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  "body": "\n## Proof\n\nWe compute the characteristic function of $X + Y$ and recognise it.\n\n1. For a real random variable $Z$, let $\\varphi_Z(t) = E[e^{itZ}]$ be its [[characteristic function|0MYDNRT7PP]], for $t \\in \\R$.\n2. By the [[characteristic function of the normal distribution|0NS0TDA4E2]], $\\varphi_X(t) = \\exp\\left(i \\mu_1 t - \\tfrac12 \\sigma_1^2 t^2\\right)$ and $\\varphi_Y(t) = \\exp\\left(i \\mu_2 t - \\tfrac12 \\sigma_2^2 t^2\\right)$.\n3. $X$ and $Y$ are [[independent|0E9502NT92]]. By the [[characteristic function of a sum of independent random variables|02JGYG0T7M]], $\\varphi_{X+Y}(t) = \\varphi_X(t) \\, \\varphi_Y(t)$ for each $t$.\n4. By the [[product of exponentials|0WM7DJCM6G]], $\\varphi_{X+Y}(t) = \\exp\\left(i (\\mu_1 + \\mu_2) t - \\tfrac12 (\\sigma_1^2 + \\sigma_2^2) t^2\\right)$.\n5. Let $\\mu = \\mu_1 + \\mu_2$ and $\\sigma^2 = \\sigma_1^2 + \\sigma_2^2$. Then $\\sigma^2 \u003e 0$, so the [[normal distribution|07RXBVTVZB]] $\\mathcal{N}(\\mu, \\sigma^2)$ is defined. Let $W \\sim \\mathcal{N}(\\mu, \\sigma^2)$.\n6. By the characteristic function of the normal distribution, $\\varphi_W(t) = \\exp\\left(i \\mu t - \\tfrac12 \\sigma^2 t^2\\right)$. By step 4, $\\varphi_W(t) = \\varphi_{X+Y}(t)$ for all $t \\in \\R$.\n7. By the theorem that [[the characteristic function determines the distribution|0T8BYAFEXX]], $X + Y$ and $W$ have the same law.\n8. Hence $X + Y \\sim \\mathcal{N}(\\mu_1 + \\mu_2, \\, \\sigma_1^2 + \\sigma_2^2)$.\n\n## Idea\n\nFor independent variables, characteristic functions multiply. The characteristic function of a normal law is the exponential of a quadratic in $t$. So in a product the means add and the variances add. Since a characteristic function fixes the law, the sum is normal.\n",
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