SOLUTION: 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)

Algebra ->  Probability-and-statistics -> SOLUTION: 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)      Log On


   

Question 1023848: IF X and Y are two random variable having joint density function f(x,y)= 1/8(6-x-y) 0
Answer by mathmate(429) About Me  (Show Source):
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Question:
IF X and Y are two random variable having joint density function f(x,y)= 1/8(6-x-y) 0
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.
∫∫f(x) dx dy over 0≤x≤2 and 2≤y≤4
=(1/8)∫[10-2y]dy for 2≤y≤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≤x≤1, 2≤y≤3} to get
∫∫f(x) dx dy over 0≤x≤1 and 2≤y≤3
=(1/8)∫[6-1/2-y]dy over 2≤y≤3
=(1/8)[5.5-4.5+2]
=3/8