因变量乘积的方差

因变量乘积方差的公式是什么？

对于自变量，公式很简单：

$${\rm var}(XY) = E(X^{2}Y^{2}) - E(XY)^{2} = {\rm var}(X){\rm var}(Y) + {\rm var}(X)E(Y)^2 + {\rm var}(Y)E(X)^2$$ 但是相关变量的公式是什么？

顺便问一下，如何根据统计数据找到相关性？

好吧，使用您指出的熟悉的身份，

$${\rm var}(XY) = E(X^{2}Y^{2}) - E(XY)^{2}$$

使用类似的协方差公式，

$$E(X^{2}Y^{2}) = {\rm cov}(X^{2}, Y^{2}) + E(X^2)E(Y^2)$$

和

$$E(XY)^{2} = [ {\rm cov}(X,Y) + E(X)E(Y) ]^{2}$$

这意味着通常，可以写成${\rm var}(XY)$

$${\rm cov}(X^{2}, Y^{2}) + [{\rm var}(X) + E(X)^2] \cdot[{\rm var}(Y) + E(Y)^2] - [ {\rm cov}(X,Y) + E(X)E(Y) ]^{2}$$

注意，在独立情况下，，这减少为${\rm cov}(X^2,Y^2) = {\rm cov}(X,Y) = 0$

$$[{\rm var}(X) + E(X)^2] \cdot[{\rm var}(Y) + E(Y)^2] - [ E(X)E(Y) ]^{2}$$

并且两个项抵消了，您得到$[ E(X)E(Y) ]^{2}$

$${\rm var}(X){\rm var}(Y) + {\rm var}(X)E(Y)^{2} + {\rm var}(Y)E(X)^{2}$$

正如您在上面指出的那样。

编辑：如果您观察到的只是而不是X和Y，那么我认为您没有办法估算c o v（X ，Y ）或c o v（X 2，Y 2），除了在特殊情况下（例如，如果X ，Y具有先验已知的均值）$XY$$X$$Y$${\rm cov}(X,Y)$${\rm cov}(X^2,Y^2)$$X,Y$

这是@Macro非常好的答案的附录，该答案准确列出了确定两个相关随机变量乘积的方差所需知道的内容。由于

$$\begin{align} \operatorname{var}(XY) &= E\left[(XY)^2\right] - \left(E[XY]\right)^2 \tag{1}\\ &= E[(XY)^2] - \left(\operatorname{cov}(X,Y)+E[X]E[Y]\right)^2\\ &= E[X^2Y^2] - \left(\operatorname{cov}(X,Y)+E[X]E[Y]\right)^2\tag{2}\\ &= \left(\operatorname{cov}(X^2,Y^2)+E[X^2]E[Y^2]\right) - \left(\operatorname{cov}(X,Y)+E[X]E[Y]\right)^2\tag{3}\\ \end{align}$$ 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).

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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 $\mathrm{Cov}\left[x,y^2\right]$.