# Proving that a random vector is not bivariate normal

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.

• Yes, I think pointing out that the bivariate normal does not have zero density anywhere is the easiest way to do this one. – Thomas Lumley Mar 10 at 4:32
• Thanks Thomas. Is there a way to contradict it explicitly? – Michael Mar 10 at 7:58
• In what sense is Thomas Lumley's point not "explicit"? – whuber Mar 10 at 16:14
• 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. – Michael Mar 10 at 18:54
• A counterexample to what assertion? – whuber Mar 11 at 14:00

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

z1<-rnorm(10000)
z2<-rnorm(10000)
keep<-(z1*z2)>0

x<-z1[keep]
y<-z2[keep]
plot(x,y)

hist(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.

• That's great explanation. Thanks Thomas! – Michael Mar 11 at 4:07