Q.6.18 - A First Course in Probability

Question

Let X1, ... , Xn and Y1, ... , Yn be independent random vectors, with each vector being a random ordering of k ones and n − k zeros. That is, their joint probability mass functions are $\begin{array}{r}P\left\{X_1=i_1,\dots ,X_n=i_n\right\}=P\left\{Y_1=i_1,\dots ,Y_n=i_n\right\}\\ =\frac{1}{\left(\begin{array}{c}n\\ k\end{array}\right)},i_j=0,1,\sum _{j=1}^n i_j=k\end{array}$ 

denote the number of coordinates at which the two vectors have different values. Also, let M denote the number of values of i for which

 Xi = 1, Yi = 0. 

(a) Relate$N to M.$

(b) What is the distribution of $M?$

(c) Find E[N]. 

(d) Find Var(N). 

Step-by-Step Solution

Answer

a)$\text{ The relation between }N\text{ to }M\text{ is }N=2M\text{. }$

b) $\text{ The distribution of }\mathrm{M}\text{ is hyper geometric random variable. }$

c) $E\left[N\right]=\frac{2k(n-k)}{n}$

d) The var(N) =$4\frac{(n-k)}{(n-1)}k\left(1-\frac{k}{n}\right)\left(\frac{k}{n}\right)$

Part (a) - Step 1: To find

The relation between N to M

Part (a) - Step 2: Explanation

Given: Given that $$\begin{gathered} X_1.....X_n \\ {Y}_1.....Y_n \end{gathered}$$ be independent random vectors, with each vector being a random ordering of k ones and n-k  zeros. That is their joint probability mass functions are$\begin{array}{r}P\left\{X_1=i_1,\dots ,X_n=i_n\right\}=P\left\{Y_1=i_1,\dots ,Y_n=i_n\right\}\\ =\frac{1}{\left(\begin{array}{c}n\\ k\end{array}\right)},i_j=0,1,\sum _{j=1}^n i_j=k\end{array}$

$\mathrm{N}=\sum _{\mathrm{i}=1}^{\mathrm{n}}\left|{\mathrm{X}}_{\mathrm{i}}-{\mathrm{Y}}_{\mathrm{i}}\right|$denote the number of coordinates at which the two vectors have different values.

Also, let M denotes the number of values of i for which$X_i=1,Y_i=0$

Calculation : From the given information it can written as

$\sum _{i=1}^n{X}_i=\sum _{i=1}^n{Y}_i$

So, M denotes the number of value of i.

it follows $N = 2M$

Hence , the relation between N to M = $2M.$

Part (b) - Step 3: To find

The distribution of M.

Part (b) - Step 4: Explanation

Consider the n-k co-ordinates chose Y - values are equal to 0 and call them the red coordinates. Because the k co-ordinates whose X - values are equal to 1 are equally likely to be any of the $\left(\begin{array}{l}n\\ k\end{array}\right)$ sets of k coordinates, It follows the number of Red coordinates among these k coordinates has the same distribution as the number of Red balls chosen when one Randomly chooses k  of a set n balls of which $\mathrm{n}-\mathrm{k}$ are Red. Therefore, M is a Hyper geometric Random variable.

Part (c) - Step 5: To find

The value of$E\left[N\right]$

Part (c) - Step 6: Explanation

Given : 

Given that $$\begin{gathered} X_1....X_n \\ {Y}_1.....Y_n \end{gathered}$$ be independent random vectors, with each vector being a random ordering of k ones and n-k  zeros. That is their joint probability mass functions are $\begin{array}{r}P\left\{X_1=i_1,\dots ,X_n=i_n\right\}=P\left\{Y_1=i_1,\dots ,Y_n=i_n\right\}\\ =\frac{1}{\left(\begin{array}{c}n\\ k\end{array}\right)},i_j=0,1,\sum _{j=1}^n i_j=k\end{array}$

$\mathrm{N}=\sum _{\mathrm{i}=1}^{\mathrm{n}}\left|{\mathrm{X}}_{\mathrm{i}}-{\mathrm{Y}}_{\mathrm{i}}\right|$denote the number of coordinates at which the two vectors have different values.

Also, let M denotes the number of values of i for which$X_i=1,Y_i=0$

Calculation :

$$\begin{gathered} E\left[N\right]=E[2 M]=2 E\left[M\right] \\ \Rightarrow E\left[N\right]=\frac{2k(n-k)}{n} \end{gathered}$$

Therefore $\frac{2k(n-k)}{n} is the E\left[N\right]$

Part (d) - Step 7 : To find

The value of $var\left(N\right)$

Part (d) - Step 8: Calculation

Calculation : $\begin{array}{r}\mathrm{Var}\left[N\right]=\mathrm{Var}\left[2M\right]\\ =2^2\mathrm{Var}\left[M\right]\\ \Rightarrow \mathrm{Var}\left[N\right]=4\frac{(n-k)}{(n-1)}k\left(1-\frac{k}{n}\right)\left(\frac{k}{n}\right)\end{array}$

Hence the value of $Var\left[N\right]= 4\frac{(n-k)}{(n-1)}k\left(1-\frac{k}{n}\right)\left(\frac{k}{n}\right)$