0
$\begingroup$

Let $X$ and $Y$ be random variables with joint PDF $f(x,y) = 2$ for $0 ≤ x ≤ y ≤ 1$, and $0$ elsewhere.

a) Find $Pr(X +Y ≤ 1)$

b) Find covariance of $X$ and $Y$

I'm not really sure how to find that probability...the bounds are confusing me a bit. Similar with the covariance. Are the bounds just the $1\times1$ square?

$\endgroup$
0
$\begingroup$

The joint support, $\{(x,y):0\leq x\leq y\leq 1\}$, is a right-triangle.   $\triangle(0,0)(0,1)(1,1)$ to be precise.

As they are jointly uniform distributed, the probability may be found directly by geometry -- just sketch a graph of where $y\leq 1-x$ intercepts the triangle (a smaller triangle) and compare.   You might also use this to determine bounds for an integration, for practice.


For the other part, the covariance will be determined by integrating the relevant functions over the given support. $$\displaystyle\int_0^1\!\!\int_0^y \ldots~\mathrm d x\mathrm d y$$

$\endgroup$
  • $\begingroup$ So, to find P(X+Y≤1) could my X-bounds of integration be 0 and 1 and Y-bounds be 0 and 1-x? $\endgroup$ – Liza Jun 27 '18 at 14:41
  • $\begingroup$ The probability is measured over: $0\leq x\leq 1$ , $0\leq y\leq 1-x$, and $x\leq y\leq 1$ . The intersection of the event description and the support. $\endgroup$ – Graham Kemp Jun 27 '18 at 14:47
  • $\begingroup$ ALso, when I calculate cov(X,Y), I am getting a negative number. I have EX=2/3, EY=1, EXY=1/4, so Cov(X,Y)= -5/12 $\endgroup$ – Liza Jun 27 '18 at 14:47
  • $\begingroup$ So, I get a probability of 1 when I integrate with thouse bounds, is that correct? $\endgroup$ – Liza Jun 27 '18 at 14:49
  • $\begingroup$ No, the correlation is positive because $\mathsf EX$ and $\mathsf EY$ are smaller than that. (Reality check: do you expect $Y$ to lie on the edge of the support?) $\endgroup$ – Graham Kemp Jun 27 '18 at 14:50

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.