We have that $N$ and $X_1, X_2, \dots$ are all independent. We also have $\operatorname{E} [X_j] = \mu$ and $\operatorname{Var}[X_j] = σ^2$.
Then, we introduce an integer–valued random variable, $N$, which is the random sum such that:
$$Z = \sum_{j=1}^{N+1}X_j.$$
Assuming that $N$ is distributed $\sim\mathrm{Poisson}(\lambda)$, what is the first moment and what is the variance of $Z$?
For a normal Poisson distribution, I know the variance is just $\lambda$, as is the mean. I’m having trouble understanding the implication of having the bounds be poisson distributed. Normally, I would just say “variance of the sum is the sum of the variance,” but I don’t think that’s how it works with random sums. Any hints/guidance appreciated.
\begin{align}
\operatorname{var}(Z) & = \operatorname{var}(\operatorname{E}(Z\mid N)) + \operatorname{E}(\operatorname{var}(Z\mid N)) & & (\text{This is the law of total variance.}) \\[10pt]
& = \operatorname{var}(N\mu) + \operatorname{E}(N\sigma^2) \\[10pt]
& = \mu^2 \operatorname{var}(N) + \sigma^2 \operatorname{E}(N) \\[10pt]
& = \mu^2 \lambda + \sigma^2 \lambda = \lambda \operatorname{E}(X_1^2).
\end{align}
Indeed, generally the $n$th cumulant of a compound Poisson distribution is the mean of the simply Poisson distribution times the $n$th raw moment of the distribution that gets compounded.
PS: The above applies if the sum is from $0$ to $N$; I’ll leave it as an exercise to figure out whether something needs to change if it’s from $1$ to $N+1$.
Hint: rewrite
$$
Z = \sum_{n=1}^\infty X_n\mathbb{1}_{\{N+1 \geq n\}}
$$ and apply your “expectation of the sum is the sum of the expectations” idea.
Following the comment below, more detail.
Write $Y_n = \mathbb{1}_{\{N+1 \geq n\}}$, which is independent of $X_n$.
Then for the expectation, you have $$\mathbb{E}[Z] = \sum_{n=1}^\infty \mathbb{E}[X_nY_n] = \sum_{n=1}^\infty \mathbb{E}[X_n] \mathbb{E}[Y_n]= \sum_{n=1}^\infty \mu \mathbb{P}\{N \geq n-1\} = \mu\sum_{n=0}^\infty \mathbb{P}\{N \geq n\} $$ using independence of $X_n$ and $Y_n$, which yields $$\mathbb{E}[Z] = \mu\lambda$$ since $\sum_{n=0}^\infty\mathbb{P}\{N \geq n\} = \mathbb{E}[N].$
If $T = X_1 + X_2 + \cdots +X_N$, where $X_i$ are iid and $N$ is
independent of $X_i$, then a standard result, usually derived
via a conditioning argument, is that $E(T) = E(N)E(X)$ and
$V(T) = E(N)V(X) + V(N)[E(X)]^2.$ (Search the Internet for
proofs under ‘Random sum of random variables’; the UNL page
seems complete.) A small adjustment will
deal with your question; see comments in the simulation code below. (Question edited: What used to be $S$ is
now $Z,$ original $S$ used below.)
In addition to its mean and variance, the $distribution$ of $S$
can be approximated through simulation. Suppose that $X_i \sim Exp(rate=1/3)$ and $N \sim Pois(\lambda = 5).$ Then a simulation
of 100,000 sums $S$ can be performed as follows. For example,
one can approximate $P(S > 20),$ which is not available from
the formulas above.
m = 10^5; s = numeric(m); rate=1/3; lam=5
for (i in 1:m) {
n = rpois(1, lam)
s[i] = sum(rexp(n+1, rate)) } # note the 'n+1'
mean(s); var(s)
## 17.93764 # approximates E(S) = (5 + 1)(3)
## 97.76707 # approximates V(S) = 6(9) + 5(9)
mean(s > 20)
## 0.36613 # approximates P(S > 20)
You can $roughly$ verify your formula for $V(S)$ by comparing
it with the simulated value above. (Variances deal with squared
units, so numerical approximations can be less precise.)
The histogram below shows the simulated distribution of $S$. The dotted density
curve of $Gamma(6, rate=1/3)$ would apply to the $deterministic$
sum of $n = 6$ random variables $X_i \sim Exp(rate=1/3)$. Using a
$random$ Poisson sum the simulated distribution of the histogram has a larger variance.
The simulated distribution is markedly right-skewed, so that
$P(S > 20)$ cannot be well approximated by the false assumption
that $S$ must be nearly normally distributed as $Norm(E(S)=18,
SD(S)=\sqrt{99}),$ solid green density curve.
Note: See a similar discussion at http://math.stackexchange.com/questions/1483316/
.