$\theta_1,\theta_2$ 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.

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    $\begingroup$ Please share your thoughts. Add the self-study tag and read the tag wiki. $\endgroup$ – 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:

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

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

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