Let $N_t$ be a Poisson process. Let $Y_i$ be i.i.d. normal RV's with mean $m$ and variance $b^2$. Let $F_t$ be the sigma algebra of information acquired by observing $N_t$. Compute: $$E\left[\exp\left(\sum_{i=1}^{N_t} Y_i\right) \mid F_s\right]$$
My attempt:
Firstly, I know how to compute $E[\exp(\sum_{i=1}^{N_t} Y_i)]$ since it is just the m.g.f. for a compound Poisson Process.
My problem arises when dealing with the conditional filtration. In particular, is it correct to write:
$$E\left[\exp\left(\sum_{i=1}^{N_t} Y_i\right) \mid F_s\right]=\sum_{k=1}^{\infty}E\left[\exp\left(\sum_{i=1}^{k} Y_i\right) \mid F_s,N_t = k\right]P(N_t=k)$$
How shall I deal with this conditional filtration here?
Let's just make some intuitive arguments and carry out the calculation. Seems to me that, at time $s$, we only need the observed $N_s$, i.e.
$E(e^{\sum_{i=1}^{N_t}Y_i}|\mathcal F_s) = E(e^{\sum_{i=1}^{N_t}Y_i}|N_s) $
The number of jumps between $s$ and $t$ $\sim$ $N_{t-s}$ is independent of what happened up to time $s$, so we have
$E(e^{\sum_{i=1}^{N_t}Y_i}|N_s) =E(e^{\sum_{i=1}^{N_s}Y_i}|N_s) E(e^{\sum_{i=1}^{N_{t-s}}Y_i}) $
Let $\alpha = E(e^{Y_i}) = e^{m+1/2b^2}$
we have
$E(e^{\sum_{i=1}^{N_{t-s}}Y_i}) = \sum_{k=1}^\infty \alpha^k P(N_{t-s}=k) = e^{-\lambda(1-\alpha)(t-s)}$
So we have
$E(e^{\sum_{i=1}^{N_t}Y_i}|\mathcal F_s) = e^{-\lambda(1-\alpha)(t-s)}\alpha^{N_s}$