Given two multivariate gaussian (say in 2D with mean $\mu$ as a 2D point and convariance marix $\Sigma$ as $2$x$2$ Matrix) $N_1(\mu_1,\Sigma_1)$ and $N_2(\mu_2,\Sigma_1)$, I would like to derive the pdf of $N_1+N_2$.

Can any one point me to the reference where i can find the pdf derivation of $N_1 + N_2$.

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The two multivariate Gaussian distributions seem to have different means but the same covariance matrix (or is that a typo?). Are they independent of each other? –  Henry Apr 22 '11 at 18:38
@whuber: The joint distribution of two separate multivariate Gaussian vectors is not necessarily distributed as a multivariate Gaussian. –  cardinal Apr 22 '11 at 20:07
@cardinal Good point. We have to assume either the joint distribution is Gaussian or else that the two variables are independent. One might, with some generosity and suitable caution, make that latter assumption for the OP. –  whuber Apr 22 '11 at 20:19
@whuber: Yes, I probably tend to be (overly) pedantic about such matters. On the other hand, I think that it can be instructive to draw attention to such matters since it can help the OP understand the boundaries of their own question. Thanks for your patience with me in this regard. Also, your second statement (independence) is just a subset of the first (multivariate Gaussianity), of course. –  cardinal Apr 22 '11 at 20:31
@Cardinal You are a mathematician, I see :-). I completely agree that it's worthwhile to draw attention to implicit assumptions and boundaries, because that heads off potential misuse and abuse. It's hard to know where to draw the line, too: when too many assumptions are left unsaid, some prodding is required to make sure we're answering the right question. On the other hand, we're not the math.se site (where rigor is the watchword and heaven help the poor person with a slightly flawed question) and, as practitioners, we tend to want to help questioners. So: perhaps "friendly pedantry"? –  whuber Apr 22 '11 at 20:39

## Method 1: characteristic functions

Referring to (say) the Wikipedia article on the multivariate normal distribution and using the 1D technique to compute sums in the article on sums of normal distributions, we find the log of its characteristic function is

$$i t \mu - t' \Sigma t.$$

The cf of a sum is the product of the cfs, so the logarithms add. This tells us the cf of the sum of two independent MVN distributions (indexed by 1 and 2) has a logarithm equal to

$$i t (\mu_1 + \mu_2) - t' (\Sigma_1 + \Sigma_2) t.$$

Because the cf uniquely determines the distribution we can immediately read off that the sum is MVN with mean $\mu_1 + \mu_2$ and variance $\Sigma_1 + \Sigma_2$.

## Method 2: Linear combinations

View the pair of MVN distributions as being a single MVN with mean $(\mu_1, \mu_2)$ and covariance $\Sigma_1 \oplus \Sigma_2$. The sum is given by a linear transformation and therefore is MVN. (See p. 2 #4 here.) The covariance again works out to $\Sigma_1 + \Sigma_2$.

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This assumes independence; it is not clear from the question whether this is the case. –  F. Tusell Apr 22 '11 at 20:18
@F That is true (and you are in good company with @cardinal): I explicitly make this assumption and even italicize it for emphasis. Which approach is constructive: to withhold or deprecate replies because the OP has not explicitly named all necessary assumptions; or to point out the natural, conventional assumptions needed for a useful answer and to provide that answer? –  whuber Apr 22 '11 at 20:24
(+1) Might be worth clarifying the meaning of $\Sigma_1 \oplus \Sigma_2$. –  cardinal Apr 22 '11 at 21:58
@Cardinal Thanks. I didn't feel like TeXing it out... :-) –  whuber Apr 22 '11 at 21:59
@F Your reminder that we don't know the underlying problem is well taken. By "natural" I meant that typically when people refer to two distributions as separate entities, as in the statement of this question, they take them to be independent. It is wise to be mindful that this is usually an important assumption: that's how I read the comments you and @cardinal have kindly provided. –  whuber Apr 23 '11 at 16:08