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  "body": "\n## Statement\n\nLet $X$ be a random variable on $(\\Omega, \\mathcal{F}, P)$ with values in $(E, \\mathcal{E})$ and with [[law|0TST4MHA3A]] $\\mu_X$. Let $g \\colon E \\to \\mathbb{R}$ be measurable.\nThen $g(X)$ is integrable if and only if $\\int_E |g| \\, d\\mu_X \u003c \\infty$. In that case,\n$$E[g(X)] = \\int_E g(x) \\, \\mu_X(dx).$$\nIn particular, if $X$ is discrete with [[mass function|0WF4EJP91W]] $p$, then $E[g(X)] = \\sum_x g(x) \\, p(x)$. If $X$ is real with [[density|04YG690N5V]] $f$, then $E[g(X)] = \\int_{-\\infty}^{\\infty} g(x) f(x) \\, dx$.\n\n## Remarks\n\nThe formula also holds for $g \\ge 0$ without any integrability condition, with both sides in $[0, \\infty]$.\nThe proof checks the formula first for indicator functions $g = 1_B$, where it says $P(X \\in B) = \\mu_X(B)$. Then it extends to simple, nonnegative and integrable $g$.\nThe point is that one does not need the law of $g(X)$ to compute $E[g(X)]$.\n\n## History\n\nThe name refers to statisticians who use the formula as the definition of $E[g(X)]$. They are \"unconscious\" that it is a theorem that needs a proof.\n\n## Explanations\n\n### L1 human\n\nTo find the average of $g(X)$, you can average $g(x)$ over the values $x$ of $X$ with their probabilities.\nFor a fair die, $E[X^2] = (1 + 4 + 9 + 16 + 25 + 36)/6 = 91/6$.\n",
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