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  "body": "\n## Statement\n\nLet $\\mu \\in \\mathbb{R}$, let $\\sigma^2 \u003e 0$ and let $X$ have the [[normal distribution|07RXBVTVZB]] $\\mathcal{N}(\\mu, \\sigma^2)$. Then $X$ is integrable and\n$$E[X] = \\int_{-\\infty}^{\\infty} x \\, \\frac{1}{\\sqrt{2\\pi\\sigma^2}} e^{-(x - \\mu)^2/(2\\sigma^2)} \\, dx = \\mu.$$\n\n## Remarks\n\nProof: substitute $x = \\mu + \\sigma z$ with $\\sigma = \\sqrt{\\sigma^2}$. The integral becomes $\\mu \\int \\varphi(z) \\, dz + \\sigma \\int z \\varphi(z) \\, dz$, where $\\varphi$ is the [[standard normal density|027JF1FBB0]].\nThe first integral is $1$. The second is $0$, because $z \\varphi(z)$ is an odd integrable function.\nSo the parameter $\\mu$ is the mean. It is also the median and the mode, because the density is symmetric about $\\mu$ and has its maximum there.\n\n## Explanations\n\n### L1 human\n\nThe bell curve of $\\mathcal{N}(\\mu, \\sigma^2)$ is symmetric about $\\mu$, so its average is $\\mu$.\n",
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