclusion

Following interpretation can be drawn from the simulation results presented in Tables 45 and also in

• (i) From Tables 45, it is observed that the proposed calibration estimator is performing better than general successive sampling estimator in the presence of non-response at both the moves under PORT-I as well as PORT-II models. This shows that the use of calibration technique to adjust the effect due to non-response is fruitful.

• (ii) Figures 12 show that the calibration estimator under proposed PORT-II model is better than the same calibration estimator under PORT-I model for all considered choices of constants and non-response rates.

• (iii) It is also observed from Figures 12 that for the fixed value of p, the PRE3 first increases up to Ψ = 0.5, then decreases for Ψ > 0.5. This shows that if more weight is attached to matched sample than PRE is higher as compared to the situation when more weight is attached to unmatched/fresh sample at current move. These results are in accordance with the theory of successive sampling.

• (iv) Figures 12 also shows that higher percent relative efficiency is observed for larger value of p i.e., PRE3 in general increased as p increases.

### 8.1. Conclusion

The estimation of sensitive population mean at current move in two move successive sampling is feasible using PORT. The calibration technique applied to adjust the effect due to non-response is proved to be fruitful under both the considered models. The proposed model PORT-II is coming out to be more efficient than the modified Sanaullah et al. (2020) model (PORT-I) on successive moves. Therefore, it is concluded that the proposed PORT model may be used for the estimation of sensitive population mean at current move in two move successive sampling.

Figures
Fig. 1. PRE of the calibration estimator under PORT-I with respect to the calibration estimator under PORT-II for different choices of non-response rates and varying for a = −15.
Fig. 2. PRE of the calibration estimator under PORT-I with respect to the calibration estimator under PORT-II for different choices of non-response rates and varying for a = −19.
TABLES

### Table 1

Inclusion probabilities

SizeFirst order inclusion probabilitySecond order inclusion probability
r1π1i = ∑isr1Pd1 (sr1)π1i j = ∑i, jsr1Pd1 (sr1)
mπi|sr1 = ∑ismPd2 (sm)πi j|sr1 = ∑i, jsmPd2 (sm)
r2πi = ∑isr2Pd3 (sr2 )πi j = Σi, jsr2Pd3 (sr2)

### Table 2

Sample design basic weights

Response SetSizeSampling design basic weight for selecting ith unit in corresponding sample
sr1r1$α1i=1π1i$
smm$αi*=1π1iπi∣sr1$
sr2r2$βi*=1(π1i)cπi∣snc$

### Table 3

Estimators of sensitive population mean and their variances

EstimatorsVariance
PORT-I$Y¯^2c=T^c*NR-(1-p)aS¯2p+(1-p)S¯1$$V[Y¯^2c]=V(T^c*NR)opt.[p+(1-p)S¯1]2$
$Y¯^2g=T^g*NR-(1-p)aS¯2p+(1-p)S¯1$$V[Y¯^2g]=V(T^g*NR)opt.[p+(1-p)S¯1]2$
$Y¯^2c1=T^c1*NR-(1-p)aS¯2p+(1-p)S¯1$$V[Y¯^2c1]=V(T^c1*NR)opt.[p+(1-p)S¯1]2$
$Y¯^2c2=T^c2*NR-(1-p)aS¯2p+(1-p)S¯1$$V[Y¯^2c2]=V(T^c2*NR)opt.[p+(1-p)S¯1]2$
$Y¯^2c3=T^c3-(1-p)aS¯2p+(1-p)S¯1$$V[Y¯^2c3]=V(T^c3*NR)opt.[p+(1-p)S¯1]2$

PORT-II$Y¯^2c*=T^c*NR$$V[Y¯^2c*]=V(T^c*NR)opt.$
$Y¯^2g*=T^g*NR$$V[Y¯^2g*]=V(T^gNR)opt.$
$Y¯^2c1*=T^c1NR$$V[Y¯^2c1*]=V(T^c1NR)opt.$
$Y¯^2c2*=T^c2NR$$V[Y¯^2c2*]=V(T^c2NR)opt.$
$Y¯^2c3*=T^c3$$V[Y¯^2c3*]=V(T^c3)opt.$

### Table 4

Percent relative efficiency of $Y¯^2g$ with respect to $Y¯^2c$ for different sets and different choices of non-response (NR) rates under PORT-I where Ψ ∈ {φg}

ΨpSET-ISET-II

NR = 10%NR = 20%NR = 30%NR = 10%NR = 20%NR = 30%

PRE1PRE1PRE1PRE1PRE1PRE1

a = −10a = −19a = −10a = −19a = −10a = −19a = −10a = −19a = −10a = −19a = −10a = −19
0.3110.7137.9178.6137.2175.8140.4100.5138.1109.1136.7136.1135.6
0.5152.7173.3152.4173.0251.9172.12156.3174.2155.0173.4151.6172.6
0.10.7362.6196.9349.1195.1336.4198.6342.6191.0315.7192.9315.1189.1
0.9319.1387.3372.9419.4406.2436.2158.0234.2201.7227.8181.4255.1

0.3120.4118.4103.5107.8120.0146.4115.6118.5115.0117.5132.3106.4
0.5140.9148.9138.8148.1137.1146.5141.1178.6141.7138.2140.1137.2
0.30.7302.5173.8292.7173.1286.7174.1295.9171.6289.8172.0281.7172.1
0.9309.9355.2332.3413.2428.3435.9185.2246.1207.1224.7239.0277.9

0.3110.4106.5116.5105.2113.8100.0100.4100.1108.3109.6116.5118.6
0.5128.6163.6126.4122.4124.7181.3129.6123.6128.6142.8127.4134.5
0.50.7244.7150.3234.5149.6231.0152.0243.0150.2237.2151.9234.9152.2
0.9405.7378.5366.0410.2454.6448.2219.4298.1231.9277.7249.5232.5

0.3112.3113.9109.1171.8108.570.4103.9183.8102.5112.7100.6101.7
0.5115.8119.0114.7186.9113.485.5117.6118.6118.0108.3116.0117.1
0.70.7184.9129.1179.3228.9174.8127.3187.0130.2185.0129.4182.3230.2
0.9424.1325.7413.4339.3423.6353.0261.6285.2293.1287.6303.4287.4

0.3100.097.294.0184.9102.0183.2100.1100.1102.0186.9100.0135.7
0.5105.3102.7104.3191.6102.5189.8107.3103.7105.9193.1105.0192.2
0.90.7134.7108.4132.0208.4131.0207.3233.4109.4135.1209.1133.7209.3
0.9355.8220.9349.9223.7341.7231.6334.2204.1327.2211.8328.7217.0

### Table 5

Percent relative efficiency of $Y¯^2g*$ with respect to $Y¯^2c*$ for different sets and different choices of non-response (NR) rates under PORT-II where Ψ ∈ {φg}

ΨpSET-ISET-II

NR = 10%NR = 20%NR = 30%NR = 10%NR = 20%NR = 30%

PRE2PRE2PRE2PRE2PRE2PRE2
0.3142.0144.7160.1130.6131.2140.6
0.5177.0208.1235.0239.0244.4257.5
0.10.7277.7339.0456.8900.6154.3396.4
0.93809.64150.15548.51578.51523.02549.4

0.3152.6161.8183.6126.1126.6129.9
0.5214.2209.0252.0254.2356.8462.4
0.30.7389.2363.9453.9460.0488.0568.5
0.93420.24141.84400.81504.11849.22485.5

0.3185.2197.5117.7136.7249.1163.4
0.5253.0259.3259.7148.7396.0188.4
0.50.7304.9326.2338.3225.1582.8228.0
0.92408.22407.62181.41582.81622.91960.6

0.3135.9137.9132.8107.9113.2106.6
0.5173.1161.4159.0137.5134.5164.0
0.70.7247.0229.0207.2213.4224.5212.8
0.91015.5931.2784.21204.81035.51012.1

0.3148.5134.4122.9167.8162.8150.0
0.57155.3142.5131.0193.9180.3155.7
0.90.7173.7165.0144.8224.6193.5176.4
0.9388.4350.2300.6496.8428.5387.6

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