0
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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*}

$\endgroup$
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.