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  "body": "\n## Proof\n\nWe substitute $x = \\mu + \\sigma z$ and integrate by parts.\n\n1. Let $\\mu \\in \\R$, $\\sigma^2 \u003e 0$, and $\\sigma = \\sqrt{\\sigma^2} \u003e 0$. Let $X$ have the [[normal distribution|07RXBVTVZB]] $\\mathcal{N}(\\mu, \\sigma^2)$.\n2. Let $f = \\varphi_{\\mu, \\sigma^2}$ and $\\varphi = \\varphi_{0,1}$ be the [[normal densities|027JF1FBB0]]. By definition, $f$ is a [[probability density function|04YG690N5V]] of $X$.\n3. By the [[mean of the normal distribution|09PR1V6RWR]], $X$ is integrable with $E[X] = \\mu$. So by definition the [[variance|007QED3E0Y]] is $\\operatorname{Var}(X) = E[(X - \\mu)^2]$.\n4. By the [[law of the unconscious statistician|01ZWX875AT]], if $\\int_\\R (x - \\mu)^2 f(x) \\, dx \u003c \\infty$, then $E[(X - \\mu)^2]$ equals this integral.\n5. The integrand in step 4 is continuous and nonnegative. So the integral is the limit of $\\int_{\\mu - \\sigma R}^{\\mu + \\sigma R} (x - \\mu)^2 f(x) \\, dx$ as $R \\to \\infty$, an [[improper integral|0VCTBJMEJF]].\n6. For all real $z$, $\\sigma f(\\mu + \\sigma z) = \\varphi(z)$. [[Substitution|0D37QFPEZE]] with $x = \\mu + \\sigma z$ gives\n$$\\int_{\\mu - \\sigma R}^{\\mu + \\sigma R} (x - \\mu)^2 f(x) \\, dx = \\sigma^2 \\int_{-R}^{R} z^2 \\varphi(z) \\, dz.$$\n7. By the [[chain rule|0RXX2PT1MQ]] and [[the derivative of the exponential function|0EX9GFJ0B3]], $\\varphi'(z) = -z \\varphi(z)$.\n8. [[Integration by parts|08Y1BBQTK9]] with $u(z) = z$ and $v(z) = -\\varphi(z)$, so that $v'(z) = z \\varphi(z)$, gives\n$$\\int_{-R}^{R} z^2 \\varphi(z) \\, dz = \\big[-z \\varphi(z)\\big]_{-R}^{R} + \\int_{-R}^{R} \\varphi(z) \\, dz = -2R\\varphi(R) + \\int_{-R}^{R} \\varphi(z) \\, dz.$$\n9. By the [[power series of the exponential function|06JZGMYTYD]], $e^{y} \\ge y$ for $y \\ge 0$. With $y = R^2/2$, this gives $R \\varphi(R) \\le 2/(R\\sqrt{2\\pi})$ for $R \u003e 0$.\n10. So $R \\varphi(R) \\to 0$ as $R \\to \\infty$.\n11. Since $\\varphi$ is the density of $\\mathcal{N}(0, 1)$, $\\int_{-R}^{R} \\varphi(z) \\, dz \\to 1$ as $R \\to \\infty$.\n12. By steps 8, 10 and 11, $\\int_{-R}^{R} z^2 \\varphi(z) \\, dz \\to 1$.\n13. By steps 5, 6 and 12, the integral in step 4 is finite and equals $\\sigma^2 \\cdot 1$.\n14. By steps 3 and 4, $\\operatorname{Var}(X) = \\sigma^2$.\n\n## Idea\n\nAfter the substitution $x = \\mu + \\sigma z$, the variance is $\\sigma^2$ times $\\int z^2 \\varphi(z) \\, dz$. Since $\\varphi' = -z\\varphi$, integration by parts turns this integral into $\\int \\varphi = 1$.\n",
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