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  "body": "\n## Statement\n\nLet $\\mu \\in \\mathbb{R}$, let $\\sigma^2 \u003e 0$ and let $X \\sim \\mathcal{N}(\\mu, \\sigma^2)$. Then for each $t \\in \\mathbb{R}$,\n$$M_X(t) = E\\left[e^{tX}\\right] = \\exp\\left(\\mu t + \\tfrac{1}{2} \\sigma^2 t^2\\right).$$\n\n## Remarks\n\nProof for $Z \\sim \\mathcal{N}(0, 1)$: complete the square, $tz - z^2/2 = t^2/2 - (z - t)^2/2$. So $E[e^{tZ}] = e^{t^2/2} \\int \\varphi(z - t) \\, dz = e^{t^2/2}$.\nThe general case follows from $X = \\mu + \\sigma Z$ with $\\sigma = \\sqrt{\\sigma^2}$: $M_X(t) = e^{\\mu t} M_Z(\\sigma t)$.\nThe moment-generating function is finite for all $t$. Its derivatives at $0$ give $E[X] = \\mu$ and $E[X^2] = \\mu^2 + \\sigma^2$.\nThe logarithm $\\mu t + \\sigma^2 t^2 / 2$ is a polynomial of degree $2$. So all cumulants of order $3$ or more are $0$.\n\n## Explanations\n\n### L1 human\n\nThe moment-generating function of a normal variable has a simple closed form.\nWith the [[multiplication rule for sums|0WRZ170YEH]], it shows quickly that a sum of independent normal variables is again normal.\n",
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