Suppose $X, Y$ are two independent variables that uniformly distributed on the interval $[-1,1]$. If $s = x^2 + y^2 < 1$, show the result of $\frac{x}{\sqrt{s}} \sqrt{-2\ln(s)}$ comes from a standard normal distribution.
I understand that $\frac{x}{\sqrt{s}}$ represents cosine of the angle that point $(x,y)$ makes, but have no clue what $\sqrt{-2\ln(s)}$ represents, and why would the product of the two satisfies a standard normal distribution.
Ref:
Marsaglia polar method
Box–Muller transform
1 Answer
$\begingroup$One approach to this is as follows. Let $U$ and $V$ be i.i.d standard normal random variables. Define $R:=\sqrt{U^2+V^2}$ and $(M,N):=(U/R,V/R)$. Then by expressing things in polar coordinates you see that (i) $(M,N)$ is uniformly distributed on the unit circle, (ii) $(M,N)$ is independent of $R$, and (iii) $P(R\le r) = 1-e^{-r^2/2}$ for $r\ge 0$. Point (iii) in turn implies that $\gamma :=e^{-R^2/2}$ is uniformly distributed on $(0,1)$. Therefore, $P(\sqrt{\gamma}\le t)=t^2$, $0\le t\le 1$, so that $\sqrt{\gamma}$ has the distribution of the radial part of a point chosen uniformly from the unit disk. That is, $(X,Y):=(M\sqrt{\gamma}, N\sqrt{\gamma})$ is uniformly distributed in the unit disk. Unraveling this transformation to express $X$ and $Y$ in terms of $U$ and $V$ you get $$ s=X^2+Y^2=e^{-R^2}=\gamma, $$ so that $R=\sqrt{-2\log(s)}$, and $$ {X\over\sqrt{s}}\sqrt{-2\log(s)}={M\sqrt{\gamma}\over\sqrt{s}}R=U, $$ has the standard normal distribution. Likewise, ${Y\over\sqrt{s}}\sqrt{-2\log(s)}$ has the standard normal distribution and is independent of ${X\over\sqrt{s}}\sqrt{-2\log(s)}$.
$\endgroup$
