Quick question about Binomial variance I am having trouble figuring a problem if we have T ~ uniform on {1,2,3,...,10} and X ~ Bin(T,0.5) and we have to find the variance of X, would that be mean*.5*.5(I found the mean to be 2.75)?
 A: Hint: To do the calculation, it is useful to know $E(X^2)$. We have
 $$E(X^2\mid T=k)=\frac{1}{4}(k+k^2).$$
It follows that
$$E(X^2)=\sum_{k=1}^{10} \frac{1}{10}\cdot\frac{1}{4}\cdot(k+k^2).$$
A: For every random variable $T$ such that $T\geqslant1$ almost surely, a random variable $X$ with conditional distribution the binomial distribution $(T,\frac12)$ conditionally on $T$, has mean $E(X)=\frac12E(T)$ and variance $\sigma^2(X)=\frac14\sigma^2(T)+\frac14E(T)$.
A direct approach to prove these identities is to consider an infinite sequence $(B_n)$ i.i.d. Bernoulli on $\{0,1\}$ independent of $T$ and to choose $X=\sum\limits_{n\geqslant1}B_n\mathbf 1_{T\geqslant n}$. Then useful intermediary results are that $E(B_n)=E(B_n^2)=\frac12$, $\sum\limits_{n\geqslant1}\mathbf 1_{T\geqslant n}=T$ and $\sum\limits_{n\geqslant1}(n-1)\mathbf 1_{T\geqslant n}=\frac12T(T-1)$. Now everything is ready to compute the expectations of $X$ and $X^2=\sum\limits_{n\geqslant1}B_n\mathbf 1_{T\geqslant n}+2\sum\limits_{1\leqslant n\lt m}B_nB_m\mathbf 1_{T\geqslant m}$.
A: Var(X) = E[Var(X|T)] + Var[E(X|T)]
Var(X|T) = $T * {1 \over 2} * {1 \over 2}$
E(${1 \over 4}T$) = $1 \over 4$E(T) = $11 \over 8$
E(X|T) = E(T) = $11 \over 2$
Var[E(X|T)] = Var[$11 \over 2$] = 0
