Variance of scalar function of 2 random variables

Question

Suppose I have a scalar function $g(X,Y)$, where $X$ and $Y$ are jointly distributed with pdf $p(x,y)$. I think the expected value of $g$ is given by

$$ \mathbb{E}[g] = \int_{-\infty}^\infty \int_{-\infty}^{\infty} g(x,y) p(x,y) ,円 dx ,円 dy $$

But what is $\operatorname{Var}[g]$?

• whuber: By definition it's $\operatorname{Var}(g) = E[g^2] - E[g]^2.$ What are you looking for in addition to that?
• Michael Hardy: @whuber Should one call that a definition, or a corollary of the definition? The mean squared deviation from the mean is what variance is. The way you've written it, sometimes called a shortcut, has a danger that when $\operatorname E(g)$ is very big compared to typical deviations from the mean, the information you seek, the variance, gets lost in the rounding error.
• whuber: @Michael I agree. But many texts do indeed define the variance in this way. This formula drops out of the mgf or cgf, for instance, indicating that at least as a mathematical proposition it's as natural as the mean squared deviation from the mean.
• Michael Hardy: @whuber : But it doesn't motivate the definition as well as mean-squared-deviation does. I suspect nobody before de Moivre used variance and standard deviation as a measure of dispersion, and de Moivre's reason for doing it was that he could find the standard deviation of the number of "heads" in 3600 coin tosses, and then apply the "bell-shaped" curve to approximate the binomial distribution. I suspect no one thought of that before he did, in the first half of the 1700s. Could he ever have thought of the "short-cut" version? Is there any reason why someone like that would?

Answer 1

Based on the definition, it works out to be

$$ \iint_{-\infty}^\infty g^2(x,y) p(x,y),円 \mathrm dx \mathrm dy - \mathbb{E}[g]^2 $$

which is equivalent to

$$ \iint_{-\infty}^\infty \left( g(x,y) - \mathbb{E}[g] \right)^2 p(x,y) ,円 \mathrm dx \mathrm dy $$

• Michael Hardy: It seems statisticians frequently write $g^2(x,y)$ for $\big( g(x,y)\big)^2.$ There is a certain danger of misunderstanding here, since sometimes $g^2(x,y)$ means $g(g(x,y)). \qquad$

Answer 2

If one were to work only with the (削除) most basic definitions (削除ここまで) more basic characterizations of expectation and variance, one could seek the density function \begin{align} f_{g(X,Y)}(w) = {} & \frac d{dw} \Pr(g(X,Y)\le w) \\[8pt] = {} & \iint\limits_{x,y,円:,円g(x,y),円\le,円w} p(x,y),円 d(x,y), \end{align} and then compute $$ \mu = \operatorname E(g(X,Y)) = \int_{-\infty}^{+\infty} w f_{g(X,Y)}(w),円 dw, $$ or, when the distribution is not absolutely continuous, $$ \mu = \operatorname E(g(X,Y)) = \int_{-\infty}^{+\infty} w ,円 dF_{g(X,Y)}(w) $$ (a Riemann–Stieltjes integral, which, in the case of discrete distributions, is merely a sum), and finally $$ \operatorname{var}(g(X,Y)) = \operatorname E\left( \big( g(X,Y) - \mu \big)^2 \right), $$ or the corresponding Riemann–Stieltjes integral.

HOWEVER, one need not work only with the most basic definitions. We also have a consequence of those definitions: $$ \mu = \operatorname E(g(X,Y)) = \iint\limits_{(-\infty,+\infty)^2} g(x,y)p(x,y) ,円 d(x,y) $$ and then $$ \operatorname{var}(g(X,Y)) = \iint\limits_{(-\infty,+\infty)^2} (g(x,y)-\mu)^2 p(x,y) ,円 d(x,y) $$ (or the corresponding Riemann–Stieltjess integral).

• whuber: Many might dispute what you characterize as "the most basic definitions," because $(X,Y):\Omega\to\mathbb R^2$ is a (bivariate) random variable on a probability space $(\Omega,\mathfrak F, \mathbb P)$ and $g:\mathbb R^2\to\mathbb R$ is a measurable function. The random variable denoted by "$g(X,Y)$" means the composition $g\circ (X,Y)$ and its expectation is, as always, $E[g(X,Y)]=\int_\Omega g(X(\omega),Y(\omega)),円\mathrm d\mathbb P(\omega).$ This does not require any distribution function nor any consideration of densities: that's why it's more basic.
• Michael Hardy: @whuber : Perhaps I should say "The more basic characterizations" rather than "The most basic definitions."