Consider points $A=(0,1),\:B=(0,0),\:C=(1,0)$. We choose a random point $(X,Y)$ on $\overline{AB} \cup \overline{BC}$ i.e. the union of the two line segments that "connect" at the origin. Find $Cov(X,Y)$.
I'm not sure if I'm interpreting it right:
1) It's asking for the "L" shape and not a region like a $1\times 1$ square. If this is the case, then $X \sim Unif(0,1)$ and $Y \sim Unif(0,1)$. Then would it be a matter of finding $X+Y$?
OR
2) The union of both segments forms an angle, a $90^{\circ}$ angle. Since standard deviation is analogous to the pythagorean theorem, then by cosine law, $Cov(X,Y)=0$
Am I misunderstanding the question or approaching it incorrectly?
There's a 1/2 probability we lie on the line segment $\overline{AB}$, in which case $X = 0$ and $Y \sim \text{Unif}(0, 1)$, and there's a 1/2 probability we line on line segment $\overline{BC}$, in which case $X \sim \text{Unif}(0,1)$ and $Y = 0$. So, $(X, Y)$ follows from the following mixture distribution: \begin{align*} (X, Y) \sim \frac{1}{2}(0, \text{Unif}(0, 1)) + \frac{1}{2}(\text{Unif}(0, 1), 0) \end{align*} If we let $B \sim \text{Ber}(\frac{1}{2})$ be a random variable denoting which component from the mixture we are sampling from, then \begin{align*} \text{Cov}(X, Y) &= \mathbb{E}[\text{Cov}(X, Y|B)] + \text{Cov}(\mathbb{E}(X|B), \mathbb{E}(Y|B)) \\ &=0 + \text{Cov}\left(\frac{1}{2}B, \frac{1}{2}(1-B)\right) \\ &= -\frac{1}{16} \end{align*}