# Variance of a product vs a product of variances

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.

• See this question and its answers for a formula to start with. Feb 19 at 0:12
• Does this answer your question? Variance of product of dependent variables
– Carl
Feb 19 at 1:36
• @Carl I already spent some time trying to use the formula given there to establish the inequality but did not manage it. Feb 19 at 1:38
• 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 $\geq0$.
– Carl
Feb 19 at 3:47

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.