1
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ This probability is going to be a random variable. Is that what you are really after? $\endgroup$
    – Learner
    Apr 15, 2013 at 14:32
  • $\begingroup$ Why? Isn't it a fixed number for fixed $x_i$ and $x_{i+1}$? $\endgroup$ Apr 15, 2013 at 14:36
  • $\begingroup$ Now that you modified the question, it is no longer random. $\endgroup$
    – Learner
    Apr 15, 2013 at 14:42

1 Answer 1

3
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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? $\endgroup$ Apr 15, 2013 at 16:14
  • $\begingroup$ Hmm.. Frankly speaking, I don't understand your comment.. Can you explain it more in details? $\endgroup$ Apr 15, 2013 at 16:36
  • $\begingroup$ See edited answer. $\endgroup$
    – Learner
    Apr 15, 2013 at 16:43

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .