# Joint distribution of normally distributed random variables

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.

## 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 $$$$\begin{split} cov(U,V) &= E((U - E(U))(V-E(V)) \\ &= E(UV)\\ &= E(3X_1-X_2-2X_3+1)(X_1+X_2+X_3)\\ &= E(3X_1^2 + 3X_1X_2 + 3X_1X_3) \\ &- E(X_1X_2 + X_2^2 +X_1X_3) \\ &- E(2X_1X_3 + 2X_2X_3 + 2X_3^2)\\ &+ E(X_1+X_2+X_3) \end{split}$$$$ Notice that all terms $$E(X_iX_j) =E(X_i)E(X_j) = 0$$ due to independence, so $$$$\begin{split} cov(U,V) &= E(3X_1^2+X_2^2+2X_3^2+X_1+X_2+X_3) \\ &= 3E(X_1^2) - E(X_2^2) - 2E(X_3^2) + E(X_1) + E(X_2) + E(X_3)\\ &= 3 - 1 - 2 + 0 + 0 + 0\\ &= 0 \end{split}$$$$

• 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 Oct 1, 2018 at 21:02
• The minus signs before $X_2^2$ and $X_3^2$ got lost. Oct 1, 2018 at 22:26
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 $$\boldsymbol \mu' = A \boldsymbol \mu + \boldsymbol b, \,M' = A M A^t$$: $$\begin{pmatrix} 3 & -1 & -2 \\ 1 & 1 & 1 \\ 2 & -3 & 1 \end{pmatrix} \boldsymbol X + \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \sim \mathcal N \!\left( \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 14 & 0 & 7 \\ 0 & 3 & 0 \\ 7 & 0 & 14 \end{pmatrix} \right).$$
• $+1$ for providing another way for linear models. Oct 1, 2018 at 22:32