Is it possible to have distributions s.t. one/both have infinite variance, but finite covariance? What about finite variance but infinite covariance?

If so, what are example distributions/what is the constraint on the pdf that makes this true?


1 Answer 1


The $t_2$-distribution has infinite variance but finite first moment. If you use a constant (which technically is a one-point distribution) as second distribution/random variable, the covariance will be zero, so finite. In general, if one variance is finite and the other one infinite, both finite and infinite covariances can happen.

For two independent $t_2$-distributions, both obviously with infinite variance, $$ Cov(X,Y)=E(X-EX)(Y-EY)=E(X-EX)E(Y-EY)=0, $$ not involving any degenerate integrals.

Finite variances and infinite covariance is impossible due to the Cauchy-Schwarz inequality, $$ Var(X)Var(Y)\ge Cov(X,Y)^2. $$

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    $\begingroup$ (+1) Your first example relies on overlooking that the covariance is the full matrix of central second moments in a multivariate distribution. It is not finite in your example: it has $+\infty$ and $0$ on the diagonal even though it has finite off-diagonal entries. Ordinarily we are unconcerned about the technical distinction between the matrix and its off-diagonal entries in the bivariate case, where it is implicitly assumed that the diagonal entries are finite. It might be better to say that a conventional formula for a "covariance" gives a finite value--but its interpretation is doubtful. $\endgroup$
    – whuber
    Sep 27, 2021 at 15:15
  • $\begingroup$ @whuber That's a fair enough comment. I was interpreting the question as asking explicitly for the off-diagonal entry, because it's clear that if a variance is infinity, one diagonal entry has to be infinity as well; the question wouldn't make much sense if that would count as "not finite". $\endgroup$ Sep 27, 2021 at 16:19
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    $\begingroup$ @DilipSarwate You're right, I didn't mean to put the "multiply" there, I meant to "correlate" it wirh a constant. Corrected it. $\endgroup$ Sep 27, 2021 at 21:22
  • $\begingroup$ @Dilip re "is infinite:" there is one exception that is crucial for the argument. When $a=0,$ $\operatorname{Cov}(aX,X)=\operatorname{Cov}(0,X)=E[0X]-E[0]E[X]=0$ is not infinite. Christian: please note that $\operatorname{Cov}(a,X)=E[aX]-E[a]E[X]=0$ is not necessarily equal to $E[X]$ as you currently claim. But it is indeed finite! $\endgroup$
    – whuber
    Sep 27, 2021 at 21:44
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    $\begingroup$ Yeah, I had the correct example in mind when writing it down in the beginning but confused myself when at the same time thinking about the next case, so messed things up when writing. Corrected now. @Hasse1987 (Amazingly the mess was still picking up votes.) $\endgroup$ Sep 27, 2021 at 21:47

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