In a probability example, there are these random variables: $X_i, N_t,$ and $Z_t=E(\sum\limits_{i=1}^{N_t} X_i | N_t)$, where $S(N_t)=\mathbb{N}_0$ and $X_i$ is a continuous random variable.
I have proved that $E(\sum\limits_{i=1}^{N_t} X_i | N_t=n)=\frac{nt}{2}$
Furthuremore, we have $E(N_t)=λt$
Now, I want to calculate $E(Z_t)$. Firstly, what is the most accurate way to calculate it? Secondly, I have calculated $E(Z_t)$ like this:
$E(Z_t)=\sum\nolimits_z zp(Z_t=z)=\sum\limits_{n=1}^{\infty} \frac{nt}{2}p(Z_t=\frac{nt}{2})=\frac{t}{2}\sum\limits_{n=1}^{\infty} np(Z_t=\frac{nt}{2})$
Here I use something that I'm not so sure. And that is $p(Z_t=\frac{nt}{2})=p(N_t=n)$. Is it true? If yes, how can I prove it? If it holds we have
$\frac{t}{2}\sum\limits_{n=1}^{\infty} np(Z_t=\frac{nt}{2})=\frac{t}{2}\sum\limits_{n=1}^{\infty} np(N_t=n)=\frac{t}{2}E(N_t)=\frac{t}{2}(λt)$
Thirdly, I think every conditional expectation like $E(A|B)$, assumes the value corresponding to $B=n$ with the probability $p(B=n)$. Is this correct?
Just answering your first question, "the most accurate way to calculate it" is using $\mathbb E(\mathbb E(X|Y))=\mathbb E (X)$. So we get $E(Z_t)=E\left(E\left(\sum_{i=1}^{N_t}X_t\big|N_t\right)\right)=E\left(\frac{N_tt}{2}\right)=\frac{\lambda t^2}{2}$.