# How to relate the distributions of these trigonometric functions of uniform variables?

$$\theta_1,\theta_2$$ are two independent random variables distributed uniformly in $$[0,2π)$$. Let $$X=\cos\theta_1,Y=\cos\theta_2.$$ Prove that $$\frac{X+Y}{2}$$ and $$XY$$ are equal in distribution,$$\frac{X+Y}{2}{\buildrel d \over =}XY$$.

I can calculate the cdf of $$X,Y$$, like PDF of cosine of a uniform random variable, but it's difficult to prove the distribution equality: the distribution function is too hard to calculate.

• Please share your thoughts. Add the self-study tag and read the tag wiki. – StubbornAtom Dec 6 '19 at 10:21

We can avoid calculating distributions altogether.

In the following I will use "$$\sim$$" to mean "has the same distribution as" while retaining "$$=$$" to mean strict equality of random variables, as usual.

The cosine is a periodic function with period $$2\pi.$$ A periodic function $$f$$ with period $$\tau$$ simply is one that satisfies $$f(x+\tau)=f(x)$$ for all $$x.$$ Equivalently, $$f$$ is a function of the numbers modulo $$\tau$$, meaning that before applying $$f$$ to any number $$x,$$ $$x$$ is replaced by its remainder after division by $$\tau.$$

Periodic functions of a uniform random variable $$\Theta$$ supported on $$[0,\tau)$$ generally enjoy some useful properties we might exploit to simplify our work. These include

1. $$f(\Theta+\omega) \sim f(\Theta)$$ for fixed $$\omega.$$

2. $$f(-\Theta) \sim f(\Theta).$$

3. When $$\Theta_1$$ and $$\Theta_2$$ are independent uniform random variables, $$(1)$$ implies $$f(\Theta_1+\Theta_2) \sim f(\Theta_1)$$ and then $$(2)$$ implies $$f(\Theta_1-\Theta_2)\sim f(\Theta_1).$$

4. Moreover, $$\Theta_1\pm\Theta_2$$ are independent when computed modulo $$\tau$$ and therefore the $$f(\Theta_1\pm \Theta_2)$$ are independent mod $$\tau,$$ too.

I leave the (easy) demonstrations to you, since this is a self-study problem.

Specializing to $$\tau=2\pi$$ and $$f=\cos,$$ use these facts and the rules of trigonometry to justify the following reasoning:

\eqalign { XY &= \cos(\Theta_1)\cos(\Theta_2) \\ &= \frac{1}{2}\left(\cos(\Theta_1-\Theta_2)+\cos(\Theta_1+\Theta_2)\right)\\ &\sim\frac{1}{2}(X+Y) .}

• Thank you, it helps a lot. – zhluo Dec 8 '19 at 1:14