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Let we have $X$, $N$ dimensional vector of independent random variables. If we multiply this vector by some matrix $V$ with size $r\times N$, with property $V*V'=I$, where $I$ is identity matrix, and ' is transpose operation, we get set of $r$ independent random variables collected in vector $Y$. If $X$ variables are with Gaussian distribution, how can we find PDF of derived elements of vector $Y$? Is it still Gaussian?

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Recall that every affine function of a normal vector is normal, and that every vector with independent normal entries is normal.

In your case, the vector $X$ has independent normal entries hence $Y=VX$ is normal, with mean $E(Y)=VE(X)$ and covariance matrix $V\Sigma V'$, where $\Sigma$ is the diagonal matrix with diagonal entries the variances of the random variables $X_k$. There is no reason to expect that the random variables $Y_j$ are independent.

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  • $\begingroup$ But, if you want to find covariance matrix. $E(YY')=E(VX(VX)')=E(VXX'V')=E(VIV')=E(VV')=I$. They are uncorrelated, you said that their mutual distribution is normal, it means that they are independent. Maybe, i forgot to mention that they are independent with same variance. $\endgroup$ – EM22 Nov 19 '14 at 15:07
  • $\begingroup$ No, E(XX') is a multiple of identity only when the X_k are centered and independent and with the same variance. Assuming only that they are centered and independent one gets E(YY')=VΣV', where Σ is a diagonal matrix, as explained in the answer. $\endgroup$ – Did Nov 19 '14 at 18:41
  • $\begingroup$ Yes, you are right. In the first post I forgot to write "they are independent with same variance", what I wrote in previous comment. Thank you very much. $\endgroup$ – EM22 Nov 20 '14 at 9:08

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