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Suppose $U$ and $V$ are independent random variables with density $f(u)$ and $g(v)$ respectively. The domain of $U$ is the interval $(0, 1)$ and the domain of $V$ is $v > 0$. After the transformation $$ X = V \sin(2 \pi U) $$ and $$ Y = V \cos(2 \pi U). $$

$X$ and $Y$ are independent, each following the standard normal distribution $N(0, 1)$.

(a) How can I find $f(u)$ and $g(v)$?

(b) Then, how do I show how to generate a normal random variable from uniform distribution without having to do integration of normal density function?

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I suspect that $2U$ should be replaced by $2\pi U$. –  sos440 Oct 30 '12 at 6:27
    
Homework? $ $ $ $ –  Did Oct 30 '12 at 6:47

1 Answer 1

up vote 1 down vote accepted

Here is a hint for (a). To compute $g$ is to be able to write, for every function $h$ measurable and bounded, $$ \mathbb E(h(V))=\int h(v)g(v)\mathrm dv. $$ But $V=\sqrt{X^2+Y^2}$ and $X$ and $Y$ are i.i.d. with density $\gamma:x\mapsto\mathrm e^{-x^2/2}/\sqrt{2\pi}$, hence $$ \mathbb E(h(V))=\mathbb E(h(\sqrt{X^2+Y^2}))=\iint h(\sqrt{x^2+y^2})\,\gamma(x)\gamma(y)\,\mathrm dx\mathrm dy. $$ The task is to write the RHS of the second displayed equation as the RHS of the first, for some function $g$. To do so, a change of variable seems mandatory. One of the new variables could be the radius $v=\sqrt{x^2+y^2}$, obviously, the other one could be the angle.

Thus, one could consider $x=v\cos\theta$ and $y=v\sin\theta$, which yields $v\geqslant0$, $0\leqslant\theta\lt2\pi$, and $\mathrm dx\mathrm dy=v\mathrm dv\mathrm d\theta$, then one would want to reach $$ \iint h(v)\,\gamma(v\cos\theta)\gamma(v\sin\theta)\,v\mathrm dv\mathrm d\theta=\int h(v)g(v)\mathrm dv. $$ Can you take it from here?

Once you will have computed the density $g$ and done the same for the density $f$, you will want to realize $U$ and $V$ from uniform random variables, that is, to find some functions $a$ and $b$ such that $a(W)$ is distributed like $U$ and $b(W)$ is distributed like $V$, for $W$ uniform on $(0,1)$. The function $a$ is trivial, $b$ a little less so...

And once all this will be done, you might want to have a look here and here.

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