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.
a:P(Y=n)=∫01P(Y=n∣p)dp=∫01CMnpn(1−p)M−ndp==CMnB(n+1,M+1−n)=CMnΓ(n+1)Γ(M+1−n)Γ(M+2)==CMnn!(M−n)!(M+1)!=1M+1Y∼Unif{0,1,...,M}b:EY=∑i=0Mi⋅1M+1=M(M+1)2(M+1)=M2EY2=∑i=0Mi2⋅1M+1=M(M+1)(2M+1)6(M+1)=M(2M+1)6Var(Y)=M(2M+1)6−(M2)2=M(M+2)12a:\\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}\\a:P(Y=n)=∫01P(Y=n∣p)dp=∫01CMnpn(1−p)M−ndp==CMnB(n+1,M+1−n)=CMnΓ(M+2)Γ(n+1)Γ(M+1−n)==CMn(M+1)!n!(M−n)!=M+11Y∼Unif{0,1,...,M}b:EY=∑i=0Mi⋅M+11=2(M+1)M(M+1)=2MEY2=∑i=0Mi2⋅M+11=6(M+1)M(M+1)(2M+1)=6M(2M+1)Var(Y)=6M(2M+1)−(2M)2=12M(M+2)
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