A specific question in conditional expectation with mixed discrete and continuous random variables

Question

In my probability class I have just met this seemingly difficult question:

Let $ \{X_n\}_{n=1}^{\infty}, \{Z_n\}_{n=1}^{\infty} $ be two i.i.d sequences of random variables such that we know $ X_n \sim \mathcal{N} (0,\sigma ^2) $ and $ Z_n \sim \mathcal{N} (0,1) $ and we define a new sequence of random variables like so: $ Y_n = X_N-X_n+Z_n $ where $ N \sim POIS(\Lambda) $ statistically independent from $ X_n,Z_n $. The question has three parts:

1. We are asked to show that $ X_N,N $ are statistically independent. Done I did this

2. We are asked to find the conditional probability $ q_{m,n} = P\{ N=m | Y_n=y \} $ (for fixed deterministic m,n) where we are asked to distinguish between whether or not $ m=n $. I am stuck here as there is a hybrid of a discrete and a continuous random variable.

3. We are to asked to find the conditional expectation $ E[X_n | Y_n] $ (for fixed deterministic n) and we are asked to express the result in terms of $ q_{m,n}. Really stuck here, no idea.

I realize this might seem like a tough question but I am still a novice in probability theory and on the last two parts where I got stuck I cannot seem to find a solution and also the third part relies on the second where I am stuck so they are connected. I truly need and appreciate the help. Thanks all.

Answer 1

2. Assuming that the P here means conditional density and not conditional probability (which would be the only definition that makes sense, since Y=y with probability 0), we have $P(N=m|Y=y)=\frac{P(N=m \cap Y=y)}{f_Y(y)}=\frac{P(N=m)f_Y(y)}{ f_Y (y)}=P(N=m)$ (since by part 1 N and $X_N$ are independent, and obviously N is independent of the other remaining terms of $Y_n$). Hence the conditional density doesn't vary from that of the Poisson distribution (we can rewrite its probability mass function as a probability density which is a step function).

3. $X_n = X_N + Z_n - Y_n \implies E[X_n|Y_n]=E[X_N + Z_n -Y_n|Y_n] = E[X_N|Y_n] + E[Z_n|Y_n]-E[Y_n|Y_n] = \sum_{m=0}^{\infty} q_{n,m} + E[Z_n|Y_n] + Y_n$ because of the independence relations from part 1. I'm not sure how to proceed further at the moment.