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You are given two independent random variables: $W \sim \mathrm{Exp}(1)$, $Q \sim U([0; 2\pi ])$.
Also, $a$ is a constant, chosen from $[-\pi/2; \pi/2]$.

You build following random variables, based on $R, W, Q, a$:
$R = \sqrt{2W}$,
$U = R \cos Q$,
$V = R \sin (Q + a)$.

The task is to prove that vector $(U, V)$ has bivariate normal distribution with correlation $\rho = \sin a$.

It is almost obvious that both $U$ and $V$ have zero expectation and odd variance:
$EU = EV = 0$, ${\sigma}_{U} = {\sigma}_{V} = 1$,
so it's not very difficult to prove, that correlation of $U$ and $V$ is $\sin a$. Though, I find it hard to prove, that the vector has normal distribution, so I'd really appreciate your help.

P.S.: Don't you find it amazing, that we can construct normal distribution based on exponential and uniform ones? :)

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    Search for Box-Muller method.2011-12-20
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    You can generate from two uniform distributions with $W$ uniform on $(0,1]$ and $R=\sqrt{-2 \log_e W}$2011-12-20
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    Yep, Box-Muller transformation is very similar to the task described above. Still, Box-Muller generates independent normally distributed values (they correspond for a=0), and I am searching for a provement for any other "a" from the interval.2011-12-20
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    If you proved already that for $a=0$ the distributions of $U$ and $V$ are normal, then you can easily show that it is also the case for $a\ne 0$. For $U$ it is the exactly the same calculation (it is independent of $a$!). For $V$ you might use $\sin(Q+a)=\cos(Q)\sin(a)+\sin(Q)\cos(a)$ So $V$ is a sum of two normal variables, which are independent (because you showed that for $a=0$, and therefore also a normal variable.2011-12-20
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    Really, it is. Thanks for help!2011-12-20

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