# $E[X_1+X_2+\cdots+X_n]=E[X_1]+E[X_2]+\cdots+E[X_n]$ Proof

I am trying to proof (from myself I have the case in my book for continuous random variable but want to find the proof for discrete random variables) that:

$$E[X_1+X_2+\cdots+X_n]=E[X_1]+E[X_2]+\cdots+E[X_n]$$

I came up with something but it seems to simple. I am only considering 2 random variables for the proof $X_1$ and $X_2$ and assume they have the same probability distribution (and that all probabilities are equal). Thus the generic definition for the expected value in this case is (where $N$ is the sample size):

$$E[X] = {\sum_{i=1} X_i \over N}$$

Now going back to the proof:

\begin{align}E[X_1 + X_2]&={\sum (X_{1i} + X_{2i}) \over N}\\[12pt] &= {\sum X_{1i} \over N }+{\sum X_{2i} \over N} \\[12pt] &=E[X_1] + E[X_2]\end{align} This seems to be too simple. Would that also mean this is only true if $X_1$ and $X_2$ have the same probability distribution?

Thank you.

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Why would it have to be complicated? Also it( the formula) doesn't care about the distribution, which makes it really useful –  Jean-Sébastien Sep 14 '13 at 23:04
But if the distribution of $X_1$ and $X_2$ are not the same, then how can I write that $E[X_1 + X_2] = {{\sum(X_{1i}+X_{2i})}\over N}$? Do I need to write: $E[X_1 + X_2] = \sum(p_{1i}X_{1i}+p_{2i}X_{2i})$ to be more generic? Does it mean my reasoning is correct though? Thank you. –  Marc Ourens Sep 14 '13 at 23:08

$$E[X] = \sum_x x \Pr(X=x).$$ This works if the random variable $X$ has a discrete distributions. For other distributions, one needs a more general formula.
If there are just finitely many possible values and the all have the same probability (so it's a discrete uniform distribution) then you can say $$E[X] = \sum_{i=1}^N x_i \frac1N$$ where $N$ is the number of possible values, and this is then the same as $$\frac{\sum_{i=1}^N x_i}{N}.$$ This works only for discrete uniform distributions.
Thank you Michael. So as I mentioned in a previous comment, if I want to be more generic and remove the condition that X has a discrete uniform distribution, is it better (and correct) to write: $E[X_1 + X_2] = \sum(p_{1i}X_{1i}+p_{2i}X_{2i})$ to be more generic? Thank you. –  Marc Ourens Sep 14 '13 at 23:13