A large lump of radioactive material has a long half life. Let $D(t)$ be the total number of decays which occur in the radioactive material in the period of $t$ hours starting at noon on a particular day. Suppose that $\{D(t) : t > 0 \}$ is a Poisson process of rate $\mu$.
At noon a counter is placed next to the radioactive material. Each decay is recorded by the counter with probability $p$ independently of whether other decays are recorded and independently of when decays occur. Let $N(t)$ be the total number of decays that have been recorded by the counter by time $t$.
Prove that $N(t)$ has a Poisson distribution with mean $\mu pt$.
Initially I thought I could use the following;
$$\mathbb{P}(N(t) = n) = \sum^\infty_{k = 0} \mathbb{P}(N(t) = n | D(t) = k)\mathbb{P}(D(t) = k)$$
And I thought $\mathbb{P}(N(t) = n | D(t) = k)$ is the probability of n recordings, and k-n non recordings which would equal $p^n(1-p)^{k-n}$, or zero if $k < n$.
I couldn't progress further, so I thought of another idea.
I think I would like to go about setting up a differential equation and showing that the pdf of a Poisson$(\mu pt)$ variable would satisfy such an equation. $$\mathbb{P}(N(t + h) = n) = \\ \mathbb{P}(N(t + h) = n | N(t) > n)\mathbb{P}(N(t)>n)\\ \mathbb{P}(N(t + h) = n | N(t) = n)\mathbb{P}(N(t)=n)\\ \mathbb{P}(N(t + h) = n | N(t) = n-1)\mathbb{P}(N(t)=n-1)\\ \mathbb{P}(N(t + h) = n | N(t) < n-1)\mathbb{P}(N(t)<n-1)$$ However, I think $\mathbb{P}(N(t + h) = n | N(t) = n)$ is the probability of no recordings in the time interval of h, and I am unable to work out this quantity either.
Edit: Prefer my original idea at the moment.
Any comments and help would be appreciated. Thanks.
This is called thinning of a Poisson Process. Suppose that $\Phi$ is the set of decay points in the real line following Poisson with intensity $\lambda$. Note that the expected value of $D(t)$ is equal to $\lambda t$.
The problem with your original idea, which I prefer too, is that the conditional distribution is indeed binomial distribution because you have to choose first $n$ decaying points out of total $k$ points. Following is the correct solution: $$ \Pr(N(t)=n)=\sum_{m=n}^\infty \Pr(N(t)=n\mid D(t)=m)\Pr(D(t)=m)\\ \sum_{m=n}^\infty \binom{m}n p^n(1-p)^{m-n}\frac{(\lambda t)^me^{-\lambda t}}{m!}\\ \sum_{m=n}^\infty \frac {m!}{n!(m-n)!} ({p}\lambda t)^n\frac{((1-p)\lambda t)^{m-n}e^{-\lambda t}}{m!}\\ =\frac {1}{n!} ({\lambda pt})^n e^{-\lambda p t}. $$