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How can I prove that the sum of $X_1, X_2, \ldots,X_n$ random variables, all of which have normal distributions $N(\mu_i, \sigma_i)$, is a random variable that is itself normally distributed with mean $$\mu =\sum_{i=1}^n \mu_i$$

and variance

$$\sigma^2 = \sum_{i=1}^n \sigma_i^2$$

Edit: I forgot to add that this was with the assumption that all $X_1, X_2,\ldots,X_n$ are independent.

  • 0
    As did said, in general your claim is wrong. But it holds if you additionally assume that the $X_i$ are independent. Note that it suffices to establish your claim for $n=2$ (you can use induction for the general case then).2012-11-16
  • 0
    Quite many proofs can be found in [Wikipedia - Sum of normally distributed random variables](http://en.wikipedia.org/wiki/Sum_of_normally_distributed_random_variables)2012-11-16
  • 0
    For the case $n=2$ (as suggested by @martini followed by induction), a proof without using moment-generating functions can be found [here](http://math.stackexchange.com/a/65871/15941) on this site.2012-11-16
  • 0
    One way is two explicitly compute the convolution of the density functions. See my answer below.2012-11-16

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