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I have little problem to show that $R^{2}$ and $\theta$ are independent in marsaglia's method and furthermore $R^{2}\sim U(0,1)$ and $\theta\sim U(0,2\pi)$.

For the first method (Box & Muller) take two random variable $U_{1}$ and $U_{2}$ such that they're iid $U(0,1)$, where:

$$X=\sqrt{-\log(U_{1})}\cos(2\pi U_{2})$$ $$Y=\sqrt{-\log(U_{1})}\sin(2\pi U_{2})$$

Then $X$ and $Y$ will be random variable indp.$N(0, 1)$, the problem of the method is the function $\sin$ and $\cos$, so for this problem, Marsaglia, Bray and MacLauren, propose taking two variables iid, in this case $U(-1,1)$, and generated this random normal variable. We note that,

$$U\sim U(0,1)\implies V\equiv 2U-1\sim U(-1,1)$$

We consider $V_{1}$ y $V_{2}$ r.v. iid $U(-1,1)$, then we can define a point in the plane of coordinate $(V_{1}, V_{2})$, with polar coordinate $(R,\theta)$, as:

$$R^{2}=V_{1}^{2}+V_{2}^2\mbox{ , }\tan(\theta)=\dfrac{V_{2}}{V_{1}}$$

Each point generate random will be inside of the square of side 2, and the point will be refused if is out of the circle of radius 1 (inside of the squared). I hope this idea will be understand, thanks a lot!!

user123456789
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  • Are you saying that the final result is independent, after you have performed enough rejections to get a result inside the circle? If so, after you have done that, you have the uniform distribution on the disk (you had the uniform distribution on the square, and rejection gave you the restriction of that to the disk). It is not hard to see that the uniform distribution on the disk factorizes between $r$ and $\theta$, this is essentially just what the change from rectangular coordinates to polar coordinates says. – Ian May 08 '16 at 21:46
  • The important thing is that the polar coordinate transformation turns $dx dy$ into $r dr d \theta$ which can be viewed as $\frac{1}{2} d(r^2) d \theta$, which is why the result says that $r^2$ and $\theta$ are independent instead of saying that $r$ and $\theta$ are independent. – Ian May 08 '16 at 21:48

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Initially, $(V_1,V_2)$ are uniform on the square. After rejection, they become uniform on the disk. The uniform distribution on the disk is the usual Lebesgue measure "$dx dy$" where the domain is the disk. Upon transformation to polar coordinates you get $r dr d \theta=\frac{1}{2} d(r^2) d \theta$ on a rectangular domain. This "factorized" form of the density and the domain tells us that $r^2$ and $\theta$ are independent. (In fact $r$ and $\theta$ are also independent, but this is easier to prove by first proving the above.)

Ian
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