# Let Q be a random point in the unit circle. Let Z=distance from Q to (-1,0) and T the distance from (1,0). What is Cov(Z,T)?

The question hints that we need to use angles since there is no joint denstity, i.e.:

Let $T=h(\theta)$ and $Z=g(\theta)$

I know in general that:

$$Cov(X,Y)= E[XY]-E[X]E[Y]$$

But I'm not sure how to use the information they gave me for this question.

• The coordinate of $Q$ is $(\cos\Theta, \sin\Theta)$ and hence you can express $T, Z$ as a function of $\Theta$.The distribution of $\Theta$ maybe assumed as uniform on $[0, 2\pi]$, I guess as it is not specified in the question. – BGM Jun 1 '16 at 16:04
• Is this ON or IN the unit circle? Does it include the points having radius < 1? – Felix Castor Jun 1 '16 at 16:39
• @FelixCastor on the unit circle – user300011 Jun 1 '16 at 16:41
• You should update your question to include that. – Felix Castor Jun 1 '16 at 16:41
• If IN the unit circle, the answer involves elliptic integrals. – Jack D'Aurizio Jun 1 '16 at 16:47

Let $L=(-1,0),R=(1,0)$, $O=(0,0)$, the center, and $M$ the random point on the circle. WLOG assume $M$ is on the upper semicircle. Note the picking the point uniformly is equivalent to sampling $\theta =\angle ROM$ uniformly over $[0,\pi]$.
Write $X$ for the length of $LM$ and $Y$ for the length of $RM$.
1. Observation: $\angle LMR$ is a right angle. Indeed, let $\alpha = \angle RLM$ and $\beta = \angle LRM$. Since the triangles $MOL$ and $MOR$ are each an isosceles, it follows that $\angle LMR=\alpha+\beta$. But then $$180^{\circ} = \angle RLM+\angle LRM+ \angle LMR=\alpha +\beta + (\alpha+\beta) = 2 (\alpha+\beta).$$
2. Corollary: $XY$ is twice the area of the triangle $LMR$.
3. Twice area of $LMR$ is also given by the length of its base $LR$ times its height. This is easy to compute. Base is $LR$ and has length $2$, and the height is equal to $\sin \theta$. Thus, $$E[XY ] =2 E[ \sin \theta ] =2 \frac{1}{\pi} \int_0^{\pi}\sin (\theta)d \theta = \frac{4}{\pi}.$$
4. Next find $E[X]$ and $E[Y]$. Because of symmetry, $E[X]=E[Y]$. By looking at the triangle $MOR$, we see that $Y= 2 \sin (\theta/2)$. Therefore, $$E [ Y ] = \frac{2}{\pi} \int_0^\pi \sin (\theta /2) d \theta = \frac{4}{\pi}\int_0^{\pi/2}\sin (t) dt = \frac 4 \pi.$$
$$\mbox{Cov} (X,Y) = E [XY] - E[X] E[Y] = \frac{4}{\pi} (1- \frac{4}{\pi}).$$