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I'm trying to show that for uniformly distributed variables $X_1$ and $X_2$ that the vector $$ (\sqrt{-2\log X_1} \cos(2\pi X_2), \sqrt{-2 \log X_1} \sin(2\pi X_2)) $$ is two-dimensional normal. Letting $Y_1$ be the first component and $Y_2$ be the second component I see that $X_1 = e^{-(Y_1^2 +Y_2^2)/2}$ and $X_2 = \frac{1}{2\pi} \arctan(Y_2/Y_1)$. Is there a way to proceed from here to show that $(Y_1, Y_2)$ is normal?

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  • $\begingroup$ Hint: do a transformation from cartesian to polar coordinates. $\endgroup$
    – orion
    Nov 14, 2014 at 18:49

1 Answer 1

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Remember the substitution rule (if we can characterize the CDF, then we are done). In particular, consider the absolute value of the determinant of the Jacobian matrix \begin{align*} \left|\det\left[\begin{array}{cc} \frac{\partial X_{1}}{\partial Y_{1}} & \frac{\partial X_{1}}{\partial Y_{2}}\\ \frac{\partial X_{2}}{\partial Y_{1}} & \frac{\partial X_{2}}{\partial Y_{2}} \end{array}\right]\right| & =\left|\det\left[\begin{array}{cc} -Y_{1}\exp\left(Y_{1}^{2}/2-Y_{2}^{2}/2\right) & -Y_{2}\exp\left(Y_{1}^{2}/2-Y_{2}^{2}/2\right)\\ -Y_{2}/\left[2\pi\left(Y_{1}^{2}+Y_{2}^{2}\right)\right] & Y_{1}/\left[2\pi\left(Y_{1}^{2}+Y_{2}^{2}\right)\right] \end{array}\right]\right|\\ & =\left|-Y_{1}^{2}\frac{\exp\left(Y_{1}^{2}/2-Y_{2}^{2}/2\right)}{2\pi\left(Y_{1}^{2}+Y_{2}^{2}\right)}-Y_{2}^{2}\frac{\exp\left(Y_{1}^{2}/2-Y_{2}^{2}/2\right)}{2\pi\left(Y_{1}^{2}+Y_{2}^{2}\right)}\right|\\ & =\left|-\frac{\exp\left(Y_{1}^{2}/2-Y_{2}^{2}/2\right)}{2\pi}\right|\\ & =\frac{\exp\left(Y_{1}^{2}/2\right)}{\sqrt{2\pi}}\frac{\exp\left(Y_{2}^{2}/2\right)}{\sqrt{2\pi}}. \end{align*}

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