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I have this random sum, $Z=\sum_{k=1}^{N}Y_{k}$ where $(Y_1,Y_2,\dots,Y_k,\dots)$ are independent and exponentially distributed random variables (mean is $\mu$). We've defined $N=M+1$ where $M$ is poisson distributed (intensity is $\lambda$).

I need to find the first and second moments of this random sum.

I have found the first moment, which I did by just messing with the bounds a little bit. However, I don't know the formula for finding the second moment. I could use help with that.

This is what I did:

$E[Z]=\sum_m^{\infty}E[Z|M=m]P_M(m)$

$= \sum_{m=0}^{\infty}(m+1)\mu e^{-\lambda}\frac{\lambda^m}{m!}$

I split the sums and obtained:

$E[Z]=\mu(\lambda+1)$.

I just need a boost getting started for $E[Z^2]$. Much appreciated. Thanks!

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    $\begingroup$ Why not find the variance and subtract $(E[Z])^2$? $\endgroup$
    – Em.
    Feb 2, 2016 at 23:20
  • $\begingroup$ I do also need to find the variance. However, that's another thing I don't know how to calculate. I was hoping to find $E[Z^2]$ in a similar manner, then use the formula for variance that depends on these two quantities. I'd be open to calculating the variance first, though, if that's easier. $\endgroup$
    – Taylor
    Feb 2, 2016 at 23:23

1 Answer 1

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$$ \mathbb{E}[Z^2\mid M] = \mathbb{E}\left[\left(\sum_{k=1}^N Y_k\right)^2\right] = \mathbb{E}\left[\sum_{k=1}^N \sum_{\ell=1}^N Y_k Y_\ell\right] = \sum_{k=1}^N \sum_{\ell=1}^N \mathbb{E}\left[Y_k Y_\ell\right] $$ by linearity of expectation. Now, $Y_k$ and $Y_\ell$ are independent if, and only if, $k\neq \ell$, so $$\begin{align} \mathbb{E}[Z^2\mid M] &= \sum_{k=1}^N \mathbb{E}\left[Y_k^2\right] + 2\sum_{k=1}^N \sum_{\ell=k+1}^N \mathbb{E}\left[Y_k\right] \mathbb{E}\left[Y_\ell\right]\\ &= \sum_{k=1}^N 2\mu^2 + 2\sum_{k=1}^N \sum_{\ell=k+1}^N \mu^2 = 2N\mu^2 + \mu^2 N(N-1) \\ &= \mu^2\left(M^2 + 3M + 2\right) \end{align}$$ (If I didn't screw up in the computations). From there, can you conclude by computing $$ \mathbb{E}[Z^2] = \mathbb{E}[\mathbb{E}[Z^2\mid M]] $$ ?

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  • $\begingroup$ Thank you so much for your reply! Do you mind explaining what $M$ actually stands for, though? $M$ is a random variable, so I didn't expect to see the moment expressed in terms of $M$. $\endgroup$
    – Taylor
    Feb 2, 2016 at 23:25
  • $\begingroup$ The first part treats it as " a constant" (formally, $\mathbb{E}[Z^2\mid M]$ is a random variable itself; that is why it is an expression that dependends on the random variable $M$ (which is, as you define it, Poisson($\lambda$) distributed). But then, you have that $\mathbb{E}[Z^2] = \mathbb{E}[\mathbb{E}[Z^2\mid M]]$, which will make $M$ "disappear" from the expression to obtain a bona fide number: the expectation of $Z^2$: $$\mathbb{E}[Z^2] = \mu^2\mathbb{E}[M^2+3M +2]$$ again provided I didn't make a computation mistake. $\endgroup$
    – Clement C.
    Feb 2, 2016 at 23:27
  • $\begingroup$ This is great. Thanks so much. $\endgroup$
    – Taylor
    Feb 3, 2016 at 5:47

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