Distribution of compound Poisson process

Question

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 $\{Y_n\}$.

1. 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.
2. 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!

Answer 1

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.

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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 $t\to\infty$.

Answer 2

Derivation of the formula for the distribution of $X_t$. The formula $$ P_{X_t } = e^{ - t\nu (\mathbb{R})} \sum\nolimits_{k = 0}^\infty {(k!)^{ - 1} t^k \nu ^k } , $$ where $\nu$ is the L\'evy measure of $X$ and $\nu^k$ is the $k$-fold convolution of $\nu$, is very easy to derive. Indeed, using the notation in the question, the law of total probability gives $$ {\rm P}(X_t \in B) = \sum\limits_{k = 0}^\infty {{\rm P}(X_t \in B|N_t = k){\rm P}(N_t = k)} . $$ Thus, $$ {\rm P}(X_t \in B) = \sum\limits_{k = 0}^\infty {{\rm P}(Y_1 + \cdots + Y_k \in B)\frac{{e^{ - \alpha t} (\alpha t)^k }}{{k!}}} = e^{ - \alpha t} \sum\limits_{k = 0}^\infty {(k!)^{ - 1} t^k \alpha ^k F_Y^k (B) } , $$ where $F_Y^k$ is the $k$-fold convolution of $F_Y$. Now, since the corresponding L\'evy measure $\nu$ is given by $\nu (dx) = \alpha F_Y (dx)$, it holds $\nu(\mathbb{R}) = \alpha$ and $\nu^k = \alpha^k F_Y ^k$, and so the original formula is established.

Remark. As should be clear, the distribution of $X_t$ is very complicated in general.

Answer 3

The distribution of $X_t$ is given by $$ P_{X_t } = e^{ - t\nu (\mathbb{R})} \sum\nolimits_{k = 0}^\infty {(k!)^{ - 1} t^k \nu ^k } , $$ where $\nu$ is the L\'evy measure of $X$, and $\nu^k$ is the $k$-fold convolution of $\nu$. This is contained in Remark 27.3 in the book [L\'evy Processes and Infinitely Divisible Distributions], by Sato.

For a compound Poisson process with rate $\alpha$ and jump distribution $F_Y$, the L\'evy measure $\nu$ is finite and given by $\nu(dx)=\alpha F_Y (dx)$.

EDIT:

For the first question, note that if $t =n \in \mathbb{N}$, then $$ X_t = X_n = (X_1 - X_0) + (X_2 - X_1) + \cdots + (X_n - X_{n-1}) $$ (note that $X_0 = 0$). Thus, $X_t$ is a sum of $n$ i.i.d. variables, each with expectation ${\rm E}(X_1)$ and variance ${\rm Var}(X_1)$. Now, as is well known and easy to show, $$ {\rm E}(X_1) = \alpha {\rm E}(Y_1) = \alpha \int {xF_Y (dx)} $$ and $$ {\rm Var}(X_1) = \alpha {\rm E}(Y_1^2) = \alpha \int {x^2 F_Y (dx)}, $$ provided that $Y_1$ has finite second moment. Thus, by the central limit theorem, $$ \frac{{X_t - n\alpha {\rm E}(Y_1 )}}{{\sqrt {\alpha {\rm E}(Y_1^2 )} \sqrt n }} \to {\rm N}(0,1), $$ as $n \to \infty$.

EDIT: Put it another way,
$$ \frac{{X_t - {\rm E}(X_t )}}{{\sqrt {{\rm Var}(X_t )} }} \to {\rm N}(0,1) $$ (shown here for the case $t \to \infty$ integer).

EDIT: Some more details in response to the OP's request.

A compound Poisson process is a special case of a L\'evy process, that is, a process $X=\{X_t: t \geq 0\}$ with stationary independent increments, continuous in probability and having sample paths which are right-continuous with left limits, and starting at $0$. In particular, for any $t \geq 0$ and any $n \in \mathbb{N}$, $X_t$ can be decomposed as a sum of $n$ i.i.d. random variables, which means that $X_t$ is infinitely divisible. There is a vast literature available online on this important topic.