Suppose X,Y are random variables and their joint pdf is given by: f(x,y)=2g(x)g(y) where x*y>0, and zero otherwise.

g(x) and g(y) are pdfs of standard normal distribution.

I was first able to prove that the marginal distribution of X and Y is standard normal.

However, I am having trouble with the question: Is the random vector (X,Y) bivariate normal?

Is it correct to say that since the support of bivariate normal RV is the entire R^2 plane, but since this random vector does not have a positive density at points wheres XY<0, then it cannot be bivariate normal?

I also tried to find a linear combination of X and Y and prove that it is not normally distribution, but to no avail.

  • 2
    $\begingroup$ Yes, I think pointing out that the bivariate normal does not have zero density anywhere is the easiest way to do this one. $\endgroup$ Commented Mar 10, 2021 at 4:32
  • $\begingroup$ Thanks Thomas. Is there a way to contradict it explicitly? $\endgroup$
    – Michael
    Commented Mar 10, 2021 at 7:58
  • $\begingroup$ In what sense is Thomas Lumley's point not "explicit"? $\endgroup$
    – whuber
    Commented Mar 10, 2021 at 16:14
  • $\begingroup$ I meant to ask for a counter example. As I mentioned, I have tried to find a linear combination that is not normally distributed without success. I thought someone will be able to assist in finding a specific counter example. $\endgroup$
    – Michael
    Commented Mar 10, 2021 at 18:54
  • $\begingroup$ A counterexample to what assertion? $\endgroup$
    – whuber
    Commented Mar 11, 2021 at 14:00

1 Answer 1

  1. Since the density is zero for some $(x,y)$ and the bivariate normal density isn't, you have a simple proof that it's not bivariate normal

  2. It's also easy to show the conditional distributions are not normal: the distribution of $Y$ given $X=1$ is half-normal, not normal.

  3. Linear combinations take a little more work. It helps to characterise the distribution as $(Z_1, Z_2)|\{Z_1Z_2>0\}$, with $Z_1$ and $Z_2$ being bivariate standard normal.

Let $l=x+y$ and think about when $l$ is close to zero. That happens if $x\approx -y$, which mostly happens when $xy<0$, which isn't allowed. We'd expect $l$ to have less probability near zero than a normal distribution.

This R code shows what's going on




density of (x,y) and histogram of x+y

Ok, how do we prove this isn't normal without looking at it? Well, by Bayes' Rule $$f_l(l)\propto f_{z_1+z_2}(l)p(z_1z_2>0|z_1+z_2=l) $$

The first term is a normal density. The second term isn't, and in fact the second term is zero if $l=0$, so the whole thing is zero at $l=0$ and has a minimum where a normal distribution would have a maximum. Or you could use actual calculation, but I hate doing that.

  • $\begingroup$ That's great explanation. Thanks Thomas! $\endgroup$
    – Michael
    Commented Mar 11, 2021 at 4:07

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