How can I prove that X and Y are standard normal variables?

Question

If $\theta$ has $U(0,2\pi)$ distribution and $r^2$ has $\operatorname{Exp}(\frac{1}{2})$ distribution, show that $X=r\cos\theta$ and $Y=r\sin\theta$ are independent and identically distributed $N(0,1)$ variables. I know the joint distribution of $r$ and $\theta$, but when I do the transformation from $(r,\theta)$ to $(X,Y)$, I’m getting the Jacobian as $\frac{2}{r}$. Shouldn’t it be $\frac{1}{r}$ ?

Answer 1

The Jacobian matrix of $(X,Y)$ to $(r,\theta)$ is

$$ \begin{pmatrix} \cos \theta & \sin \theta \\ - r \sin \theta & r \cos \theta \end{pmatrix} $$

which has determinant $r$. Therefore the inverse transformation has determinant $1/r$. (It's easier to compute the Jacobian of the inverse than to compute the Jacobian.)

Although, since it is $r^2$ which is exponential distributed, you might be interested in the transformation $(r^2, \theta) \to (X, Y)$. This will be the product of the Jacobians $(r^2,\theta) \to (r,\theta)$ and $(r,\theta) \to (X, Y)$. The Jacobiam matrix of $(r^2,\theta) \to (r,\theta)$ is

$$\begin{pmatrix} 2r & 0 \\ 0 & 1 \end{pmatrix}.$$

So the total Jacobian from $(r^2, \theta) \to (X, Y)$ is $2r/r = 2$.

Finally, we can transform the PDF to obtain

$$ \frac{1}{2\pi}\left( \frac12 \exp(-\tfrac12 r^2) \right) \to \frac{1}{2\pi} \exp[-\tfrac12 (x^2 + y^2)]. $$