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?

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 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