# How do you calculate the mean and variance of a random var with a distribution function that has a parameter with its own distribution function?

I am busy with ruin theory.

$$S(t) = \sum_{i=1}^{N(t)} X_i$$

$S(t)$ is the aggregate claim size after $t$ years, where $X_i$ is the individual claim size (with mean and variance given) and $N(t)$ is the number of claims that follow a Poisson distribution with parameter $\lambda$, and it is assumed that $\lambda$ has an exponential distribution with given mean.

Now since $N(t)$ has a Poisson distribution, $S(t)$ has a Compound Poisson distribution with parameter $\lambda$, right?

Then is the expected value of $S(t)$:

$E[S(t)] = E[N(t)]\cdot E[X_i]$
$\,\:\quad\qquad = (E[\lambda]\cdot t)\cdot E[X_i]\,$ ?

I am specifically confused about the $E[N(t)]$ part, does it include the $t$ variable even though it is only distributed Poisson($\lambda$) or not? And then do you use $E[\lambda]$ or only $\lambda$ in calculating the $E[N(t)]$?

Also, the variance of $N(t)$, is it equal to the variance of $\lambda$ or the expected value of $\lambda$? And again should it be multiplied by $t$ even though $t$ is not given as part of the parameter in the question?

• Your description is missing the connection between $S(t)$, $N(t)$ and the $X_i$'s. Can you please include it? It could also help you in understanding the inequality. (The Wikipedia entry on Compound Poisson distributions is quite helpful towards resolving your interrogations.) – Xi'an Oct 26 '15 at 8:20
• Thank you @Xi'an! I edited that into the question, S(t)=Sum of the Xi's from i=1 to N(t) – user94 Oct 26 '15 at 18:10

Generally, the parameter of a Poisson distribution is defined as the distribution mean. However, for ruin theory, the expected number of claims increases with time, and should not be equal for different times, as having a time-independent $\lambda$ would imply. It is more likely that $\lambda$ is the rate parameter of the Poisson process $N(t)$, then $S(t)$ is a compound Poisson process with rate $\lambda$.
Following the linked article, $E[S(t)]$ is as given by the OP. The last step requires another application of the law of total expectation: $$E[N(t)]=E[E[N(t)|\lambda]]=E[\lambda t]=E[\lambda]t$$ since $\lambda$ is not constant but random.
The variance of $N(t)$ is given by the law of total variance: $$Var[N(t)]=E[Var[N(t)|\lambda]]+Var[E[N(t)|\lambda]]=E[\lambda t]+Var[\lambda t]=E[\lambda]t+(E[\lambda]t)^2$$ since $\lambda$ is exponentially distributed.
In the definition of the compound Poisson process, there is an underlying assumption that, as hinted by @Xian, the random variables $$X_{i}$$'s are i.i.d. as well as independent of $$N(t)$$. To compute $$E[S(t)]$$, consider $$\begin{eqnarray*} % \nonumber to remove numbering (before each equation) E\left[S(t)|N(t)=n\right] &=& E\left[X_{1}+X_{2}+\cdots +X_{N(t)}|N(t)=n\right] \\ &=& E\left[X_{1}+X_{2}+\cdots +X_{n}\right]\\ & & \quad\quad\quad\quad\quad\mbox{ since X's and N(t) are independent RV's}\\ &=& n E[X], \quad\quad\mbox{ since X_{i}'s are iid RV's} \end{eqnarray*}$$ Multiplying on both sides by $$P\left\{N(t)=n\right\}$$ and taking summation over all possible values of $$n$$, we get, $$\begin{eqnarray*} E\left[S(t)\right] &=& \sum_{n}\underbrace{E\left(S(t)|N(t)=n\right)}_{n\cdot E[X]} P\left\{N(t)=n\right\}\\ &=& \sum_{n} n\cdot E[X]\cdot P\left\{N=n\right\}\\ &=&E(X)\cdot \sum_{n}n\cdot P\left\{N=n\right\}\\ E\left[S(t)\right] &=& E[X]\cdot E[N(t)] = (\lambda t) \cdot E[X] \end{eqnarray*}$$ since $$\{N(t), t\geq 0\}$$ is Poisson process, $$E[N(t)] = \lambda t$$.
In order to find an expression for $$Var[S(t)]$$, first find the second conditional moment of $$S(t)$$. $$\begin{eqnarray*} E(S(t)^{2}|N(t)=n) &=& E[(X_{1}+X_{2}+\cdots +X_{N(t)})^{2}|N(t)=n] \\ &=& E[(X_{1}+X_{2}+\cdots +X_{n})^{2}]\\ &=& E\left\{\sum_{i=1}^{n}X_{i}^{2}+2\sum_{i Now $$\begin{eqnarray*} Var(S(t)) &=& E(S(t)^{2}) - \left[E(S(t))\right]^{2} \\ &=& \sum_{n}E(S(t)^{2}|N(t)=n)P\{N(t)=n\}-[E(X)E(N(t))]^{2}\\ &=&\sum_{n}\left[nVar\left(X\right)+n^{2}\left(E\left(X\right)\right)^{2}\right] P\left\{N(t)=n\right\} -\left[E(X)E(N(t))\right]^{2}\\ &=& Var(X) \sum_{n}nP\left\{N(t)=n\right\} + \left(E\left(X\right)\right)^{2}\sum_{n}n^{2}P\left\{N(t)=n\right\} - \left[E(X)E(N(t))\right]^{2}\\ &=&E(N(t))Var(X) + \left(E\left(X\right)\right)^{2} \left[E(N(t)^{2})-(E(N(t)))^{2}\right]\\ & = & E[N(t)\cdot Var[X] + Var[N(t)]\cdot (E[X])^{2}\nonumber\\ & =& \lambda t\cdot Var[X] + \lambda t\cdot (E[X])^{2}\nonumber\\ &=& \lambda t\cdot [Var[X] + (E[X])^{2}]\nonumber\\ Var[S(t)]&=& \lambda t\cdot E[X^{2}] \end{eqnarray*}$$