Given $X$ Poisson, and $f_{Y}(y\mid X = x)$, find $\mathbb{E}[X\mid Y]$

Question

I am given a problem:

Suppose $X \sim \text{Poisson}(\lambda)$, and $f_{Y}(y\mid X = x) = \binom{x}{y}p^y(1-p)^{x- y}$ the conditional PDF of $Y$ given $X = x$, where $p$ is some probability. Find $\mathbb{E}[X\mid Y]$.

My approach is:

\begin{align*} \mathbb{E}[X\mid Y = y] &= \sum\limits_{x = 0}^\infty x f_{X}(x\mid Y = y)\\ &= \sum\limits_{x = 0}^\infty x f_{X,Y}(x, y)/ f_Y(y)\\ & = \sum\limits_{x = 0}^\infty x f_{Y}(y\mid X = x) f_X(x)/f_Y(y) \\ & = \sum\limits_{x = 0}^\infty \dfrac{x f_{Y}(y\mid X = x) f_X(x)}{\sum\limits_{x = 0}^\infty f_Y(y\mid X = x) f_X(x)} \end{align*}

It seems that I have gotten myself a really ugly expression and I can't see how to get myself out of this. For instance:

$$xf_Y(y\mid X = x) f_X(x) = x\binom{x}{y}p^y(1-p)^{x- y}\frac{\lambda^x}{x!}e^{-\lambda} = ??$$

Can someone either show me another way, or if not, provide a hint as to how to simplify the above expression

Answer 1

You have $\displaystyle\quad\sum_{x=0}^\infty\mathrm{Pr}(Y=y|X=x)\mathrm{Pr}(X=x)$

$\displaystyle\qquad\qquad\quad=\sum_{x=0}^\infty\binom{x}{y}p^y(1-p)^{x-y}.e^{-\lambda}\frac{\lambda^x}{x!}$

$\displaystyle\qquad\qquad\quad=e^{-\lambda}p^y\sum_{x=y}^\infty\binom{x}{y}q^{x-y}\frac{\lambda^x}{x!}$ ,where $q=1-p$

$\displaystyle\qquad\qquad\quad=\frac{e^{-\lambda}p^y\lambda^y}{y!}\sum_{x=y}^\infty\frac{(\lambda q)^{x-y}}{(x-y)!}$

Thus, $\displaystyle\mathrm{Pr}(Y=y)=\frac{e^{-\lambda p}(\lambda p)^y}{y!}\mathbf{1}_{\{y=0,1,\cdots\}}$ where $\lambda p>0$

So the conditional pmf of $X|Y=y$ is given by

$\displaystyle\mathrm{Pr}(X=x|Y=y)=\frac{\binom{x}{y}p^y(1-p)^{x-y}.e^{-\lambda}\frac{\lambda^x}{x!}}{\dfrac{e^{-\lambda p}(\lambda p)^y}{y!}}\mathbf{1}_{\{x=y,y+1,\cdots\}}$