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If $X \sim N(0,1)$, does $Y=(X,X)^\prime$ have a bivariate normal distribution? If so (actually also if not), what is the joint density?

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  • $\begingroup$ @Glen_b I'm very sorry for my poor explanatory skills. By (X,X)' I mean a column vector that has two rows and one column (I tried to express this using the ' to mean transpose the row vector). I'm wondering about the corner case of a bivariate normal (although this is my question, is it technically still a "bivariate normal"?) where there is perfect correlation between the elements. In this case, since X=X, there is perfect correlation. I know what the density of a bivariate normal is when there is not perfect correlation, but when there is perfect correlation I'm curious if the density exists $\endgroup$
    – Xu Wang
    Oct 17, 2016 at 19:57
  • $\begingroup$ Sorry to have initially misunderstood. I have made some small edits which make it clearer what you meant. It was actually the "multivariate" that threw me off, since I assumed X must have been a vector. $\endgroup$
    – Glen_b
    Oct 17, 2016 at 20:27
  • $\begingroup$ @Glen_b ah yes I understand now. bivariate is more specific and clear. $\endgroup$
    – Xu Wang
    Oct 18, 2016 at 14:58
  • $\begingroup$ Related: stats.stackexchange.com/q/263506/119261. $\endgroup$ Jun 25, 2020 at 18:47

1 Answer 1

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Two copies of the same normal variable stacked up in a vector yield a degenerate bivariate normal.

See wikipedia on the degenerate case of the multivariate normal

While it is a special case of the multivariate normal it doesn't have a bivariate density.

The variance-covariance matrix of $Y=(X,X)^\prime$ ($\text{Var}(Y)=\Sigma$) for a standard normal $X$ will be all ones.

As such you can't invert $\Sigma$ and must instead use a generalized-inverse in the exponent; you'll also need to redefine the determinant -- the ordinary determinant will be 0 -- to a pseudo-determinant, which in this case will give $1$.

The density will be zero everywhere but the line $y_1=y_2$ (but if you condition on being on that line it looks like a standard normal density)

If you consider the bivariate distribution in the orthogonal direction to that line you're looking at the density of the variable $Y_1-Y_2=X-X=0$ -- you have a degenerate "normal" with mean 0 and variance 0; it's in this direction we see that the density disappears (it's just a spike at 0).

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  • $\begingroup$ Thank you! Interesting. Reading the wiki page basically says "it has a density but does not have a density", or more technically it has a density with respect to one measure but not with respect to another. $\endgroup$
    – Xu Wang
    Oct 18, 2016 at 14:59
  • $\begingroup$ Yeah, it doesn't have a bivariate density except in a degenerate sense (you can write it as a limit of a sequence of bivariate densities). $\endgroup$
    – Glen_b
    Oct 19, 2016 at 1:13

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