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  "body": "\n## Statement\n\nLet $U_1$ and $U_2$ be independent random variables, each with the [[uniform distribution|0AREFQ9Z65]] on $(0, 1)$.\nLet\n$$Z_1 = \\sqrt{-2 \\ln U_1}\\, \\cos(2\\pi U_2), \\qquad Z_2 = \\sqrt{-2 \\ln U_1}\\, \\sin(2\\pi U_2).$$\nThen $Z_1$ and $Z_2$ are [[independent|0GQ938GDE0]], and each has the standard [[normal distribution|05AQE0QR0D]], with mean $0$ and variance $1$.\n\n## Remarks\n\nLet $R = \\sqrt{-2 \\ln U_1}$ and $\\Theta = 2\\pi U_2$, so that $(Z_1, Z_2)$ has [[polar coordinates|0BSW4T7THS]] $(R, \\Theta)$.\nFor $t \\ge 0$, $P(R^2 \u003e t) = P(U_1 \u003c e^{-t/2}) = e^{-t/2}$, so $R^2$ has the [[exponential distribution|0E99F51GVB]] with mean $2$, and $R$ has density $r e^{-r^2/2}$ on $(0, \\infty)$.\nAlso $\\Theta$ is uniform on $(0, 2\\pi)$ and independent of $R$.\nBy the [[change of variables formula|00N5FHP1CV]] with Jacobian $r$, the pair $(Z_1, Z_2)$ has density $\\frac{1}{2\\pi} e^{-r^2/2} = \\frac{1}{\\sqrt{2\\pi}} e^{-z_1^2/2} \\cdot \\frac{1}{\\sqrt{2\\pi}} e^{-z_2^2/2}$, which is the product of two standard normal densities.\nExample: $U_1 = e^{-2}$ and $U_2 = 1/8$ give $R = 2$, $\\Theta = \\pi/4$, and $(Z_1, Z_2) = (\\sqrt{2}, \\sqrt{2})$.\nIndependence of $U_1$ and $U_2$ is necessary: with $U_2 = U_1$ the point $(Z_1, Z_2)$ lies on one curve in the plane, so the pair is not a standard normal pair.\nTo obtain a normal variable with mean $\\mu$ and standard deviation $s$, use $\\mu + s Z_1$.\n\n## Explanations\n\n### L1 human\n\nThe transform turns two uniform random numbers into two independent bell-curve random numbers.\nThe first uniform number decides the distance from the origin, through a logarithm, and the second decides the direction.\nThe trick works because the two-dimensional standard normal distribution is the same in every direction, and its squared radius has a simple exponential law.\n\n### L2 human\n\nOne cannot invert the normal distribution function in closed form, so plain inverse transform sampling does not give normal variables cheaply; going to two dimensions and polar coordinates solves this.\nMarsaglia's polar method avoids the sine and cosine by sampling a uniform point in the unit disc by rejection and rescaling it.\nIn floating point, $U_1$ must avoid $0$, and the tail of $Z_1$ is truncated by the smallest representable $U_1$.\nFor speed, many libraries now prefer the ziggurat algorithm, which uses mostly one multiplication and one comparison per sample.\n",
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