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Let $X_1, X_2, X_3 \sim N(0,1)$ be independent random variables. I needed to figure out the distribution of $U=3X_1-X_2-2X_3+1$, $V=X_1+X_2+X_3$ and $W=2X_1-3X_2+X_3$ which resulted in $U \sim N(1,1)$, $V \sim N(0,3), W \sim N(0,2)$. Is that correct?

Now I need to calculate the joint distribution of $(U,V,W)$ which I think is done by drawing up the covariance matrix.

But I am not sure how to calculate $E(UV)$ for example which I need vor $cov(U,V)$. Any hints or approaches are much appreciated.

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2 Answers 2

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Weighted Sum of Normal Distributions

Let $$Y = \alpha_1 X_1 + \ldots +\alpha_n X_n + \beta$$ where $X_k \sim N(0,1)$ then $Y$ is also normally distributed with mean $$E(Y) = E(\alpha_1 X_1 + \ldots +\alpha_n X_n) = \alpha_1 E(X_1) + \ldots +\alpha_n E(X_n) = 0 + \ldots + 0+ \beta = \beta$$ and $$\operatorname{var}(Y) = \operatorname{var}(\alpha_1 X_1 + \ldots +\alpha_n X_n+ \beta) = \alpha_1^2 \operatorname{var}(X_1) + \ldots +\alpha_n^2 \operatorname{var}(X_n)+0 = \alpha_1^2 + \ldots + \alpha_n^2$$ So your means are correct, but your variances are not. For example, $$\operatorname{var}W=2^2+3^2 +1^2=14$$ and the same applies for the rest.


Joint Distribution

The joint distribution of Normal distributions is a multi-variate normal with mean vector containing the means of each individual normal random variable, therefore you don't need $E(UV)$, you need $E(U), E(V),E(W)$, which, if placed in a vector, is $$\mu = [1, 0,0]$$

As for the covariance, you have $\operatorname{var}(U),\operatorname{var}(V)$ and $\operatorname{var}(W)$, which serve as the diagonal elements of the covariance matrix, now the off-diagonal element, which are $cov(U,V),cov(U,W),cov(V,W)$, you would have to compute them by definition, here's one example \begin{equation} \begin{split} cov(U,V) &= E((U - E(U))(V-E(V)) \\ &= E(UV)\\ &= E(3X_1-X_2-2X_3+1)(X_1+X_2+X_3)\\ &= E(3X_1^2 + 3X_1X_2 + 3X_1X_3) \\ &- E(X_1X_2 + X_2^2 +X_1X_3) \\ &- E(2X_1X_3 + 2X_2X_3 + 2X_3^2)\\ &+ E(X_1+X_2+X_3) \end{split} \end{equation} Notice that all terms $E(X_iX_j) =E(X_i)E(X_j) = 0$ due to independence, so \begin{equation} \begin{split} cov(U,V) &= E(3X_1^2+X_2^2+2X_3^2+X_1+X_2+X_3) \\ &= 3E(X_1^2) - E(X_2^2) - 2E(X_3^2) + E(X_1) + E(X_2) + E(X_3)\\ &= 3 - 1 - 2 + 0 + 0 + 0\\ &= 0 \end{split} \end{equation}

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  • $\begingroup$ helped my understanding a lot. I never used the definition of the covariance to actually calculate it, I always thought that $E(XY)-E(X)E(Y)$ is sufficient to know $\endgroup$
    – Tesla
    Oct 1, 2018 at 21:02
  • $\begingroup$ The minus signs before $X_2^2$ and $X_3^2$ got lost. $\endgroup$
    – Maxim
    Oct 1, 2018 at 22:26
  • $\begingroup$ @Tesla glad it was helpful. $\endgroup$ Oct 1, 2018 at 22:32
  • $\begingroup$ @Maxim thanks for the note .. i have edited. $\endgroup$ Oct 1, 2018 at 22:32
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You have $\boldsymbol X =(X_1, X_2, X_3)^t$ jointly normal with the mean vector $\boldsymbol \mu = 0$ and the covariance matrix $M = (\delta_{ij})$. $A \boldsymbol X + \boldsymbol b$ will also be jointly normal with $\boldsymbol \mu' = A \boldsymbol \mu + \boldsymbol b, \,M' = A M A^t$: $$\begin{pmatrix} 3 & -1 & -2 \\ 1 & 1 & 1 \\ 2 & -3 & 1 \end{pmatrix} \boldsymbol X + \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \sim \mathcal N \!\left( \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 14 & 0 & 7 \\ 0 & 3 & 0 \\ 7 & 0 & 14 \end{pmatrix} \right).$$

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  • $\begingroup$ $+1$ for providing another way for linear models. $\endgroup$ Oct 1, 2018 at 22:32

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