# A Rant About Uncorrelated Normal Random Variables

### by Jeffrey S. Rosenthal, 2005

On my department's PhD Comprehensive Examinations this year, the following question was asked: Suppose X and Y are two jointly-defined random variables, each having the standard normal distribution N(0,1). Suppose further that X and Y are uncorrelated, i.e. that Cov(X,Y) = 0. Does this necessarily imply that X and Y are independent?

Sadly, no student got this question correct (though I had intended it to be easy). In fact, I suspect that many practicing statisticians would also get it wrong, due to the tremendous influence of the bivariate (or multivariate) normal distribution on statistical thinking.

What is true is that if the random variable pair (X,Y) follows the bivariate normal distribution, and Cov(X,Y) = 0, then X and Y must be independent.

But what is not true is that if each of X and Y is normally distributed, and Cov(X,Y) = 0, then X and Y must be independent.

Consider a simple example. Let X have a standard normal distribution. Let Z be independent of X, with Z equally likely to be +1 or -1 (i.e., Pr[Z=+1] = Pr[Z=-1] = 1/2). Let Y = X Z (the product of X and Z). Then it is easy to see that Y also has a standard normal distribution, and that Cov(X,Y) = 0. On the other hand, clearly X and Y are not independent: indeed, it always holds that |X| = |Y|. The point is that, just because each of X and Y has a normal distribution, that does not mean that the pair (X,Y) has a bivariate normal distribution, nor even that (X,Y) is jointly absolutely continuous, nor does it mean that zero covariance implies independence.

(As an aside, in the above example, Pr[X + Y = 0] = 1/2, so X + Y isn't normally distributed, i.e. the sum of normals isn't necessarily normal, either.)

So, when discussing (or teaching) statistics, do remember that just because two random variables are each normally distributed, that does not necessarily mean that they jointly follow a bivariate normal distribution -- nor that zero covariance implies independence. It simply isn't so!