Probability of seeing $Y$ objects given there are $n$ objects.

Question

Suppose that I have $M$ objects, where $M$ is a Poisson(λ) random variable and reveal them to you briefly. For each of these objects, independently of what happens with the others, you observe it with probability $p$. Call the total number of objects that you observe $Y$.

I am asked to show, conditional on the event that $M = n$, $Y$ has the Binomial(n, p) mass function.

This seems obvious to me as it is the probability of seeing $Y$ objects out of $n$ objects with a probability of $p$. But to explicitly show it, I get confused as I'm not sure if I can work on the basis that $M$ and $Y$ are independent. My concern comes from the fact the maximum number we see is limited by $M$, therefore affects $Y$ in this way. But if they are independent then I can simply write out the conditional probably equation and cancel the probability that $M=n$.

Answer 1

It is indeed obvious.

Under condition $M=n$ there are $n$ objects and for each them the probability that I observe the object (a success) equals $p$. Further there is independence regarding the others objects ("...independently of what happens with the others").

That is enough for the conclusion that under $M=n$ the distribution of $Y$ is binomial with parameters $n$ and $p$.

Here $M$ and $Y$ are indeed not independent, as you noticed, but we come to the conclusion above without any appeal on this (absent) independence.

Answer 2

Let $X_i$ be the 0-1 random variable telling you whether you observe object $i$ (you observe $i$ if and only if $X_i=1$), for all $i=1,\ldots,M$. Then, $Y=X_1+X_2+\cdots+X_M$ and you're asked to show that $$ P(X_1+X_2+\cdots+X_M = k | M=n) = { n \choose k } p^k (1-p)^{n-k} $$ Notice that $$P(X_1+X_2+\cdots+X_M = k | M=n) = P(X_1+X_2+\cdots+X_n = k | M=n) = \frac{P(X_1+X_2+\cdots+X_n = k \cap M=n)}{P(M=n)}.$$ Since $X_1+X_2+\cdots+X_n$ has the desired binomial mass function, you can arrive to the desired conclusion if $X_1+X_2+\cdots+X_n$ and $M$ are independent, otherwise it may not be true.