Let $X_1,X_2,...$ be independent random variables each with expectation $\mu$ and $N$ a random variable that..

Question

Let $X_1,X_2,...$ be independent random variables each with expectation $\mu$ and $N$ a random variable that..... takes the value of the set of natural numbers. $N$ and $X_i$ are independent random variables $\forall i=1,2...$. Prove that:

$$E(\sum_{i=1}^{N}X_i)=\mu EN$$

• I will put a star above the equalities that are problematic

$$E(\sum_{i=1}^{N}X_i)=EE(\sum_{i=1}^{N}X_i|N)=^{\{*1\}}\sum_{n=1}^{\infty}E(\sum_{i=1}^{N}X_i|N=n)P\{N=n\}=\sum_{n=1}^{\infty}E(\sum_{i=1}^{n}X_i|N=n)P\{N=n\}=^{\{*2\}}\sum_{n=1}^{\infty}E(\sum_{i=1}^{n}X_i)P\{N=n\}=...$$ the rest is easy, Im not too sure about these two inequalities. Thoughts?

Answer 1

Starting with the second one.$\sum_{i}^{n}X_i$ and $N$ are independent, therefore the conditional expectation and expectation are equal. Generally speaking $E(X|Y) = E(X)$ if X and Y are independent.

And regarding the first one $Y=E(\sum_{i}^{N}X_i|N)$ where $Y$ is N-mesaurable and N is discrete, we can write $E(Y)$ as sum of Y conditioned to {N=n} times the probability of this event. So $E(Y) = \sum_{n=1}^{\infty} E(Y|N=n)P(N=n)$ .

Answer 2

First off, you don't need to write the expression right before $\{*1\}$. So directly from the first expression to the expression immediately after $\{*1\}$, this is due to conditional expectation.

The equality $\{*2\}$ is true because $\sum_{i=1}^{n}X_i$ is independent of $N$.