Assume in a certain year the number of credit obligors of a bank who are at risk of default is Poisson distributed with intensity $\lambda = 10$. Each individual obligor may actually default in that year with probability $p = 0.2$. Assume mutual independence of the obligors. Let $X$ be the number of obligors that default in that year. Show that $X$ is Poisson distributed and determine its intensity.
I denote $N$ the number of obligors at risk. And by using the identity $$P(X = k) = \sum_E P(X = k\mid N = n)P(N = n)$$ to find the pmf I think I am supposed to arrive at the Poisson distribution. $P(N=n)$ is already Poisson. I am having trouble figuring out the $P(X = k\mid N = n)$ and coming up with the probabilities of the pmf which need to be summed up to yield the final Poisson.
Hintage: $\mathsf P(X=k\mid N=n)$ is the probability of $k$ 'successes' in a sequence of $n$ independent trials, where 'success' means "obligator defaults" and has rate $p$.
Do you recognise this distribution now?
PS: $E\equiv \{n\mid n\in\{k,\ldots,\infty\}\}$, because you cannot have more successes than trials.
Just substitute the p.m.f. to get: $~\sum\limits_{n=k}^\infty \dfrac {n!~p^k~(1-p)^{n-k}}{ k!~(n-k)!}\cdotp\dfrac{\lambda^n e^{-\lambda}}{n!}~$, cancel common factors, move unbound factors to the left of the sigma sign, and simplify the series with a change of variables.
The "trick" then lies in recalling the Taylor series expansion $~e^t = \sum\limits_{m=0}^\infty \dfrac{t^m}{m!}~$.