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Suppose $U, W, V, S$ are four independent normal random variables with mean $0$ and variance $1$. Let $X=W+U$, $Y=2W+S$, $Z=3W+V$. What is $f(X, Y, Z)$?

Thanks!

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  • $\begingroup$ What attempts have you made so far and what have the results been? For instance, have you figured out what kind of distribution $(X,Y,Z)$ must be? Have you computed its (multivariate) expectation? What about higher (multivariate) moments? Also, what does "f" stand for: a probability distribution function? Its pdf? Something else? $\endgroup$ – whuber Dec 12 '12 at 18:29
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A linear transformation of (independent) normals is always normal and fully determined by its mean and variance (resp. covariance matrix in the multivariate case).

The vector $(X, Y, Z)$ will be multivariate normal. To determine its mean and covariance matrix, you need only apply the rules for the variances and covariances of linear combinations of random variables—which you should have previously studied—combined with the information that $U$,$W$, $V$ and $S$ are independent (hence, the covariance between any pair is zero) with means 0 and variances 1.

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  • $\begingroup$ cardinal edited my original post by adding the word "independent". The original was correct, since affine transformations of multivariate normals remain normal. If $Y=AX+c$ and $X$ is $N(\mu, \Sigma)$, then $Y$ is normal and its variance matrix is $A \Sigma A'$. Check Mardia, Kent and Bibby on Multivariate Analysis. $\endgroup$ – Placidia Dec 12 '12 at 20:46
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    $\begingroup$ You did not say "multivariate" normal, so I added the "independent" in parentheses (i) to fit within the context of the question and (ii) to remove the possibility that someone would point out that, as stated, your claim was incorrect. :-) $\endgroup$ – cardinal Dec 12 '12 at 21:27

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