Conditional expectaction with probabilities for a sum of independent random variable

Question

I have a r.v $S_N$ built as a sum of Bernoulli with parameter $p$. So $S_N = X_1 + X_2 + \ldots + X_N$. There is a second variable N, such that $N \sim Poisson(\lambda) $.

I have to compute:

1. $P(S_N=0)$
2. $\mathop{\mathbb{E}}(S_N \ | \ N = 4 )$
3. $\mathop{\mathbb{E}}(S_N \ | \ N )$

Now, for the first point, I need to think my sum of Bernoulli as a Binomial. Thus:

$$ P(S_N = 0) = \binom{n}{0} p^k (1-p)^{n-k} = (1-p)^n $$

However, i'm stuck with the two expectations. But i remember that for the conditional expectaction for two discrete random variables it holds:

$$ \mathop{\mathbb{E}}(X \ | \ Y = k ) = \sum_x x \ f_{X|Y}(x|y) $$

Answer 1

In Problem 2, $N=4$ is a constant. Thus, $S_N$ is a binomial variable with the mean equals $4p$ (assuming the Bernoulli takes value either 1 or 0).

For Problem 3: In general for the binomial distribution with $N=n$, the mean is $np$. Now for $N$ is a random variable, the mean then is $E[np]=\bar{n}p$. In particular when $N$ is of Poisson distribution, $\bar{n}=\lambda$, thus the mean is $\lambda p$.