TITLE: Distribution of compound Poisson process
QUESTION [9 upvotes]: Suppose a compound Poisson process is defined as $X_{t} = \sum_{n=1}^{N_t} Y_n$, where $\{Y_n\}$ are i.i.d. with some distribution $F_Y$, and $(N_t)$ is a Poisson process with parameter $\alpha$ and also independent from $\{Y_n\}$.

Is it true that as 
$t\rightarrow \infty, \, \frac{X_{t}-E(X_{t})}{\sigma(X_t)
    \sqrt(N_t)}  \rightarrow \mathbf{N}(0, 1)$ in distribution, where the limit is a standard Gaussian distribution? I am considering
using Central Limit Theorem to show it, but the
theorem I have learned only applies when $N_t$ is
fixed and deterministic instead of
being a Poisson process. 
A side question: is it possible to derive the
distribution of $X_{t}$, for each $t\geq 0$? Some book that has the derivation?

Thanks!

REPLY [12 votes]: Let $Y(j)$ be i.i.d. with finite mean and variance, and set 
$\mu=\mathbb{E}(Y)$ and $\tau=\sqrt{\mathbb{E}(Y^2)}$. 
If $(N(t))$ is an independent Poisson process with rate $\lambda$,
then the compound Poisson process is defined as 
$$X(t)=\sum_{j=0}^{N(t)} Y(j).$$
The characteristic function of $X(t)$ is calculated as follows:
for real $s$ we have 
\begin{eqnarray*}
\psi(s)&=&\mathbb{E}\left(e^{is X(t)}\right)\cr
       &=&\sum_{j=0}^\infty \mathbb{E}\left(e^{is X(t)} \ | \ N(t)=j\right) \mathbb{P}(N(t)=j)\cr
       &=&\sum_{j=0}^\infty \mathbb{E}\left(e^{is (Y(1)+\cdots +Y(j))} \ | \ N(t)=j\right) \mathbb{P}(N(t)=j)\cr 
       &=&\sum_{j=0}^\infty \mathbb{E}\left(e^{is (Y(1)+\cdots +Y(j))}\right) \mathbb{P}(N(t)=j)\cr
       &=&\sum_{j=0}^\infty \phi_Y(s)^j {(\lambda t)^j\over j!} e^{-\lambda t}\cr
       &=& \exp(\lambda t [\phi_Y(s)-1])
       \end{eqnarray*}
where $\phi_Y$ is the characteristic function of $Y$.
From this we easily calculate $\mu(t):=\mathbb{E}(X(t))=\lambda t \mu$ 
and $\sigma(t):=\sigma(X(t))= \sqrt{\lambda t} \tau$.
Take the expansion $\phi_Y(s)=1+is\mu -s^2\tau^2 /2+o(s^2)$ and substitute it into 
the characteristic function of the normalized random variable  ${(X(t)-\mu(t)) /\sigma(t)}$ to obtain
\begin{eqnarray*}
\psi^*(s) &=& \exp(-is(\mu(t)/\sigma(t))) \exp(\lambda t [\phi_Y(s/\sigma(t))-1]) \
       &=& \exp(-s^2/2 +o(1))
\end{eqnarray*}
where $o(1)$ goes to zero as $t\to\infty$. This gives the central limit theorem
$${X(t)-\mu(t)\over\sigma(t)}\Rightarrow N(0,1).$$  
We may replace $\sigma(t)$, for example, with $\tau \sqrt{N(t)}$ to get
$${X(t)-\mu(t)\over\tau \sqrt{N(t)}}= {X(t)-\mu(t)\over\sigma(t)} \sqrt{\lambda t \over N(t)} \Rightarrow N(0,1),$$
by Slutsky's theorem, since  $\sqrt{\lambda t \over N(t)}\to 1$ in probability by the law of large numbers. 

Added: Let $\sigma=\sqrt{\mathbb{E}(Y^2)-\mathbb{E}(Y)^2}$ be the standard deviation of $Y$,
and define the sequence of standardized random variables 
$$T(n)={\sum_{j=1}^n Y(j) -n\mu\over\sigma\sqrt{n}},$$
so that 
$${X(t)-\mu N(t)\over \sigma \sqrt{N(t)}}=T(N(t)).$$
Let $f$ be a bounded, continuous function on $\mathbb{R}$. By the usual
central limit theorem we have $\mathbb{E}(f(T(n)))\to \mathbb{E}(f(Z))$ where
 $Z$ is a standard normal random variable. 
We have for any $N>1$,
$$\begin{eqnarray*}
|\mathbb{E}(f(T(N(t)))) - \mathbb{E}(f(Z))|
  &=& \sum_{n=0}^\infty |\mathbb{E}(f(T(n)) - \mathbb{E}(f(Z))|\ \mathbb{P}(N(t)=n)  \cr
  &\leq& 2\|f\|_\infty \mathbb{P}(N(t)\leq N) +\sup_{n>N} |\mathbb{E}(f(T(n)))- \mathbb{E}(f(Z)) |.
\end{eqnarray*} 
$$
First choosing $N$ large to make the right hand side small, then letting $t\to\infty$ so
that $\mathbb{P}(N(t)\leq N)\to 0$, shows that 
$$  \mathbb{E}(f(T(N(t)))) \to \mathbb{E}(f(Z)). $$
This shows that $T(N(t))$ converges in distribution to a standard normal as $t\to\infty$.