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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:


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.

[to be continued]

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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
@MichaelHardy This doesn't answer the question (so far!) I think what Marc is looking for is a proof that linearity of expectation applies for arbitrary random variables, each from a different distribution. It is easy to show that it applies for random variables which have jointly continuous distributions, I think he wants the general case. – Bitrex Sep 14 '13 at 23:27
@Bitrex yes you are right. Thank you very much. – Marc Ourens Sep 14 '13 at 23:29
Michael, Bitrex, could anyone help me with this please. It would be great as Bitrex said to have a formal and generic answer to this. It would make me feel better as right now I am really unsatified with how much I understand of this problem (and I have no one to ask the question to). Thank you. – Marc Ourens Sep 15 '13 at 12:47

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