# random variables

• Jan 22nd 2009, 04:18 PM
ashimb9
random variables
Suppose $\displaystyle U_{1}$ and $\displaystyle U_{2}$ are i.i.d. U(0,1) random variables. Let $\displaystyle X=\sqrt{-2.log(U_{1})}.cos(2\Pi.U_{2})$ and $\displaystyle Y=\sqrt{-2.log(U_{1})}.sin(2\Pi.U_{2})$
Show that X and Y are i.i.d. N(0,1)
• Jan 23rd 2009, 04:26 AM
Isomorphism
Quote:

Originally Posted by ashimb9
Suppose $\displaystyle U_{1}$ and $\displaystyle U_{2}$ are i.i.d. U(0,1) random variables. Let $\displaystyle X=\sqrt{-2.log(U_{1})}.cos(2\Pi.U_{2})$ and $\displaystyle Y=\sqrt{-2.log(U_{1})}.sin(2\Pi.U_{2})$
Show that X and Y are i.i.d. N(0,1)

If X and Y are iid N(0,1), then try showing $\displaystyle \sqrt{X^2 + Y^2}$ is Rayleigh distributed and the angle $\displaystyle \theta = \tan^{-1} \frac{Y}{X}$ is uniformly distributed. Also note that this means, given any rayleigh, one can construct two i.i.d normally distributed random variables such that their length has the distribution of the given rayleigh, and the angle is unif. distributed.Call this a "decomposition result".

Now it is straight forward (Use Jacobian!) to show that $\displaystyle \sqrt{-2.log(U_{1})}$ is Rayleigh distributed. Now observe that $\displaystyle X = \sqrt{-2.log(U_{1})}.\cos(2\Pi.U_{2}) = R \cos \theta$ where R is rayleigh and the angle is uniformly distributed. Now by decomposition result, we are done.