Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. It only takes a minute to sign up.
Sign up to join this community
What is the formula for variance of product of dependent variables?
In the case of independent variables the formula is simple:
$$ {\rm var}(XY) = E(X^{2}Y^{2}) - E(XY)^{2} = {\rm var}(X){\rm var}(Y) + {\rm var}(X)E(Y)^2 + {\rm var}(Y)E(X)^2 $$ But what is the formula for correlated variables?
By the way, how can I find the correlation based on the statistical data?
Well, using the familiar identity you pointed out,
$$ {\rm var}(XY) = E(X^{2}Y^{2}) - E(XY)^{2} $$
Using the analogous formula for covariance,
$$ E(X^{2}Y^{2}) = {\rm cov}(X^{2}, Y^{2}) + E(X^2)E(Y^2) $$
and
$$ E(XY)^{2} = [ {\rm cov}(X,Y) + E(X)E(Y) ]^{2} $$
which implies that, in general, ${\rm var}(XY)$ can be written as
$$ {\rm cov}(X^{2}, Y^{2}) + [{\rm var}(X) + E(X)^2] \cdot[{\rm var}(Y) + E(Y)^2] - [ {\rm cov}(X,Y) + E(X)E(Y) ]^{2} $$
Note that in the independence case, ${\rm cov}(X^2,Y^2) = {\rm cov}(X,Y) = 0$ and this reduces to
$$ [{\rm var}(X) + E(X)^2] \cdot[{\rm var}(Y) + E(Y)^2] - [ E(X)E(Y) ]^{2} $$
and the two $[ E(X)E(Y) ]^{2}$ terms cancel out and you get
$$ {\rm var}(X){\rm var}(Y) + {\rm var}(X)E(Y)^{2} + {\rm var}(Y)E(X)^{2} $$
as you pointed out above.
Edit: If all you observe is $XY$ and not $X$ and $Y$ separately, then I don't think there is a way for you to estimate ${\rm cov}(X,Y)$ or ${\rm cov}(X^2,Y^2)$ except in special cases (for example, if $X,Y$ have means that are known a priori)
This is an addendum to @Macro's very nice answer which lays out exactly what needs to known in order to determine the variance of the product of two correlated random variables. Since \begin{align} \operatorname{var}(XY) &= E\left[(XY)^2\right] - \left(E[XY]\right)^2 \tag{1}\\ &= E[(XY)^2] - \left(\operatorname{cov}(X,Y)+E[X]E[Y]\right)^2\\ &= E[X^2Y^2] - \left(\operatorname{cov}(X,Y)+E[X]E[Y]\right)^2\tag{2}\\ &= \left(\operatorname{cov}(X^2,Y^2)+E[X^2]E[Y^2]\right) - \left(\operatorname{cov}(X,Y)+E[X]E[Y]\right)^2\tag{3}\\ \end{align} where $\operatorname{cov}(X,Y)$, $E[X]$, $E[Y]$, $E[X^2]$, and $E[Y^2]$ can be assumed to be known quantities, we need to be able to determine the value of $E\left[X^2Y^2\right]$ in $(2)$ or $\operatorname{cov}(X^2,Y^2)$ in $(3)$. This is not easy to do in general, but, as pointed out already, if $X$ and $Y$ are independent random variables, then $\operatorname{cov}(X,Y) = \operatorname{cov}(X^2,Y^2) = 0$. In fact, dependence, not correlation (or lack thereof) is the key issue. That we know that $\operatorname{cov}(X,Y)$ equals $0$ instead of some nonzero value does not, by itself, help in the least in our efforts are determining the value of $E\left[X^2Y^2\right]$ or $\operatorname{cov}(X^2,Y^2)$ even though it does simplify the right sides of $(2)$ and $(3)$ a little.
When $X$ and $Y$ are dependent random variables, then in at least one (fairly common or fairly important) special case, it is possible to find the value of $E\left[X^2Y^2\right]$ relatively easily.
Suppose that $X$ and $Y$ are jointly normal random variables with correlation coefficient $\rho$. Then, conditioned on $X = x$, the conditional density of $Y$ is a normal density with mean $E[Y] + \rho\left.\left.\sqrt{\frac{\operatorname{var}(Y)}{\operatorname{var}(X)}} \right(x-E[X]\right)$ and variance $\operatorname{var}(Y)(1-\rho^2)$. Thus, \begin{align}E[X^2Y^2 \mid X] &= X^2E[Y^2 \mid X]\\ &= X^2\left[\operatorname{var}(Y)(1-\rho^2) + \left(E[Y] + \rho\left.\left.\sqrt{\frac{\operatorname{var}(Y)}{\operatorname{var}(X)}} \right(X-E[X]\right)\right)^2\right] \end{align} which is a quartic function of $X$, say $g(X)$, and the Law of Iterated Expectation tells us that $$E[X^2Y^2] = E\left[E[X^2Y^2\mid X]\right] = E[g(X)]\tag{4}$$ where the right side of $(4)$ can be computed from knowledge of the 3rd and 4th moments of $X$ -- standard results that can be found in many texts and reference books (meaning that I am too lazy to look them up and include them in this answer).
Further addendum: In a now-deleted answer, @Hydrologist gives the variance of $XY$ as $$\mathrm{Var}\left[xy\right] = \left(\mathrm{E}\left[x\right]\right)^2\mathrm{Var}\left[y\right] + \left(\mathrm{E}\left[y\right]\right)^2\mathrm{Var}\left[x\right] + 2\mathrm{E}\left[x\right]\mathrm{Cov}\left[x,y^2\right] + 2\mathrm{E}\left[y\right]\mathrm{Cov}\left[x^2,y\right]\\ + 2\mathrm{E}\left[x\right]\mathrm{E}\left[y\right]\mathrm{Cov}\left[x,y\right] +\mathrm{Cov}\left[x^2,y^2\right] - \left(\mathrm{Cov}\left[x,y\right]\right)^2 \tag{5}$$ and claims that this formula is from two papers published a half-century ago in JASA. This formula is an incorrect transcription of the results in the paper(s) cited by Hydrologist. Specifically, $\mathrm{Cov}\left[x^2,y^2\right]$ is a mistranscription of $E[(x-E[x])^2(y-E[y])^2]$ in the journal article, and similarly for $\mathrm{Cov}\left[x^2,y\right]$ and $\mathrm{Cov}\left[x,y^2\right]$.
