Let $S_{N}=\sum_{i=1}^{N}X_{i}$, where $X_{i}$ are i.i.d random variables with pdf $f$ and distribution $F$ and $N$ is a r.v. following a Poisson($\lambda$) which is independent of the $X_{i}$'s. Let $Y$ be a r.v. with distribution $F$ and independent of $S_{N}$. Let $h$ be a measurable real function.
I need to show that $$E[S_{N}h(S_{N})]=\lambda E[Yh(S_{N}+Y)]$$
I have tried to compute the right-hand side by conditioning to $N=k$, say, but I don't get anyhing. Also I try it starting from the left-hand side but I don't know how to get the r.v. $Y$ in the formula.
If anybody could help me I would be very thankful.
$\newcommand{\e}{\operatorname E}$In the special case in which $\Pr(X_i=1)=1,$ the proposition says $$ \e(Nh(N)) = \lambda \e(h(N+1)). $$ This is called the Robbins Lemma, after Herbert Robbins, who introduced it in the 1950s for use in empirical Bayes methods in statistics.
\begin{align} & \e(S_N h(S_N)) \\[8pt] = {} & \e(\e(S_Nh(S_N)\mid N)) \\[8pt] = {} & \sum_{n=1}^\infty \e(S_n h(S_n))\Pr(N=n) \end{align} (We don't need a term for $n=0,$ since in that case the term is $0.$) \begin{align} = {} & \sum_{n=1}^\infty \e(S_n h(S_n)) \cdot \frac{e^{-\lambda}\lambda^n}{n!} \\[8pt] = {} & \sum_{n=0}^\infty \e(S_{n+1} h(S_{n+1})) \cdot \frac{e^{-\lambda} \lambda^{n+1}}{(n+1)!} \\[8pt] = {} & \lambda \sum_{n=0}^\infty \e\left(\frac{S_{n+1} h(S_{n+1})}{n+1} \right) \cdot\frac{e^{-\lambda} \lambda^n}{n!} \\[8pt] = {} & \lambda \sum_{n=0}^\infty \sum_{k=1}^{n+1} \e\left( \frac{X_k h(S_{n+1})}{n+1} \right) \cdot\frac{e^{-\lambda} \lambda^n}{n!} \\[8pt] = {} & \lambda \sum_{n=0}^\infty \sum_{k=1}^{n+1} \e\left( \frac{X_{n+1} h(S_{n+1})}{n+1} \right) \cdot\frac{e^{-\lambda} \lambda^n}{n!} \end{align} This last step holds since every term in this sum has the same expected value. Since there are $n+1$ terms, the sum is $n+1$ times the value of any one term, so we get \begin{align} = {} & \lambda \sum_{n=0}^\infty \e\left( X_{n+1} h(S_{n+1}) \right) \cdot\frac{e^{-\lambda} \lambda^n}{n!} \end{align} Can you finish this in a few seconds?