Conditional probability given only an average

I’ve been working on this question, which I found on physics.SE. Unfortunately it was closed because it’s a homework question, but I’d like to get more of a hint than the original poster got.

I know that I’m supposed to use Bayes’ Theorem, but I don’t see how I’m supposed to use the fact that $$\bar{N}$$ is known. Every effort so far has yielded an unwieldy fraction that can’t be simplified.

Edit: Apparently I'm supposed to copy/paste the question. Here it is:

This is one of the exercises of Barnett's book on quantum information.

A particle counter records counts with an efficiency $$\eta$$. This means that each particle is detected with probability $$\eta$$ and missed with probability $$1-\eta$$. Let $$N$$ be the number of particles present and $$n$$ be the number of detected. Then: $$$$P(n|N)=\frac{N!}{n!(N-n)!}\eta^n (1-\eta)^{N-n}$$$$

I know the mean number of particle present is:

$$$$\bar{N}=\sum N P(N)$$$$

I want to calculate $$P(N|n)$$. I'm stuck here by a while, so I do not know how to proceed.

Edit 2: I'm adding a screenshot of the question in question:

• Please don't link a closed question. Copy the relevant (only math) parts here. Commented Feb 3, 2019 at 2:28
• @leonbloy I didn't know that was a rule, sorry. I've copied it now.
– NNN
Commented Feb 3, 2019 at 4:43
• "I want to calculate 𝑃(𝑁|𝑛)" To do so, you are missing some information, for example the (unconditional) distribution of N would do.
– Did
Commented Feb 3, 2019 at 9:19

1 Answer

You surely know that

$$P(N|n) = \frac{P(n|N) P(N)}{P(n)}=\frac{P(n|N) P(N)}{\sum_N P(n|N) P(N)}$$

This tells you that you $$P(n|N)$$ is not enough, you need $$P(N)$$. To understand this should be your starting point.

Now, here, you only give us the mean of $$N$$ (actually you wrote the general formula of the expected value, but I guess you meant that you know $$\bar{N}$$). Still not enough.

Perhaps there is some implicit statistical model for $$N$$ that you've missed? Perhaps (just guessing) it follows a Poisson distribution? If so, you are done, because the Poisson distribution depends on a single parameter (which is the mean), hence you indeed can write $$P(N)$$.

• I've attached a screenshot of the exact question from the textbook. Unfortunately, doing this same calculation for a Poisson distribution was part (a) of the question. Do you have any other thoughts, given this?
– NNN
Commented Feb 3, 2019 at 15:31
• No, I don't see how one could do with the mean alone. Commented Feb 4, 2019 at 15:10