There is this problem that I think I have solved. I need feedback if I have solved it correctly. I also have some questions regarding the intuition on the solution I have obtained.
Problem Statement: Radioactive decay
Assume that a radioactive sample emits a random number of $\alpha$ particles in any given hour, and that the number of $\alpha$ particles emitted in an hour is Poisson distributed with parameter $\lambda$. Suppose that a faulty Geiger-Muller counter is used to count these particle emissions. In particular, the faulty counter fails to register an emission with probability $p$, independently of other emissions.
(a) What is the probability that the faulty counter will register exactly $k$ emissions in an hour?
(b) Given that the faulty counter registered $k$ emissions in an hour, what is the PMF of the actual number of emissions that happened from the source during that hour?
My Attempt: So with a discrete random variable having Poisson distribution,
$$p_X(k) = \frac{e^{-\lambda}\lambda^{k}}{k!}$$
where $k \in \mathbb{N} \cup \{0\}$. In the given experiment, there are three random variables that are implicitly defined. Let $Y$ be the random variable which is the value that the counter registers after the first hour, let $Z$ be the random variable which denotes the number of failures in the counter during the hour, and let $X$ be the random variable which is the number of $\alpha$ particles emitted in one hour.
(a) We need to find $\mathbb{P}(Y=k)$. \begin{eqnarray*} \mathbb{P}(Y=k) &=& \sum_{n=0}^{\infty}\mathbb{P}(Z=n) \times \mathbb{P}(X=n+k) \\ &=& \sum_{n=0}^{\infty}{n+k \choose n} (1-p)^k p^n \frac{e^{-\lambda}\lambda^{n+k}}{(n+k)!} \\ &=& \frac{ \left \{ (1-p)\lambda \right \}^k e^{-\lambda(1-p)}}{k!} \end{eqnarray*}
This is as if the Poisson distribution has changed with a new constant of $(1-p)\lambda$.
(b) Here, we need $$\mathbb{P}\left \{ (X=n+k)|(Y=k) \right \}= \frac{\mathbb{P} \left \{ (X=n+k)\cap(Y=k) \right \}}{\mathbb{P}(Y=k)}$$ and we know that $$\mathbb{P} \left \{ (X=n+k)\cap(Y=k) \right \}=\mathbb{P}(Z=n) \times \mathbb{P}(X=n+k)$$ Hence we finally get $$\mathbb{P}\left \{ (X=n+k)|(Y=k) \right \} = e^{-\lambda p} \frac{(\lambda p)^n}{n!}$$ We can replace $n+k$ with $N$ to get $$\mathbb{P}\left \{ (X=N)|(Y=k) \right \} = e^{-\lambda p} \frac{(\lambda p)^{N-k}}{(N-k)!}$$ if $N \geq k$ and $0$ otherwise.
I wish to know the following.
- Is my solution correct?
- If my solution is correct, what is the intuitive explanation to the fact that if $p=0$, answer to (b) is $0$? Specifically, when $p=0$, then the counter is not making any mistakes. Why then is the conditional probability $0$?
- If my solution is correct, what is the intuitive explanation to the fact that if $p \to 1$, answer to (b) is a Poisson process with coefficient $\lambda$? Specifically, when $p \to 1$, the counter is surely making a mistake each time. Why then is the conditional probability Poisson distributed when we know that the counter errs each time?
Yes, that's correct.
The key observation here to makes sense of our intuition is that the observed decays and the unobserved decays are independent random variables.
Intuitively, as $p\to 0$, the detector becomes practically infallible, and we're virtually certain that the decays we observed are all of them. The conditional probability you get there in the limit isn't simply zero - it's the point mass $\delta(N-k)$, equal to $1$ if $N=k$ and zero otherwise.
As $p\to 1$, we lean on the independence. It doesn't matter how many decays we observed (although that isn't likely to be many) - the distribution of the unobserved decays always looks the same. Since, in the limit, all of the decays are unobserved, this approaches the Poisson distribution with parameter $\lambda$ that describes all of the decays.
The distribution of total decays given $k$ observed decays will be the appropriate Poisson distribution plus $k$. That's what the $N-k$ in your formulas is doing; it's shifting the argument. The answer to (b) is not a Poisson distribution but instead a translation of a Poisson distribution.