Question 1023848
 
Question:
IF X and Y are two random variable having joint density function f(x,y)= 1/8(6-x-y) 0<x<2 2<y<4 =0 elsewhere FIND 1)P(x<1, y<3)
 
Solution:
 
When we are given a PDF or JPDF, it is always a good idea to make sure the integral over the domain equals one.  If not, the given information is false.
 
In this given problem,
f(x)=(6-x-y)/8
first find the integral over the domain.
&int;&int;f(x) dx dy over 0&le;x&le;2 and 2&le;y&le;4
=(1/8)&int;[10-2y]dy  for 2&le;y&le;4
=(1/8)[40-16+4-20]
=1
So f(x) is a valid JPDF.
 
Next step is to repeat the same procedure, but limited to the given area of
{0&le;x&le;1, 2&le;y&le;3} to get
&int;&int;f(x) dx dy  over 0&le;x&le;1 and 2&le;y&le;3 
=(1/8)&int;[6-1/2-y]dy  over 2&le;y&le;3
=(1/8)[5.5-4.5+2]
=3/8