Index: The Book of Statistical ProofsGeneral Theorems ▷ Probability theory ▷ Expected value ▷ Law of the unconscious statistician

Theorem: Let $X$ be a random variable and let $Y = g(X)$ be a function of this random variable.

1) If $X$ is a discrete random variable with possible outcomes $\mathcal{X}$ and probability mass function $f_X(x)$, the expected value of $g(X)$ is

$\label{eq:mean-lotus-disc} \mathrm{E}[g(X)] = \sum_{x \in \mathcal{X}} g(x) f_X(x) \; .$

2) If $X$ is a continuous random variable with possible outcomes $\mathcal{X}$ and probability density function $f_X(x)$, the expected value of $g(X)$ is

$\label{eq:mean-lotus-cont} \mathrm{E}[g(X)] = \int_{\mathcal{X}} g(x) f_X(x) \, \mathrm{d}x \; .$

Proof: Suppose that $g$ is differentiable and that its inverse $g^{-1}$ is monotonic.

1) The expected value of $Y = g(X)$ is defined as

$\label{eq:mean-lotus-disc-s1} \mathrm{E}[Y] = \sum_{y \in \mathcal{Y}} y \, f_Y(y) \; .$

Writing the probability mass function $f_Y(y)$ in terms of $y = g(x)$, we have:

$\label{eq:mean-lotus-disc-s2} \begin{split} \mathrm{E}[g(X)] &= \sum_{y \in \mathcal{Y}} y \, \mathrm{Pr}(g(x) = y) \\ &= \sum_{y \in \mathcal{Y}} y \, \mathrm{Pr}(x = g^{-1}(y)) \\ &= \sum_{y \in \mathcal{Y}} y \sum_{x = g^{-1}(y)} f_X(x) \\ &= \sum_{y \in \mathcal{Y}} \sum_{x = g^{-1}(y)} y f_X(x) \\ &= \sum_{y \in \mathcal{Y}} \sum_{x = g^{-1}(y)} g(x) f_X(x) \; . \end{split}$

Finally, noting that “for all $y$, then for all $x = g^{-1}(y)$” is equivalent to “for all $x$” if $g^{-1}$ is a monotonic function, we can conclude that

$\label{eq:mean-lotus-disc-s3} \mathrm{E}[g(X)] = \sum_{x \in \mathcal{X}} g(x) f_X(x) \; .$

2) Let $y = g(x)$. The derivative of an inverse function is

$\label{eq:der-inv} \frac{\mathrm{d}}{\mathrm{d}y} (g^{-1}(y)) = \frac{1}{g'(g^{-1}(y))}$

Because $x = g^{-1}(y)$, this can be rearranged into

$\label{eq:dx-dy} \mathrm{d}x = \frac{1}{g'(g^{-1}(y))} \, \mathrm{d}y$

and subsitituing \eqref{eq:dx-dy} into \eqref{eq:mean-lotus-cont}, we get

$\label{eq:mean-lotus-cont-s1} \int_{\mathcal{X}} g(x) f_X(x) \, \mathrm{d}x = \int_{\mathcal{Y}} y \, f_X(g^{-1}(y)) \, \frac{1}{g'(g^{-1}(y))} \, \mathrm{d}y \; .$

Considering the cumulative distribution function of $Y$, one can deduce:

$\label{eq:Y-cdf} \begin{split} F_Y(y) &= \mathrm{Pr}(Y \leq y) \\ &= \mathrm{Pr}(g(X) \leq y) \\ &= \mathrm{Pr}(X \leq g^{-1}(y)) \\ &= F_X(g^{-1}(y)) \; . \end{split}$

Differentiating to get the probability density function of $Y$, the result is:

$\label{eq:Y-pdf} \begin{split} f_Y(y) &= \frac{\mathrm{d}}{\mathrm{d}y} F_Y(y) \\ &\overset{\eqref{eq:Y-cdf}}{=} \frac{\mathrm{d}}{\mathrm{d}y} F_X(g^{-1}(y)) \\ &= f_X(g^{-1}(y)) \, \frac{\mathrm{d}}{\mathrm{d}y} (g^{-1}(y)) \\ &\overset{\eqref{eq:der-inv}}{=} f_X(g^{-1}(y)) \, \frac{1}{g'(g^{-1}(y))} \; . \end{split}$

Finally, substituing \eqref{eq:Y-pdf} into \eqref{eq:mean-lotus-cont-s1}, we have:

$\label{eq:mean-lotus-cont-s2} \int_{\mathcal{X}} g(x) f_X(x) \, \mathrm{d}x = \int_{\mathcal{Y}} y \, f_Y(y) \, \mathrm{d}y = \mathrm{E}[Y] = \mathrm{E}[g(X)] \; .$
Sources:

Metadata: ID: P138 | shortcut: mean-lotus | author: JoramSoch | date: 2020-07-22, 08:30.