Conditional expectation with binomial random variables

Question

Suppose $X \sim Bin(n,p)$ and $Y \sim Bin(X,d)$, what is $\mathbb{E}[X|Y]$? I tried computing $P(Y=y)$ without succes.

Answer 1

I tried computing P(Y=y) without succes.

It should be noted that you are actually presented with the conditional distribution of $Y$ given $X$.

$\begin{align}X&\sim\mathcal{Bin}(n,p)\\ Y\mid X&\sim\mathcal{Bin}(X,d)\\[2ex]\mathsf P(Y=y) &=\sum_{x=y}^n\mathsf P(X=x, Y=y)\\&=\sum_{x=y}^n \mathsf P(Y=y\mid X=x)~\mathsf P(X=x)\\&=\mathbf 1_{y\in [[0:n]]}~\sum_{x=y}^n \dbinom xy d^y (1-d)^{x-y}\dbinom nx p^x(1-p)^{n-x}\\&= \binom ny(pd)^y\mathbf 1_{y\in [[0:n]]}~\sum_{x=y}^n\dbinom{n-y}{x-y}\bigl(p(1-d)\bigr)^{x-y}(1-p)^{(n-y)-(x-y)}\\&= \binom ny(pd)^y\mathbf 1_{y\in [[0:n]]}~\sum_{z=0}^{n-y}\dbinom{n-y}{z}\bigl(p (1-d)\bigr)^z(1-p)^{(n-y)-z}\\&=\dbinom ny (pd)^y\mathbf 1_{y\in [[0:n]]}~(p-pd+1-p)^{n-y}\\&=\dbinom ny (pd)^y(1-dp)^{n-y}\mathbf 1_{y\in[[0:n]]}\end{align}$

Which is that $Y\sim\mathcal{Bin}(n, pd)$

That is the easy bit...

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Let $(Y,U,V) \sim\mathcal{Multinom}(n,\{pd, (p-pd), (1-p)\})$ (so that $X=U+Y$).

Then show that $U\mid Y\sim\mathcal{Bin}\left(n-Y, \dfrac{p-pd}{1-pd}\right)$ in the same way that $Y\mid V\sim\mathcal{Bin}\left(n-V, \dfrac{pd}{1-(1-p)}\right)$ that is $Y\mid X \sim \mathcal{Bin}(X, d)$

And so find $\mathsf E(X\mid Y)$