Calculate expected values E(x) & E(y) & variance of x & y of joint PDF, which was previously transformed from Polar to Cartesian

Question

Given two independently uniform distributed random variables angle $\theta \in [0,2\pi]$ and radius $r \in [0,1]$.

I obtain for the joint density function with polar coordinates: $$ f_{r,\theta}(r,\theta)= \frac{1}{2\pi} $$

and thus for the transformed density function in cartesian coordinate system (with multiplying with the Jacobian determinant $\frac{1}{r}$): $$ f_{x,y}(x,y)= \frac{1}{2\pi \sqrt{x^2+y^2}} $$

Now to calculate the expected values and variances of x & y, how do I set the integral limits of the marginal distribution? Does the following make sense? $$ f_x(x)=\int^{\sqrt{1-x^2}}_0 \frac{1}{2\pi \sqrt{x^2+y^2}} dy $$

Thanks.

Answer 1

A generalization brings out the essence of the problem.

Suppose the radius $r$ has a distribution function $F_R$ (supported on $[0,\infty)$) and the angle $\theta$ remains uniformly distributed. That uniformity implies the distribution of $(r,\theta)$ is circularly symmetric. In particular,

1. The distribution of $x$ and the distribution of $-y$ are the same (rotate by $\pi/2$).

2. The distributions of $x$ and $-x$ are the same (rotate by $\pi$).

Because these imply $E[x]=E[-y]=E[-x]=E[y],$ all these expectations are either $0,$ infinite, or undefined. Let's assume they are zero, as in the question. Thus

$$\operatorname{Var}(x) = E[x^2]-E[x]^2 = E[x^2] = E[y^2] = E[y^2] - E[y]^2 = \operatorname{Var}(y)$$

and

$$\operatorname{Cov}(x,y) = E[xy]-E[x]E[y] = E[xy] = E[(-y)x] = -E[xy].$$

Assuming the variances are finite, this implies the covariance is zero. Moreover, expressing $x=r\cos\theta$ and $y=r\sin\theta$ in polar coordinates, compute

$$\operatorname{Var}(x) + \operatorname{Var}(y) = \int_0^\infty \int_0^{2\pi} \left[(r\cos\theta)^2 + (r\sin\theta)^2\right]\frac{1}{2\pi}\mathrm{d}\theta\,\mathrm{d}F(r) = \int_0^\infty r^2 \,\mathrm{d}F(r).$$

Since the variances are equal, the solution is

$$\operatorname{Var}(x) = \operatorname{Var}(y) = \frac{1}{2}\int_0^\infty r^2 \,\mathrm{d}F(r) = E\left[r^2/2\right].$$

(A similar analysis is effective for spherically symmetric distributions with more variables: all variances will be equal and computing the sum of the variances is simple to do because it reduces to an ordinary integral over the radial coordinate; all the angular coordinates disappear.)

In the question $\mathrm{d}F(r) = I(0\le r\le 1)$ yielding

$$\frac{1}{2}\int_0^\infty r^2 \,\mathrm{d}F(r) = \frac{1}{2}\int_0^1 r^2\,\mathrm d r = \frac{1}{2}\left(\frac{1}{3} - \frac{0}{3}\right) = \frac{1}{6}.$$

This will help you answer your original question concerning the limits of integration in Cartesian coordinates: if your limits are correct, you should obtain the same results.