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let's say I have two random variables, both have a mean of 0, one has a variance of 2, the other has a variance of 3. How can you determine the distribution of their sum?

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You can't (without a lot more information). – cardinal Nov 13 '11 at 23:35
You seem to be assuming that the mean and variance determine the distribution, or perhaps you're implicitly thinking of normally distributed variables. All you can determine from the mean and variance of the variables is the mean and variance of the sum. – joriki Nov 13 '11 at 23:37
Two examples: (1) Let $X = 2$ with probability $1/2$ and $= -2$ with probability $1/2$. Let $Y = 3X / 2$. Then $X + Y$ is equal to $5$ with probability $1/2$ and $-5$ with probability $1/2$. (2) Let $X$ be as before and let $Y$ be independent of $X$ such that $Y = \pm 3$ with probability $1/2$ each. Then $X+Y \in \{-5,-1,1,5\}$, taking on each possible value with probability $1/4$. – cardinal Nov 13 '11 at 23:40
@joriki: Minor note (maybe to give you enough time to make a quick edit): The mean and variance are not in general enough to determine the variance of the sum. – cardinal Nov 13 '11 at 23:41
@cardinal: Ha -- I was myself implicitly thinking of independent normally distributed variables :-) Yes, the quick edit thing almost worked -- I was in time to open the comment for editing, but not to save the change :-) – joriki Nov 13 '11 at 23:46
up vote 1 down vote accepted

All you can say is that the mean of the sum is the sum of the means, i.e. $0$, and that the variance of the sum is between $5-\sqrt{24}$ and $5+\sqrt{24}$ (about 0.1 and 9.9).

If the two random variables are independent or at least uncorrelated then the variance of the sum is the sum of the variances, i.e. $5$.

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