Answer to Question #322948 in Statistics and Probability for Dinor123

Question

Consider Y, the number of successes in Mindependent Bernoulli trials each with success

probability X. Suppose that X itself is a r.v which is uniformly distributed over (1, 0)

(a) Find the p.m.f of Y and identify the distribution

(b) What is the mean and variance of Y.

Answer 1

$a:\\P\left( Y=n \right) =\int_0^1{P\left( Y=n|p \right) dp}=\int_0^1{C_{M}^{n}p^n\left( 1-p \right) ^{M-n}dp}=\\=C_{M}^{n}B\left( n+1,M+1-n \right) =C_{M}^{n}\frac{\varGamma \left( n+1 \right) \varGamma \left( M+1-n \right)}{\varGamma \left( M+2 \right)}=\\=C_{M}^{n}\frac{n!\left( M-n \right) !}{\left( M+1 \right) !}=\frac{1}{M+1}\\Y\sim Unif\left\{ 0,1,...,M \right\} \\b:\\EY=\sum_{i=0}^M{i\cdot \frac{1}{M+1}}=\frac{M\left( M+1 \right)}{2\left( M+1 \right)}=\frac{M}{2}\\EY^2=\sum_{i=0}^M{i^2\cdot \frac{1}{M+1}}=\frac{M\left( M+1 \right) \left( 2M+1 \right)}{6\left( M+1 \right)}=\frac{M\left( 2M+1 \right)}{6}\\Var\left( Y \right) =\frac{M\left( 2M+1 \right)}{6}-\left( \frac{M}{2} \right) ^2=\frac{M\left( M+2 \right)}{12}\\$