Say $Y$ is the nummer of accidents by a car, Poisson distributed ($\lambda$). People hit by the car have a probability $p=\frac{1}{2}$ to survive. Let $Z$ be the nummer of people killed by a car accident.
Now I want to determine $P(Z=k)$ for $k=0,1,2...$ .
Is it true that $P(Z=k) = \frac{1}{2} \cdot P(Y=k)$. Such that: \begin{align} P(Z=k) = \frac{1}{2} \cdot e^{- \lambda} \cdot \frac{\lambda ^k}{k!}? \end{align}
No, that is not correct. Because $Z\leq Y$, $$P(Z = k) = \sum_{n = k}^\infty P(Z = k \text{ and } Y = n) = \sum_{n = k}^\infty P(Z = k\mid Y = n) P(Y = n).$$ When $n$ people are hit there are $2^n$ total survival outcomes and $\binom{n}{k}$ ways for $k$ people to survive, so $P(Z = k\mid Y = n) = \binom{n}{k}/2^n$. This gives $$P(Z = k) = \sum_{n = k}^\infty e^{-\lambda} \binom{n}{k} \frac{\lambda^n}{2^n n!} = \frac{e^{-\lambda}}{k!}\biggl(\frac{\lambda}{2} \biggr)^k \sum_{n = k}^\infty \frac{\bigl(\frac{\lambda}{2}\bigr)^{n-k}}{(n - k)!} = \frac{e^{-\lambda/2}}{k!}\biggl(\frac{\lambda}{2} \biggr)^k,$$ which is just a long way of saying that $Z$ is Poisson with parameter $\lambda/2$.
