3
$\begingroup$

How would you solve the below

Let X and Y be random variables with joint probability density function (PDF) given by:

$$f_{X,Y}(x,y) = \begin{cases}cxy &: x \ge 0, y \ge 0, x + y \le 2\\ 0&: \text{otherwise}\end{cases}$$

(a) Find $c$.

(b) Find $\Bbb E [XY ]$.

Thanks in advance for your assistance.

$\endgroup$
  • 2
    $\begingroup$ Seems like a homework question! tell us what you've figured out so far and where you are stuck. $\endgroup$ – Ehsan M. Kermani Oct 11 '13 at 22:24
  • 2
    $\begingroup$ Hello, welcome to Math.SE. Thank you for your question! It is good practice on this site to add a bit of information on the context your question came up in, and to share your own work on it. It's also fine if you state that you're completely lost -- the information is helpful for answerers to gauge their answer. For more information on asking a good question on this site, see here. $\endgroup$ – Lord_Farin Oct 11 '13 at 22:51
  • 1
    $\begingroup$ Also, to get to know how I nicely typeset your question, and other information about writing maths at this site, see e.g. here, here, here and here. $\endgroup$ – Lord_Farin Oct 11 '13 at 22:55
  • $\begingroup$ Please read the FAQ and learn the proper etiquette for posting here. $\endgroup$ – Stefan Smith Oct 12 '13 at 0:58
2
$\begingroup$

Draw a picture. Let $T$ be the triangle with corners $(0,0)$, $(2,0)$, $(0,2)$. We want to choose $c$ so that $$\iint_T cxy \,dy\,dx=1.$$ To evaluate the integral, express it as the iterated integral $$\int_{x=0}^2\left(\int_{y=0}^{2-x} cxy\,dy\right)\,dx.$$

For $E(XY)$, we want $$\iint_T (xy)(cxy)\,dy\,dx.$$

$\endgroup$
  • $\begingroup$ Thank you... Now I can see where I went wrong... in solving I find that $\endgroup$ – Eric Oct 12 '13 at 1:11
  • $\begingroup$ @Eric: You are welcome. But your comment got cut off, probably inadvertently pressed the Enter key. It is a flaw in the system, used to get me a lot. So I do not know what number you got. $\endgroup$ – André Nicolas Oct 12 '13 at 1:18
  • $\begingroup$ In solving for C I find that C=1 , However I am at a lost of how to solve for the numerical value of E(XY). Since that are dependent variables would I need to solve for them separately?? Would I evaluate on the same integral as C was evaluated. $\endgroup$ – Eric Oct 12 '13 at 1:25
  • $\begingroup$ The answer says what to do. Integrate $xyf_{X,Y}(x,y)$ over the plane, which comes down to integrating $(xy)(cxy)$ over the triangle. By the way, I do not get $1$ for $c$. The first integral is $cx(2-x)^2/2)=c(2x-2x^2 +x^3/2)$. Integrating with respect to $x$ from $0$ to $2$ gives $2c/3$, so $c=3/2$. Do check your work, and please note that I could have made an arithmetical error. $\endgroup$ – André Nicolas Oct 12 '13 at 1:32
  • $\begingroup$ can you please inform where the 2c/3 derived from ... when I integrate for c I keep coming up with 2c/2 which would make c= 1 $\endgroup$ – Eric Oct 12 '13 at 2:36

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.