View Full Version : P(X + Y < 1/2) Last problem

03-24-2008, 11:57 AM
I thought I had done this question right, but I get a probability of 0.
If the joint probability density of X and Y is given by f(x,y) = { 24xy for 0<x<1, 0<y<1, x+y<1
0 elsewhere
find P(X + Y < 1/2).

I did the integral from 0 to 1/2 of the integral from x to (1/2 - x) of f(x) dy dx. Are my integrals correct? After the first integration, I get the integral from 0 to 1/2 of 12 xy^2 dx evaluated from x to (1/2 - x), which is the integral from 0 to 1/2 of (3x - 24x^3)dx. That equals (3/2x^2 - 6x^4) evaluated from 0 to 1/2 which = 0. What am I doing wrong? Sorry I don't know how to write it out using the actual math symbols.

03-24-2008, 01:23 PM
1) Official Reminder of Fundamental Principle. ALWAYS check things when you are able to do so.

In this case, you can check the ENTIRE distribution to make sure it is all there. If you get something other than unity (1) in your checking, then you have set something up incorrectly. So, show us the integral expression that results in Unity. Then, we can discuss the tiny setup error if you do not see it yourself.

Personally, if I were writing a statistics exam, I would have at least one problem with a fake distribution, just to see how many students buy it. Of course, I would have to be very clear in class that I would expect every student to check every distribution every time. Failure to do so would result in lost credit on the exam and I would not be very friendly toward pleas to the contrary.

03-24-2008, 02:08 PM
Doing that helped a lot. Thank you very much. My new integrals were from 0 to 1/2 and from 0 to (1/2 - x). Thanks!

03-24-2008, 05:26 PM
Good work. You'll never skip the checking step again, right?

03-24-2008, 07:14 PM
:) Not planning on it.