$\text{Cov}[X,Y]$ if $\mathbb{E}[X^2]<\infty$ but $\mathbb{E}[Y^2]=\infty$

Question

What can be said about $\text{Cov}[X,Y]$ if $\mathbb{E}[X^2]<\infty$ but $\mathbb{E}[Y^2]=\infty$?

Intuitively, I've always thought of $\text{Cov}[X,Y]$ as the measure by which two R.V.s "change together". I'd like to say, then, that the covariance of $X$ and $Y$ is bounded by the finite variance of $X$, though that doesn't sound right to me...

Any ideas?

Answer 1

You can't bound covariance by the variance of one of the variables in general.

For an example, consider $X$ with density $$f(x) = (\alpha-1)/x^\alpha \quad \mbox{ for }x\geq 1$$ and some $\alpha\in(7/2,4)$. Clearly $\mathbf E X^k < \infty$ iff $\alpha - 1 > k$, so $\mathbf E X^2 < \infty$. Take now $Y = X^{3/2}$. Then $\mathbf E Y = \mathbf E X^{3/2} < \infty$ and $\mathbf E Y^2 = \mathbf E X^3 = \infty$ and $$ Cov(X,Y) = \mathbf E XY - \mathbf E X \, \mathbf E Y = \mathbf E X^{5/2} - \mathbf E X \, \mathbf E X^{3/2} = \infty. $$ On the other hand if $Z$ is taken independent of $X$ but having the same law as $Y$, since $\mathbf E Z = \mathbf E X^{3/2} < \infty$ we have that $$ Cov(X,Z) = \mathbf E X \, \mathbf E Z - \mathbf E X \, \mathbf E Z = 0.$$