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  "body": "\n## Proof\n\nWe compute directly from the definition of the variance.\n\n1. Let $X$ be a real random variable with finite [[variance|007QED3E0Y]], and let $a \\in \\R$.\n2. By the definition of the variance, $X$ is integrable. Let $m = E[X]$. Then $E[(X - m)^2] = \\operatorname{Var}(X) \u003c \\infty$.\n3. By [[linearity of expectation|0P7PTC8DDE]], $aX$ is integrable and $E[aX] = a m$.\n4. Pointwise, $(aX - am)^2 = a^2 (X - m)^2$.\n5. The variable $(X - m)^2$ is integrable by step 2. By linearity of the [[expected value|0DCT0QZVQ1]], $E[a^2 (X - m)^2] = a^2 E[(X - m)^2]$.\n6. By steps 3 to 5 and the definition of the [[variance|007QED3E0Y]], $\\operatorname{Var}(aX) = E[(aX - am)^2] = a^2 \\operatorname{Var}(X)$.\n\n## Idea\n\nScaling $X$ by $a$ scales its mean by $a$, so the deviation from the mean is scaled by $a$. Squaring turns the factor $a$ into $a^2$.\n",
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        "locator": "Addition and multiplication by a constant",
        "url": "https://en.wikipedia.org/wiki/Variance"
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  "name": "Proof of variance under scaling",
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