Proof of Marsaglia polar method I studied Polar method and I can use it very well to simulate to Standard Normal Variable.
But I can't figure it out that how it works!
So is there any proof/theorem to learn reasoning behind Polar method?
Thanks! Regards,
 A: First I'll quickly go through the Box-Muller method and then show how the Polar method takes advantage of it.
Suppose that 
$$
\left(
\begin{array}{c}
X\\
Y\\
\end{array}
\right) \sim N(0, I_2).
$$
Note that by multiplying by $\Sigma^{1/2}$ and adding $\mu$ we can get any normal distribution at all, so the important part is getting iid normals.
Define $R^2 = X^2 + Y^2$ and $\Theta = arctan(Y/X)$. It can be shown (via Jacobian theorem) that $R^2 \sim Exp(2)$, $\Theta \sim Unif(0, 2\pi)$, and $R^2 \perp \Theta$. Using polar coordinates we have that
$$
\left(
\begin{array}{c}
X\\
Y\\
\end{array}
\right) = \left(
\begin{array}{c}
R \cos\Theta \\
R \sin \Theta\\
\end{array}
\right) 
$$
so if we can sample $R^2$ and $\Theta$ we're good to go. This turns out to be easy: Let $V_1, V_2 \sim \ \textrm{iid} \ Unif(0,1)$. Then by the inverse CDF method $-2\log V_1 := R^2 \sim Exp(2)$ and $2\pi V_2 := \Theta \sim Unif(0, 2\pi)$ with both independent. Plugging these in to the above equation gives us our $X$ and $Y$.
Now I'll show how the Polar method uses this.
Let $\mathcal U = \{(u, v) : u^2 + v^2 \leq 1\}$, i.e. $\mathcal U$ is the unit disk. Take $(X, Y) \sim Unif(\mathcal U)$. We can get this by taking $U_1, U_2 \sim \ \textrm{iid} \ Unif[-1, 1]$ and rejecting until we're in $\mathcal U$.
Now define $T = X^2 + Y^2$, i.e. $T$ is the squared distance from the origin to our point $(X, Y)$. 
$$
P(T \leq t) = P(X^2 + Y^2 \leq t) = \frac{\pi (\sqrt t)^2}{\pi (1)^2} = t,
$$
i.e. the probability that $T \leq t$ is the ratio of the area of the disk with radius $\sqrt t$ over the area of the entire disk, which is just $t$. This means that $T \sim Unif(0, 1)$.
From the Box-Muller bit we know that 
$$
\left(
\begin{array}{c}
R \cos\Theta \\
R \sin \Theta\\
\end{array}
\right) \sim N(0, I_2)
$$
when $R^2 \sim Exp(2) \perp \Theta \sim Unif(0, 2\pi)$. $T \sim Unif(0, 1)$ so we can take $R^2 := -2 \log T$.
Now note that for a particular $(X, Y)$ we'll have a triangle like so:

This means that $\sin \Theta = \frac{Y}{\sqrt T}$ and $\cos \Theta = \frac{X}{\sqrt T}$.
We can just plug these into the Box-Muller equations and there we go. It also turns out that the independence assumption is satisfied. This frees us from computing any trig functions which can be time consuming.
