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  "body": "\n## Proof\n\nWe expand the square and use linearity of expectation.\n\n1. Let $X$ be a real random variable with $E[X^2] \u003c \\infty$.\n2. For each real $x$, $(\\abs{x} - 1)^2 \\ge 0$, so $\\abs{x} \\le (1 + x^2)/2$.\n3. Hence $0 \\le \\abs{X} \\le (1 + X^2)/2$. The integral of nonnegative functions is monotone, so the [[expected value|0DCT0QZVQ1]] $E[\\abs{X}]$ is at most $(1 + E[X^2])/2 \u003c \\infty$.\n4. So $X$ is integrable. Let $m = E[X]$, a real number. The [[variance|007QED3E0Y]] of $X$ is $E[(X - m)^2]$.\n5. Pointwise, $(X - m)^2 = X^2 - 2mX + m^2$.\n6. The variables $X^2$, $X$ and the constant $m^2$ are integrable. By [[linearity of expectation|0P7PTC8DDE]], $E[(X - m)^2] = E[X^2] - 2m E[X] + m^2$.\n7. Since $E[X] = m$, the right side is $E[X^2] - 2m^2 + m^2 = E[X^2] - m^2$.\n8. So $\\operatorname{Var}(X) = E[X^2] - (E[X])^2$.\n\n## Idea\n\nSquare out $(X - m)^2$ and take the expectation of each term. The cross term gives $-2m^2$, which cancels the constant $m^2$ down to $-m^2$.\n",
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