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  "body": "\n## Statement\n\nLet $X_1, \\dots, X_n$ be [[independent|0XCE070V8B]] real random variables with $E[X_i^2] \u003c \\infty$ for each $i$. Then\n$$\\operatorname{Var}(X_1 + \\dots + X_n) = \\operatorname{Var}(X_1) + \\dots + \\operatorname{Var}(X_n).$$\n\n## Remarks\n\nThe proof uses the [[variance of a sum|0D86GJEZ39]]: independent variables are [[uncorrelated|077GN9HDR6]], so each covariance term is $0$.\nSo the formula holds more generally for pairwise uncorrelated $X_i$.\nFor i.i.d. $X_i$ with variance $\\sigma^2$, the sample mean $\\bar{X}_n = (X_1 + \\dots + X_n)/n$ has variance $\\sigma^2 / n$.\nFor $X \\sim \\operatorname{Bin}(n, p)$, the formula gives $\\operatorname{Var}(X) = np(1 - p)$, as $X$ is a sum of $n$ independent Bernoulli variables.\n\n## History\n\nThe formula is also called the Bienaymé formula, after Irénée-Jules Bienaymé.\n\n## Explanations\n\n### L1 human\n\nFor independent variables, variances simply add.\nThis is why an average of many independent measurements is more precise than one measurement: its variance gets smaller like $1/n$.\n",
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