# Likelyhood of Poisson Distribution

The number of accidents in a week follows a poisson distribution with mean $\lambda$.

Likelyhood is given as $$L(\lambda)=\frac{ \lambda^{\sum_1^n x_i } e^{-n\lambda}} { \prod x_i!}$$

However only a proportion p, are reported and each accident is reported with probability p, independent of all others.

How would you modify the likelyhood function to take account of this?

## 1 Answer

The probability without the report issue is $$P(X = n) = \frac{\lambda^n e^{-\lambda}}{n!}$$

including the report issue it becomes for $n\neq 0$: $$P(X = n) = \sum_{m=n}^\infty \frac{\lambda^m e^{-\lambda}}{m!} p^n(1-p)^{m-n}$$ (this is the probability of $m$ accidents, $n$ of which being reported).

Then just write $$L(\lambda) = \prod_{k=1}^n \sum_{m=n}^\infty \frac{\lambda^m e^{-\lambda}}{m!} p^n(1-p)^{m-n}$$

I'm not sure you can get any better expression.

• So for n, the number of reported accidents, do they still follow a poisson distribution? I can't really tell just by looking at the expression. – Dylan Morgan Oct 20 '14 at 22:45
• I don't think the expression does have the right shape: it has the form $[p/(1-p)]^n F(n)$ where F(n) is the rest of an exponential series. But this is not a proof... – mookid Oct 20 '14 at 22:49