What is the expected value of $UX$ for $X \sim Pos(U)$?

Question

A real number $U$ is chosen uniformly from the interval $(0,10)$. Next, an integer $X$ is chosen according to the Poisson distribution with parameter $U$. What is the expected value of $UX \, $?

Here is my work.

By the law of total expectation, $E[UX]=E[E[UX|X]]$

$E[UX|X]=XE[U]$ by conditioning on a known variable.

$E[U]=5$, since $U \sim UNIF(0,10)$.

Hence, $E[Y]=E[5X]=5E[X].$

$E[X]=E[E[X|U]]; \hspace{1cm} X|U \sim Pos(U); \hspace{1cm} E[X|U]=U; \hspace{1cm} E[X]=E[U]=5.$

Hence, $E[Y]=5 \times 5=25.$

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The answer key says $E[UX]=\frac{100}{3}$. I'm not sure where I went wrong.

Answer 1

Why are you trying to condition over $X$?

Since you have the conditional distribution for $X$ when given $U$, you should use that conditioning in the expectation too.

$$\begin{align}\mathsf E(UX)&=\mathsf E(\mathsf E(UX\mid U))\\[1ex]&=\mathsf E(U~\mathsf E(X\mid U))\end{align}$$

Now, because $X\mid U\sim\mathcal{Poiss}(U)$ then ...$$\mathsf E(X\mid U)=U$$

So, you can continue from here.

Answer 2

You need to condition on what you know. You have $$\mathbb E(UX)=\mathbb E \left[ \mathbb E(UX | U) \right]=\mathbb E\left[ U \mathbb E (X| U) \right]$$ where first you used tower property (the small subsigma algebra wins) and then pullout.

Now since you know $\mathbb P_{X|U}$ is regular and in $L^1$, you have

\begin{aligned} \mathbb{E}\left(X | U=u\right) &=\int_{\Omega} 1_{\{U=u\}}(\omega) X(\omega) \mathbb{P}(d \omega) \\ &=\int_{\mathbb{N}_{0}} k \mathbb{P}\left(X=k | U=u\right) \\ &=\sum_{k=0}^{\infty} k e^{-u} \frac{u^{k}}{k !} \\ &=ue^{-u} \sum_{k=1}^{\infty} \frac{u^{k-1}}{(k-1) !}=u. \end{aligned} so $U$ is a version of $\mathbb E(X|U)$ and so $\mathbb E(UX)=\mathbb E(U^2)$. Lastly $$\mathbb E U^2=\int_\mathbb R x^2 \mathbb P_U( dx)=\int_{-\infty}^\infty x^2 \frac{1}{10} 1_{[0,10]}(x) ~dx=\frac{1}{10}\int_0^{10} x^2 dx=\frac{1}{10} \frac{1000}{3}=\frac{100}{3}.$$