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  "body": "\n## Statement\n\nLet $X$ and $Y$ be real random variables on one probability space with $E[X^2] \u003c \\infty$ and $E[Y^2] \u003c \\infty$. Then $XY$ is integrable and\n$$\\left(E[XY]\\right)^2 \\le E[X^2] \\, E[Y^2].$$\nEquality holds if and only if $aX + bY = 0$ almost surely for some reals $a$ and $b$, not both $0$.\n\n## Remarks\n\nProof: for each real $s$, $0 \\le E[(X + sY)^2] = E[X^2] + 2s E[XY] + s^2 E[Y^2]$. A quadratic polynomial in $s$ that is never negative has a discriminant that is not positive.\nThis is the Cauchy–Schwarz inequality for the inner product $\\langle X, Y \\rangle = E[XY]$ on the space $L^2$ of square-integrable random variables.\nApplied to $X - E[X]$ and $Y - E[Y]$, it gives $\\operatorname{Cov}(X, Y)^2 \\le \\operatorname{Var}(X) \\operatorname{Var}(Y)$. So the correlation coefficient is in $[-1, 1]$.\n\n## Explanations\n\n### L1 human\n\nThe average of a product cannot be larger than what the sizes of the two factors allow.\nThis simple inequality is behind the fact that a correlation is always between $-1$ and $1$.\n",
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