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If $X_1$ is $N(0,1)$ and $X_2 = -X_1$ if $-1 \leq X_1 \leq 1$ or $X_2 = X_1$ otherwise, how do you prove that $X_1$ and $X_2$ do not have a bivariate normal distribution?

Attempt: Using the Jacobian method I proved that the distribution of $X_2$ is the standard normal distribution. To prove that $X_1$ and $X_2$ do not have a bivariate normal distribution there is a hint that says to consider the linear combination $X_1-X_2$ where $P(X_1-X_2) = P(|X_1|>1) = 0.3174$.

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up vote 3 down vote accepted

If $X = (X_1, X_2)$ were bivariate normal, each linear combination $\alpha_1 X_1 + \alpha_2 X_2$ would be normally distributed. But $X_1 - X_2$ has \[ 0 < P(X_1 - X_2 = 0) = P(|X_1| > 1) < 1 \] so cannot be normally distributed as a non-degenerate normal distributed variable $Y$ has $P(Y = y) = 0$ for all $y \in \mathbb R$ and a degenerate one has $P(Y = y) \in \{0,1\}$ for all $y \in \mathbb R$.

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since this is continuous the probability of one single value is 0 but in this case it is not – lord12 Oct 10 '12 at 13:42

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