Q.6.17 - A First Course in Probability

Question

7. Find the probability that X1, X2, ... , Xn is a permutation of 1, 2, ... , n, when X1, X2, ... , Xn are independent and

 (a) each is equally likely to be any of the values 1, ... , n;

 (b) each has the probability mass function P{Xi = j} = pj, j = 1, ... , n 

Step-by-Step Solution

Answer

a) The probability is $p=\frac{n!}{n^n}$

b) The probability mass function $P\left(X_i=j\right)=n!p_1{p}_2\dots \dots .p_{n-1}{p}_n$

Part (a) - Step 1: To find

$The probability of X_1....X_2$

Part (a) - Step 2: Explanation

Given: $X_1....Xn$

Formula to used: ${\mathrm{F}}_{\mathrm{X}}\left(\mathrm{X}\right)=\mathrm{P}(\mathrm{X}\le \mathrm{x})$

Calculation : 

X be a continuous random variable

$\mathrm{X}~Uni(1,2,3,\dots n)$

The total combination ${\mathrm{n}}^{\mathrm{n}}$

The random vector

${\mathrm{X}}_1,{\mathrm{X}}_2,\dots \dots \dots {\mathrm{X}}_{\mathrm{n}}$

Every random variable assumes one and only one variable The probability is

$p=\frac{n!}{n^n}$

Part (b) - Step 3: To find

The probability of mass function of $X_1,X_2....X_n$

Part (b) - Step 4: Explanation

Given: X be a continuous random variable

$\mathrm{X}~Uni(1,2,3,\dots n)$

The total combination ${\mathrm{n}}^{\mathrm{n}}$

The random vector ${\mathrm{X}}_1,{\mathrm{X}}_2,\dots \dots \dots {\mathrm{X}}_{\mathrm{n}}$

Every random variable assumes one and only one variable The probability is 

$\begin{array}{r}P\left(\underset{\sigma \{1,2,3,\dots n\}}{\cup }\underset{i=1}{\overset{n}{\cap }} X_i=\sigma \left(i\right)\right)=\sum _{\sigma \{1,2,3,\dots n\}} \prod _{i=1}^n P\left(X_i=\sigma \left(i\right)\right)\\ \prod _{i=1}^n P\left(X_i=\sigma \left(i\right)\right)=\prod _{i=1}^n p_{\sigma \left(i\right)}=p_1{p}_2\dots \dots .p_{n-1}{p}_n\\ \sum _{\sigma \{1,2,3,\dots n\}} \prod _{i=1}^n P\left(X_i=\sigma \left(i\right)\right)=\sum _{\sigma \{1,2,3,\dots n\}} p_1{p}_2\dots \dots .p_{n-1}{p}_n\\ \sum _{\sigma \{1,2,3,\dots n\}} \prod _{i=1}^n P\left(X_i=j\right)=n!p_1{p}_2\dots \dots .p_{n-1}{p}_n\\ P\left(X_i=j\right)=n!p_1{p}_2\dots \dots .p_{n-1}{p}_n\end{array}$

Therefore the probability of mass function of $P\left(X_i=j\right)=n! p_1{p}_2\dots \dots .p_{n-1}{p}_n$