Suppose that particles are emitted by a radioactive source and that the number of particles emitted over a period of one hour is given by a Poisson distribution with parameter λ. Admit that a metering device, which registers these emissions, occasionally fails to register an emitted particle. Specifically, suppose any emitted particle has a probability p of being registered. a) If Y is defined as the number of registered particles, what is the expression for the probability distribution of Y?
My approach to this question was the standard way to handle functions of two random variables, by making $Y=N-R$, where $N$ is the number of particles emitted and $R$ is the number of particles registered, then finding an auxiliary function $Y'$ to form the joint distribution $p(y, y')$, calculating the determinant of the Jacobian $J(n,r)$ and then finding the following distribution:
$p_{Y,Y'} = \frac{1}{|J(n,r)|}p_{N,R}(G_1(y,y'), G_2(y, y'))$
where $y = H_{1}(n, r) $, $y' = H_{2}(n, r), $ $n = G_{1}(y, y')$ and $r = G_{2}(y, y')$.
And then, to find $p_Y$, I sum over all possible values of $Y'$.
What I want to know is if this is correct. Since N and R are given by a Poisson and a Binomial, it got kind of complicated to sum over all values of $Y'$ in the joint distribution $p_{N,R}$. I'm wondering if this approach is the only one (and correct) or if there is a simpler way to handle this problem.
This seems overly complicated, but maybe I am misunderstanding what you're trying to do.
The way I see it, you should marginalize over the number of emitted particles, since the number of registered particles $R$ conditioned on the number of emitted particles $N=n$ is distributed as Bin($n,p$), while $N$ itself is Poisson($\lambda$). Thus by the total probability formula,
$$P(R=r)=\sum_{i=r}^\infty f_{\mathrm{Bin}}(r;i,p) f_{\mathrm{Poisson}}(i;\lambda)$$
where $f_{\mathrm{Bin}}$ is the binomial PMF and $f_{\mathrm{Poisson}}$ is the Poisson PMF. This can actually be simplified:
$$P(R=r)=\sum_{i=r}^\infty {i \choose r} p^r q^{i-r} e^{-\lambda} \frac{\lambda^i}{i!} \\ = \frac{e^{-\lambda} (p\lambda)^r}{r!} \sum_{i=r}^\infty \frac{(q\lambda)^{i-r}}{(i-r)!} \\ = \frac{e^{-p\lambda} (p\lambda)^r}{r!}.$$
Thus you find that you have just the Poisson($p\lambda$) PMF, which is nice and intuitive.