Let Xi, X2,X3,... be i. i.d. Bernoulli trials with success probability p and Sk=X1++Xk. Let m< n.

Find the conditional probability mass function PSm]Sn(l\k) of Sm, given Sn=k.

(a) Identify the distribution by name. Can you give an intuitive explanation for the answer?

(b) Use the conditional probability mass function to find E[Sm|Sn].

Detailed step wise solution is given below:

Step-by-step explanation:

If X_i,i=1,2,3,... are Bernoulli random variables, then its PMF is

P\left (X_i =1 \right )=p, P\left (X_i =0 \right )=1-p,i=1,2,3,...

Define S_k=X_1+X_2+...+X_k . When S_n=k,0\leqslant k\leqslant n. Then k out of n random variables equals to 1. There are \binom{n}{k} possible combinations of k 1's and n-k 0's. So we have

P\left ( S_n=k \right )=\binom{n}{k}p^k\left ( 1-p \right )^{n-k},k=0,1,2,...,n . That is S_n has Binomial distribution.

a)The joint probability mass function of random vector \left ( X_1,X_2,...,X_m \right ) given S_n=X_1+X_2+...+X_n=k    defined as \left (n\geqslant m \right )

P\left ( X_1=a_1,X_2=a_2,...,X_m=a_m|S_n=k \right ) can be calculated as below.

P\left ( S_m=l,S_n=k \right )=\binom{m}{l}p^l\left ( 1-p \right )^{m-l}\binom{n-m}{k-l}p^{k-l}\left ( 1-p \right )^{n-m-k+l}\\ P\left ( S_m=l,S_n=k \right )=\binom{m}{l}\binom{n-m}{k-l}p^k\left ( 1-p \right )^{n-k};l=0,1,2,..,m;k=l,..,n

The conditional distribution,

P\left ( S_m=l|S_n=k \right )=\frac{P\left ( S_m=l,S_n=k \right )}{P\left ( S_n=k \right )}\\ P\left ( S_m=l|S_n=k \right )=\frac{\binom{m}{l}\binom{n-m}{k-l}p^k\left ( 1-p \right )^{n-k}}{\binom{n}{k}p^k\left ( 1-p \right )^{n-k}}\\ {\color{Blue} P\left ( S_m=l|S_n=k \right )=\frac{\binom{m}{l}\binom{n-m}{k-l}}{\binom{n}{k}};l=0,1,2,..,m;k=l,..,n}

This distribution is Hyper geometric distribution. We have to get l successes in first m trials and k-l successes in the next n-m trials. The total ways of happening this is \binom{n}{k} . Hence Hyper geometric.

b) The conditional expectation is

E\left ( S_m=l|S_n=k \right )=\sum_{l=0}^{m}lP\left ( S_m=l|S_n=k \right )\\ E\left ( S_m=l|S_n=k \right )=\sum_{l=0}^{m}l\times \frac{\binom{m}{l}\binom{n-m}{k-l}}{\binom{n}{k}}\\

Use the formula for expectation of hyper geometric distribution, {\color{Blue} E\left ( S_m=l|S_n=k \right )=\frac{k m}{n}}

one that is bigger than 3 1/2