If we define 2 independent variables $Y_1$ and $Y_2$ as follows:
\begin{align}
Y_1 &= (Y_{11},Y_{12},Y_{13})^T \sim\mathcal N_3(\mu_1,\Sigma_{11}), \\
Y_2 &= (Y_{21},Y_{22})^T \sim\mathcal N_2(\mu_2,\Sigma_{22})
\end{align}
where,
\begin{align}
\mu_1 &= (2, 2, 2)^T &\Sigma_{11} &=
\left[\begin{array}{ccc} 3 &1 &0 \\ 1 &2 &0 \\ 0 &0 &3 \end{array}\right] \\
\mu_2 &= (3, 4)^T &\Sigma_{22} &=
\left[\begin{array}{cc} 4 &2 \\ 2 &4 \end{array}\right]
\end{align}
Then how can I find the joint distribution of $Y_{11}-Y_{13}+Y_{22}$ and $Y_{21}-Y_{12}$?
I know its a simple question but I could find if it was asked for $Y_1-Y_2$ or something. How am I supposed to solve it when it is like that?
Best Answer
Assuming that the word
independentin the opening statement is used in the way that probabilists use the word and not in the sense of independent versus dependent variable as is common in regression analysis, the joint distribution of the five random variables $Y_{11}, Y_{12}, Y_{13}, Y_{21},Y_{22}$ is the product of the joint distributions of $Y_{11}, Y_{12}, Y_{13}$, and $Y_{21},Y_{22}$, both of which are multivariate normal. This $5$-variate joint distributions is also a multivariate normal distribution in which the mean vector is just the concatenation $(\mu_1, \mu_2)^T$ of the two mean vectors and the covariance matrix is $$\Sigma = \left[\begin{matrix}\Sigma_{11} & 0\\0 & \Sigma_{22}\end{matrix}\right].$$ Thus, the joint distribution of $Y_{11}-Y_{13}+Y_{22}$ and $Y_{21}-Y_{12}$ is a bivariate normal distribution which can be found by the standard methods involving setting up a linear transformation mapping $(Y_{11}, Y_{12}, Y_{13}, Y_{21},Y_{22})$ to $Y_{11}-Y_{13}+Y_{22},Y_{21}-Y_{12})$ and doing matrix calculations. More simply, the means and variances of $Y_{11}-Y_{13}+Y_{22}$ and $Y_{21}-Y_{12}$ as well as their covariance can be computed more directly and used in writing down the mean vector and covariance matrix of this bivariate normal distribution.