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Assume that $w$ is a $M$ dimensional random vector, such that: $w \sim N(w|0,\alpha^{-1} I)$. Now I have a $N \times M$ matrix $\Phi$, which is not random. I want show that the vector $Y= \Phi w$ is another Gaussian vector. What I tried is the following: Each component of $w_i$ is a scalar Gaussian, with zero mean and variance $1/\alpha$. Moreover, these components are independent due to the form of the covariance matrix of the joint distribution. I tried to form the distribution of the following linear combination of deterministic vectors: $$ Y = \sum_{m=1}^{M} w_m\Phi_m$$. $\Phi_m$ is the $m.$ column of the matrix $\Phi$. By using this approach, I can show that each component of the vector $Y_n$ is a Gaussian, but I cannot show that they are jointly Gaussian as well. What should I do instead here?

Edit: Note that I just want to show $Y$ is another Gaussian; I am not after showing its mean and covariance, which I can find indeed.

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  • $\begingroup$ Multiplying by a matrix is a linear transformation, so this is duplicate of math.stackexchange.com/questions/332441/… $\endgroup$ – kjetil b halvorsen May 13 '15 at 16:56
  • $\begingroup$ I want to see that the new distribution is Gaussian as well. I do not ask how to calculate the mean and covariance. I can do them indeed. $\endgroup$ – Ufuk Can Bicici May 13 '15 at 17:04
  • $\begingroup$ Look at the answer to the inked question. $\endgroup$ – kjetil b halvorsen May 13 '15 at 17:07
  • $\begingroup$ Yes, I looked at that. It just packages the mean and covariance terms into a Gaussian pdf at the end. It does not show how the density $f_y(.)$ obtains its Gaussian form. $\endgroup$ – Ufuk Can Bicici May 13 '15 at 17:10
  • $\begingroup$ The simplest way to prove it would be via calculating the moment-generating function. $\endgroup$ – kjetil b halvorsen May 13 '15 at 17:41

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