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Let $\mathbf{Y} = (Y_1, Y_2)'$ where $\mathbf{Y} \sim N(\mu, \Sigma)$. Specifically, $\mathbf{Y}$ is a multivariate normal random variable: \begin{align*} \begin{bmatrix} Y_1 \\ Y_2 \end{bmatrix} \sim N\left(\begin{bmatrix} \mu_1 \\ \mu_2 \end{bmatrix}, \begin{bmatrix} \sigma_1^2 & cov(Y_1, Y_2) \\ cov(Y_1, Y_2) & \sigma_2^2 \end{bmatrix} \right). \end{align*} Let $B = \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}$. Show that for $\mathbf{Z} = B\mathbf{Y} = (Z_1, Z_2)'$ that $Z_1$ and $Z_2$ are univariate normal random variables but their joint distribution is not bivariate normal. Explain why.

My question: I am able to show that $Z_1$ and $Z_2$ are univariate normal (which follows simply by the definition of a multivariate normal random, i.e., $a_1Y_1 + a_2Y_2$ is univariate normal for all real numbers $a_1$ and $a_2$), but how do I show it is not bivariate normal?

I know this theorem that if $B$ has full row rank then $Z \sim N(B\mu, B\Sigma B')$, but here, clearly $B$ is singular, so it does not have full rank, but does this "prove" that $Z_1$ and $Z_2$ are not jointly bivariate normal? I am not convinced since just because the premises of the theorem does not hold, it does not necessarily mean the conclusion is false. So, how can I actually prove this?

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  • $\begingroup$ If $B$ is not full rank, then $\boldsymbol Z$ can't really be said to be "bivariate" any more than one can say that $(X, -X)$ is "bivariate." What you are looking at is just a univariate normal distribution. $\endgroup$ – heropup Oct 18 '17 at 21:13
  • $\begingroup$ Ah, that makes sense, thank you. $\endgroup$ – user40333 Oct 19 '17 at 3:17

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