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 $X+Y$?


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*}

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    $\begingroup$ Is the Bernoulli part necessary? With the mixture distribution you gave doesn't it just follow that $E[XY]-E[X]E[Y]=0-(.25)\cdot(.25)=-.0625$. $\endgroup$ – Tomás Palamás Apr 29 at 21:16
  • $\begingroup$ @TomásPalamás Yes, both approaches are valid. Your's is a bit more direct, though. $\endgroup$ – Graham Kemp Apr 29 at 23:30

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