I am trying to get a closed form expression for the expected value of the following summation of RVs: $\sum_{i=1}^{Y} X_{i}$, where $Y$ is Poisson distributed with parameter $\lambda$ and $ X_{i} $ follows some known distribution $f_X(x)$. Are there any means to drop the RV $Y$ using $\lambda$?
Thank you for your time and patience.
Answer
Yes if your $X_i$'s are iid, then you can use iterated conditioning to prove the Wald's identity.
\begin{align}
\mathbb{E}_X[\sum_{i=1}^Y X_i] &= \mathbb{E}_Y[\mathbb{E}_X[\sum_{i=1}^y X_i | Y=y]]\\
&=\mathbb{E}_Y[ y \mathbb{E}_X[X_1]]\\
&= \sum_{y=0}^{\infty} y \mathbb{E}_X[X_1] e^{-\lambda} \frac{\lambda^y}{y !}\\
&= \lambda \mathbb{E}[X_1] e^{-\lambda} \sum_{y=1}^{\infty} \frac{\lambda^{y-1}}{(y-1) !}\\
&= \lambda \mathbb{E}X_1
\end{align}
since the sum is an series expansion for $e^{\lambda}$. We're using the iid property to go from the first to the second line. Then it's just the expectation of the Poisson distribution.
No comments:
Post a Comment