Trying to understand better $\mathbb E[Y\mid N]$. Let $N$ a r.v. (let say that $\mathbb P\{N\in\mathbb N\}=1.$) Let $X_1,X_2,...$ iid random variables and we set $\mathbb E[X_i]=\mu$. Set $$Y=X_1+X_2+...+X_N,$$
Does $\mathbb E[Y]$ has a sense ? Because it looks to be $$\mathbb E[Y]=\sum_{i=1}^N \mathbb E[X_i]=\mu N,$$ and thus it looks to be a r.v. and not a number. And what would be $\mathbb E[Y\mid N]$ here ? When we write $\mathbb E[Y,\mid N]$ do we consider $N$ a fixed element in $Y$ or not ?
 A: Here, $\mathbb E [ Y \vert N ]$ is simply $\mu N$. Intuitively speaking, this is what you expect the value of $Y$ to be on average, given that you know what $N$ is.
Note that you don't really need this to calculate $\mathbb E[Y]$, since
\begin{align*}
\mathbb E[Y] &= \mathbb E\left [\sum_{i=1}^N X_i \right] \\
&= \sum_{n\in \mathbb N}\mathbb E\left[\left.\sum_{i=1}^N X_i \right\vert N=n \right] \mathbb P (N=n)\\
&= \sum_{n\in \mathbb N}\mathbb E\left[\sum_{i=1}^n X_i \right] \mathbb P (N=n) \\
&= \sum_{n\in \mathbb N}n\mu \mathbb P (N=n) \\
&= \mu \mathbb E[N].
\end{align*}
Note that in the third equality, I implicitly make use of the assumption that $N$ is independent of the $X_i$'s. Otherwise, this calculation is not valid.
However, if you'd like to see how to calculate $\mathbb E [ Y \vert N ]$, you can see my answer here.
A: The problem is that the sumation is itself random, so you can't compute $\mathbb E[Y]$ without any information on $N$. In your calculation you suppose $N$ fixed (but unfortunately it's not). In fact, $\mu N$ is precisely $\mathbb E[Y\mid N]$. Therfore, 
$$\mathbb E[Y]=\mathbb E[\mathbb E[Y\mid N]]=\mathbb E[\mu N]=\mu\mathbb E[N].$$
Indeed, when you conditionate, it's if you suppose that $N$ is fixed (in a certain way).
