Problem Consider the mixed binomial distribution $Bin(n, U)$, where U has a standard uniform(0,1) distribution.
(i) Compute E[X | U]
(ii) Compute Var [X | U]
(iii) Find the mean and variance of X
Following is my thought process, I'm not sure if they are even remotely close to be on the right track.
(i) E[X|U] = E[$\frac{f(X,U)}{f_U(u)}$] = $\frac{\sum_{n = 0}^{\infty} x\space{n \choose x} \; p^x (1-p)^{n-x} U(0,1)}{E[U(0,1)]}$ = $\frac{np}{\frac{1}{2}}$ , since p is uniform distribution probability will converge to 1 and np = n. Therefore E[X|U] = 2n?
I assume that p= probability of standard Uniform distribution, which is 1?
(ii) Var [X | U]=E[$X^2 | U] – (E[X | U])^2$ = $np(1-p) – (2n)^2$
(iii) $$E[X] = E(E[X|U]) = E[np] = np \quad (since \space X|U \sim Binomial(n,U)) $$
$$Var[X]=Var(E[X|U]) + E(Var[X|U]) $$
$$Since \space Var(E[X|U])=Var(nP) = n^2U(0,1)= n^2(1)$$
$$E(Var[X|U]) = nE[p(1-p)] = 0?$$
$$Therefore \space Var[X]=n^2?$$
Best Answer
No, they are not remotely close. There is no $p$ involved anywhere.
You have been told that $X\mid U\sim\mathcal{Bin}(n, U)$ and $U\sim\mathcal{U}(0;1)$.
(i) $\mathsf E(X\mid U)$ will be a function of $U$, and further ...
(ii) Similarly for $\mathsf{Var}(X\mid U)$
(iii) This relies on the tower properties.$$\mathsf E(X)=\mathsf E(\mathsf E(X\mid U))\\ \mathsf{Var}(X) = \mathsf E(\mathsf {Var}(X\mid U))+\mathsf {Var}(\mathsf E(X\mid U))$$
Well, you should be able to complete that now.