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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?


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@Shai: Sources: books.google.com/… and answers.yahoo.com/question/index?qid=20110208135141AAoP4ZO . In the former, it does not claim the convergence, but in the latter, it claims it when the state space of $(X_t)$ is finite. I think I should consider the restricted case. –  Ethan Apr 6 '11 at 15:17
Thanks for the sources. –  Shai Covo Apr 6 '11 at 15:19
The denominator in your expression in question 1 is not right. You want either $\sigma(X_t)$ or a multiple of $\sqrt{N_t}$, not both multiplied together. In your first reference, they divide by $\sigma \sqrt{M}$; the correct analogy in your problem is $\sigma(Y) \sqrt{N_t}$ –  Byron Schmuland Apr 6 '11 at 16:23

3 Answers 3

up vote 5 down vote accepted

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 $t\to\infty$.

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@Byron: Thanks! Let me try to understand your reply. There are two different ratios here, and you have shown the first ratio converges in distribution, while you are working on the second one? –  Ethan Apr 6 '11 at 20:38
@Ethan You're welcome! Actually I have three ratios. I've shown that the first two converge to $N(0,1)$ and I'm sure that the third one converges to a normal distribution with some other variance. But I haven't checked the details yet. –  Byron Schmuland Apr 6 '11 at 20:54
@Byron: Thanks! (1) About "the expansion $\phi_Y(s)=1+is\mu -s^2\tau^2 /2+o(s^2)$", I seem to have seen this in Kailai Chung's A course in probability theory, but now cannot find it. Do you know where I can find some explanation about the expansion of the characteristic function? (2) Is "the cardinality of the state space of $(X_t)$ is finite" required for the three ratios to converge to N(0,1)? –  Ethan Apr 6 '11 at 21:17
@Ethan (1) I don't have that book of Chung, but this expansion can be found in many probability texts. For instance, it is Theorem 3.8 (section 2.3) in Durrett's Probability: Theory and Examples (2nd edition) and Theorem 4 in section 5.7 of Grimmett and Stirzaker's Probability and Random Processes (3rd edition). As for point (2), we make no assumption on the state space of $X(t)$. We only need the $Y$ random variables to have finite variance. –  Byron Schmuland Apr 6 '11 at 22:10
@Ethan: Kai Lai Chung. –  Did Apr 6 '11 at 22:33

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 $\nu(dx)=\alpha F_Y (dx)$.


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.

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@Ethan: What do you mean by "Also same question except by using the distribution given by the book by Sato?" –  Shai Covo Apr 6 '11 at 15:11
@Shai: I meant if one could show the convergence, respectively by using Central limit theorem, and by the distribution of $X_t$ from the book you cited. –  Ethan Apr 6 '11 at 15:13
@Ethan: Only by using CLT... –  Shai Covo Apr 6 '11 at 16:24
@Shai: Thanks! I was wondering why "$X_t$ is a sum of n i.i.d. variables", i.e. why those increments of compound Poisson process are i.i.d.? –  Ethan Apr 6 '11 at 20:34
@Ethan: Indeed. –  Shai Covo Apr 6 '11 at 20:35

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.

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