Conditional Expectation Discrete and Continuous

Question

Find $E[X]$ and $Var[X]$

So for the expectation so far I got that:

$$E[X] = E[X|N=n]P(N=n) = \large\frac{n+1}{\lambda} \frac{\lambda^{n}}{n!}e^{-\lambda}$$ but for conditioning on both a discrete and continuous random variable I am not sure whether to use the summation or integration. For integration it comes out to be just $\frac{n+1}{\lambda}$ which does not seem right since that would indicate that the expectation is independent of each other.

Answer 1

I think your first step should be $$E[X] = \sum_{n\ge 0} E[X \mid N=n] P(N=n).$$

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Adding some details:

We first find a nice result using the Taylor series of the exponential.

\begin{align*} e^x &= \sum_{n \ge 0} \frac{1}{n!} x^n\\ xe^x &=\sum_{n \ge 0} \frac{1}{n!} x^{n+1}\\ e^x(x+1) &=\sum_{n \ge 0} \frac{n+1}{n!} x^{n} & \text{take the derivative of both sides}\\ \end{align*}

We apply this to where you got stuck. $$E[X]= \frac{e^{-\lambda}}{\lambda} \sum_{n \ge 0} \frac{n+1}{n!} \lambda^n=\frac{e^{-\lambda}}{\lambda} e^{\lambda}(\lambda+1)=\frac{\lambda+1}{\lambda}.$$

Answer 2

The mean of a Gamma or Erlang distribution $\Gamma(k,\lambda)$ is $k/\lambda.$

Then $$E(X)=E(E(X | N))=E\left(\frac{N+1}{\lambda}\right)=\frac{\lambda+1}{\lambda}.$$