Let $U$ and $V$ be two independent random variables uniformly distributed on $(0,1)$ and let $R= \sqrt {-2log (U)}$.

I want to show that $X := R \cos (2\pi V) $ and $Y := R\sin (2\pi V)$ are independent $N (0,1)$ random variables.

My problem is, I can't find the joint density of $X $ and $Y $. Any hint?

I could compute that $R^2$ is Exp($\frac {1}{2}$) distributed, but I don't know if this is helpful.


To get the joint probability distribution for $X$ and $Y$, it's good to know the change of variable formula for multidimensional integrals: $$ dX \ dY = \left| \det \left[ \begin{array}{cc} \frac{\partial X}{\partial R} & \frac{\partial X}{\partial V} \\ \frac{\partial Y}{\partial R} & \frac{\partial Y}{\partial V} \end{array}\right]\right| \ dR \ dV = 2\pi R \ dR \ dV.$$ [I encourage you to check the algebra. See here or here for more info.]

Also, $$ dU = \left| \frac{dU}{dR} \right| \ dR = \exp(-\tfrac {1}{2} R^2) \ R \ dR.$$ [Again, please check the algebra!]

Combining these two observations, we get a formula that tells us how to relate the volume element $dU \ dV$ to the volume element $dX \ dY$: $$ dU \ dV = \tfrac 1 {2\pi} \exp(-\tfrac{1} 2 R^2) dX \ dY = \tfrac 1 {2\pi} \exp(-\tfrac{1} 2 (X^2 + Y^2)) \ dX \ dY .$$

Therefore, if $f(U,V)$ is the probability density function for $U$ and $V$, the probability density function for $X$ and $Y$ must be $$\tfrac 1 {2\pi}\exp(-\tfrac{1} 2 (X^2 + Y^2)) \times f(U(X,Y), V(X,Y))$$ [To see that this is true, you could perhaps think about how we usually compute probabilities by integrating the probability density function over appropriate regions, and look at our above formula for converting $dU \ dV $ to $dX \ dY$.]

But as $(U,V)$ varies over $[0,1]\times [0,1]$, $(X,Y)$ varies over $\mathbb R^2$. [Check this!] And $$f(U,V) = 1$$ for $(U,V)$ in $[0,1]\times [0,1]$. So the joint probability density function for $X$ and $Y$, valid over $\mathbb R^2$ is $$\tfrac 1 {2\pi} \exp(-\tfrac{1} 2 (X^2 + Y^2)). $$

Hopefully you recognise this as the joint distribution for two independent ${\rm N}(0,1)$ variables!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.