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!
Best Answer
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 $t\to\infty$.