# Cumulative distribution function or density for Compound Poisson distribution

I have the Compound Poisson distribution $$\xi = \sum_{n=1}^N X_n$$ where N has Poisson ($\lambda$) distribution and $X_i$ are independent and identically distributed and have normal distribution. I need to calculate the theoretical probability (for fixed numbers $x_i$ and $x_{i+1}$) of $$P(\xi\in[x_i,x_{i+1}])$$ According to this question, Compound Poisson distribution have no density because it isn't cumulative distibution. It means, I can't take definite integral from density... Please, help, how can this probability be calculated?

P.S. Sorry for my English.

• This probability is going to be a random variable. Is that what you are really after? Apr 15, 2013 at 14:32
• Why? Isn't it a fixed number for fixed $x_i$ and $x_{i+1}$? Apr 15, 2013 at 14:36
• Now that you modified the question, it is no longer random. Apr 15, 2013 at 14:42

Let's say that the $X_i$ are normal $N \left( \mu, \sigma^2 \right)$ and $N$ has a truncated Poisson distribuiton. Then, for $n>0$ fixed, $\sum_{i = 1}^n X_i$ is $N \left( \mu n, \sigma^2 n \right)$. Let $\Phi_{\mu m, \sigma^2 n}$ be its CDF, then, for $a$ and $b$ fixed \begin{eqnarray*} \Pr \left\{ \xi \in \left[ a, b \right] \right\} & = & E_N \left[ \text{Pr}\{\xi \in \left[ a, b \right]\} |N = n \right]\\ & = & \frac{ \mathrm{e}^{- \lambda}}{1-\mathrm{e}^{- \lambda}} \sum_{n = 1}^{\infty} \frac{\lambda^n}{n!} \left[ \Phi_{\mu n, \sigma^2 n} \left( b \right) - \Phi_{\mu n, \sigma^2 n} \left( a \right) \right] \end{eqnarray*}
If $N=0$ with positive probability (and thus N is just a Poisson random variable), then $\xi$ will have a mass at 0 with positive probability and thus the probability that $\xi$ falls in set $A$ is given by \begin{eqnarray*} \Pr \left[ \xi \in A \right] & = & \mathrm{e}^{- \lambda} 1 \left( 0 \in A \right) + \mathrm{e}^{- \lambda} \sum_{n = 1}^{\infty} \frac{\lambda^n}{n!} \int_A \phi_{\mu n, \sigma^2 n} \left( x \right) \mathrm{d} x \end{eqnarray*} where $\phi_{\mu m, \sigma^2 n}$ are the densities of $N \left( \mu n, \sigma^2 n \right)$ random variables.
• Thanks. But if consider a=-infinity we get CDF for this random variable. And if I try to take derivative by $b$ I get (un)expected result: this series and series of derivatives converges uniformly on $x\in R$ and all functions are continuous => there are exists continuous function "density of probability" which is derivative of CDF. But it contradicts the question I posted link to. Where are an error? Apr 15, 2013 at 16:14