Is the variance of a product $Var(XY)$ of possibly dependent variables necessarily larger or smaller than the product of variances $Var(X)Var(Y)$? Looking at examples I only see it being larger but I don't know if it is in general.

  • $\begingroup$ See this question and its answers for a formula to start with. $\endgroup$
    – B.Liu
    Feb 19 at 0:12
  • 1
    $\begingroup$ Does this answer your question? Variance of product of dependent variables $\endgroup$
    – Carl
    Feb 19 at 1:36
  • $\begingroup$ @Carl I already spent some time trying to use the formula given there to establish the inequality but did not manage it. $\endgroup$
    – sayda
    Feb 19 at 1:38
  • $\begingroup$ Yes, if covariance is zero. $$ {\rm var}(XY) = {\rm var}(X){\rm var}(Y) + {\rm var}(X)E(Y)^2 + {\rm var}(Y)E(X)^2 $$ The last two terms are $\geq0$. $\endgroup$
    – Carl
    Feb 19 at 3:47

1 Answer 1


There is no general relationship.

To see this, consider the extreme situations below: in one case the product of variances is, on a relative scale, arbitrarily larger than the variance of the product; while in the other case the product of variances is arbitrarily smaller than the variance of the product. You can tweak these examples to create any ratio you like, from $0$ through $\infty,$ of $\operatorname{Var}(XY):\operatorname{Var}(X)\operatorname{Var}(Y).$

  1. Consider the uniform distribution on the four points in the set $\{(\pm 1,0), (0,\pm 1)\}.$ The marginal variances are $1$ but since the products of the components are always zero, the variance of the product is zero. That is, $$0 = \operatorname{Var}(XY) \lt \operatorname{Var}(X)\operatorname{Var}(Y) = (1)(1)=1.$$

  2. Let $X$ have a cumulative distribution function $F(x) = 1-1/x^3$ for $x\ge 1.$ Consequently it has a density function $f(x) = F^\prime(x) = 3/x^4$ for $x\ge 1,$ whence for any $k\lt 3$ $$E[X^k]=E[Y^k] = \int_1^\infty x^k\left(\frac{3}{x^4}\right)\,\mathrm{d}x = \frac{3}{3-k}.$$ For any larger $k,$ the integral diverges: it is infinite. From the cases $k=1,2$ we obtain $$\operatorname{Var}(X) = E[X^2]-E[X]^2 = \frac{3}{3-2} - \left(\frac{3}{3-1}\right)^2=\frac{3}{4}.$$ Suppose $Y=X.$ Compute $$\operatorname{Var}(XY) = \operatorname{Var}(X^2) = E(X^4) - E(X^2)^2.$$ This is infinite. Thus, $$\infty = \operatorname{Var}(XY) \gt \operatorname{Var}(X)\operatorname{Var}(Y) =\left(\frac{3}{4}\right)^2 = \frac{9}{16}.$$

Roughly, the first case is one in which although the individual components $X$ and $Y$ vary appreciably, they do so without changing $XY$ much. The second is one in which large values of $X$ and $Y$ tend to co-occur, thereby greatly magnifying the variance of either one in the product.

  • 1
    $\begingroup$ not the first time, probably won't be the last to have my hopes dashed by your counterexamples $\endgroup$
    – sayda
    Feb 19 at 22:15

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