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I am looking for a proof for convolution of two multivariate Gaussians (where each Gaussian has multi-dimensional mean and co-variance). I found a proof in here: http://www.tina-vision.net/docs/memos/2003-003.pdf where it provides a proof for convolution of two uni-variate Gaussians and also it provides the mean and variance of the convolved PDF. Now I am looking forward to find a proof for (or extend it to) multivariate Gaussian PDFs.

Does anybody know any solution already available, or can help me with it?

Thank you.

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  • $\begingroup$ I need to add that I am looking for the mean and co-variance of the convolved PDF. So, maybe knowing the whole proof is not necessary for me. $\endgroup$ – PickleRick Oct 9 '15 at 8:49
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Perhaps the easiest way to understand convolution, in the context of probability distributions, is in terms of the sum of independent random variables. Suppose that independent random variables $X_1$ and $X_2$ have distributions $d_1$ and $d_2$ respectively. Then $X_1+X_2$ has distribution given by the convolution of $d_1$ and $d_2$. For more details on this, see for example these notes (they only deal with the univariate case but the same concepts apply equally in multivariate situations.)

In the case of two multivariate Gaussians, it is well known (e.g. by considering characteristic functions) that the sum of independent $X\sim\mathcal N(\mu_X,\Sigma_X)$ and $Y\sim\mathcal N(\mu_Y,\Sigma_Y)$ is just $X+Y\sim\mathcal N(\mu_X+\mu_Y,\Sigma_X+\Sigma_Y)$. So all you need to do is add the mean vectors and covariances matrices.

An alternative (more direct but less illuminating) approach can be found in this document.

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  • $\begingroup$ Thank you S. Catterall, Your answer was very helpful and explanatory. What I am looking for is to find the parameters of a PDF which is consist of convolution of two PDFs. I am not sure exactly what it means from their random variables point of view. If d1 is a PDF with mu1 and Sigma1 and d2 is another PDF with mu2 and Sigma2, what would be the parameters of d1 * d2, where * is the convolution? $\endgroup$ – PickleRick Oct 25 '15 at 17:12
  • $\begingroup$ Hi, if the PDFs are both Gaussian then the parameters of the convolution would be mean=mu1+mu2 and variance=Variance1+Variance2, so if your Sigma1 and Sigma2 are standard deviations then the convolution standard deviation would be the square root of Sigma1^2+Sigma2^2. $\endgroup$ – S. Catterall Oct 25 '15 at 18:25
  • $\begingroup$ But why for the univariate case (you can find the proof in the link I proided) its Sigma is so different? the SD of two convolved Gaussian PDFs is (Sigma1^2+Sigma2^2)^1/2 and I expected for multivariate case the sigma becomes something similar. $\endgroup$ – PickleRick Oct 27 '15 at 6:01
  • $\begingroup$ I'm not sure exactly what you mean. The multivariate case looks very similar to the univariate case. Section 4 of the document you linked to proves, in the univariate case, that Variance (of the convolution)=Variance1+Variance2. In the multivariate case, this becomes Covariance matrix (of the convolution)=Covariance1+Covariance2, with the covariance matrix being the multivariate analogue of the variance in the univariate case. $\endgroup$ – S. Catterall Oct 28 '15 at 22:26
  • $\begingroup$ Sorry, for some reason I assumed in my comment on 25th October that you were asking about the univariate case. For the multivariate case, you simply add the means and the covariance matrices. $\endgroup$ – S. Catterall Oct 28 '15 at 22:35

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