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  "body": "\n## Proof\n\nWe substitute $x = \\mu + \\sigma z$ and complete the square in the exponent.\n\n1. Let $\\mu \\in \\R$, $\\sigma^2 \u003e 0$, and $\\sigma = \\sqrt{\\sigma^2} \u003e 0$. Let $X \\sim \\mathcal{N}(\\mu, \\sigma^2)$ be [[normal|07RXBVTVZB]], and fix $t \\in \\R$.\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 [[law of the unconscious statistician|01ZWX875AT]], if $\\int_\\R e^{tx} f(x) \\, dx \u003c \\infty$, then $e^{tX}$ is integrable and $E[e^{tX}]$ equals this integral.\n4. The integrand is continuous and positive. So the integral is the limit of $\\int_{\\mu - \\sigma R}^{\\mu + \\sigma R} e^{tx} f(x) \\, dx$ as $R \\to \\infty$, an [[improper integral|0VCTBJMEJF]].\n5. For all real $z$, $\\sigma f(\\mu + \\sigma z) = \\varphi(z)$. [[Substitution|0D37QFPEZE]] with $x = \\mu + \\sigma z$ gives\n$$\\int_{\\mu - \\sigma R}^{\\mu + \\sigma R} e^{tx} f(x) \\, dx = \\int_{-R}^{R} e^{t\\mu + t\\sigma z} \\varphi(z) \\, dz.$$\n6. Completing the square gives $t\\sigma z - z^2/2 = \\sigma^2 t^2/2 - (z - \\sigma t)^2/2$.\n7. By step 6 and the [[exponential of a sum|0E8HNVBCSP]], $e^{t\\mu + t\\sigma z} \\varphi(z) = e^{\\mu t + \\sigma^2 t^2/2} \\, \\varphi(z - \\sigma t)$.\n8. Substitution with $u = z - \\sigma t$ gives $\\int_{-R}^{R} \\varphi(z - \\sigma t) \\, dz = \\int_{-R - \\sigma t}^{R - \\sigma t} \\varphi(u) \\, du$.\n9. Since $\\varphi$ is the density of $\\mathcal{N}(0, 1)$, its integral over $\\R$ is $1$. So the integral in step 8 tends to $1$ as $R \\to \\infty$.\n10. By steps 4, 5, 7 and 9, $\\int_\\R e^{tx} f(x) \\, dx = e^{\\mu t + \\sigma^2 t^2/2} \u003c \\infty$.\n11. By step 3, the [[moment-generating function|0NNRNGKMFM]] is $M_X(t) = E[e^{tX}] = \\exp\\left(\\mu t + \\tfrac{1}{2}\\sigma^2 t^2\\right)$.\n\n## Idea\n\nMultiplying the standard normal density by $e^{sz}$ shifts the centre of the Gaussian bell to $s$ and multiplies it by $e^{s^2/2}$. This is completing the square. The shifted bell still has total mass $1$.\n",
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