Given a compound Poisson distribution $$S(t):=\sum_{k=1}^{N(t)} X_{k}$$ with

  1. $N(t)\in\mathbb{N},\,t\geq0$ a Poisson process with rate $\lambda.$
  2. $X_{k}\in L^{2}$ are iid random variables, i.e. $\mathbb{E}\left[\vert X_{k}\vert\right]<\infty$ and $\sigma^{2}:=\operatorname{Var}(X_{k})<\infty$.
  3. $N$ and $(X_{1},X_{2},\ldots)$ are independent.

Then $$\mathbb{E}\left[ S\right]= \mathbb{E}\left[ N\right]\mathbb{E}\left[ X_{1}\right]$$ as well as $$\operatorname{Var}\left[ S\right]= \operatorname{Var}\left[ N\right]\,\mathbb{E}\left[ X_{1}\right]^{2} + \mathbb{E}\left[ N\right]\operatorname{Var}\left[ X_{1}\right].$$ Then we have, by a version of the central limit theorem, that

$$\frac{S(t)-\mathbb{E}\left[S(t) \right]}{\sqrt{\operatorname{Var}\left[S(t) \right]}}\stackrel{d}{\to} \mathcal{N}(0,1)\,\text{as } t\to\infty.$$

Suppose that $\mathbb{E}\left[ X_{k}\right]=0$, e.g. by replacing $X_{k}$ above by the translation $X_{k}-\mathbb{E}\left[ X_{k}\right]$, and $N(t), t \geq 0$ is a family of positive, integer valued random variables, such that there is a $\theta >0$ $$ \frac{N(t)}{t}\stackrel{\mathbb{P}}{\rightarrow}\theta,\,\text{ as } t\to\infty. $$

According to Renyi's or Anscombe's Theorem, we then have \begin{align*} \frac{S_{N(t)}}{\sigma\sqrt{N(t)}} \xrightarrow[]{d} \mathcal{N}(0,1)\,\text{ as } t\to\infty \\ \frac{S_{N(t)}}{\sigma\sqrt{\lambda\cdot t}} \xrightarrow[]{d} \mathcal{N}(0,1)\,\text{ as } t\to\infty, \\ \end{align*} which is different from the above normal approximation (using Wald's identity).

My questions are now:

  1. What is the key difference between the two approximations? And which one is when preferrable?

  2. Given a realization of a compound Poisson process $$ s=\sum_{k=1}^{N} x_{k},$$ with unknown parameters. How can one estimate the parameters in order to afterwards apply the central limit theorem (with Wald's identity)?

For example, if one uses sample means, we would get $$\frac{s-N\cdot \frac{\sum_{k=1}^{N}x_{k}}{N}}{\tilde{\sigma}}=\frac{s-s}{\tilde{\sigma}}=0,$$ which is not useful.

  • $\begingroup$ At the top, we have $X_k$ is non-negative; in the middle, you added that $\mathbb{E}[X_k]=0$, so $X_k$ would be identically zero; is this what you meant? $\endgroup$ – Bill Bradley Jul 6 '18 at 19:21
  • $\begingroup$ Ooops, not really. I replaced the condition in the beginning by $X_{k}\in L^{2}.$ For the second statement, one has then just to "center" $X_{k}$ around 0. $\endgroup$ – Strickland Jul 8 '18 at 18:48

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