# Math Help - Joint Probability

1. ## Joint Probability

Hi,

I need help integrating this problem.

Annie and Alvie have agreed to meet for lunch between noon (0:00 pm) and 1:00 pm. Denote Annies arrival time by X, Alvies arrival by Y, and suppose X & Y are independent with pdfs

f(x) = 3x^2 for 0<=x<=1 or 0 otherwise
f(y)= 2y for 0<=y<=1 or 0 otherwise

What is the expected amount of time that the one who arrives first must wait for the other person?

Since independence is observed, the cdf is given by f(x,y) = f(x) * f(y)
which is f(x,y) = 6yx^2

To find expected value the formula is

let S = integration sign
and
|x-y| = h(X,Y)

E[h(X,Y)] = S (from 0 to 1) S (from 0 to y) (x-y)(6yx^2) dydx + S (from 0 to 1) S (from y to 1) (x-y)(6yx^2) dydx

Each time I integrate the first double integral I get a negative area. Any help would be appreciated.

2. Originally Posted by shogunhd
Hi,

I need help integrating this problem.

Annie and Alvie have agreed to meet for lunch between noon (0:00 pm) and 1:00 pm. Denote Annies arrival time by X, Alvies arrival by Y, and suppose X & Y are independent with pdfs

f(x) = 3x^2 for 0<=x<=1 or 0 otherwise
f(y)= 2y for 0<=y<=1 or 0 otherwise

What is the expected amount of time that the one who arrives first must wait for the other person?

Since independence is observed, the cdf is given by f(x,y) = f(x) * f(y)
which is f(x,y) = 6yx^2

To find expected value the formula is

let S = integration sign
and
|x-y| = h(X,Y)

E[h(X,Y)] = S (from 0 to 1) S (from 0 to y) (x-y)(6yx^2) dydx + S (from 0 to 1) S (from y to 1) (x-y)(6yx^2) dydx

Each time I integrate the first double integral I get a negative area. Any help would be appreciated.
|x - y| = y - x for x < y. So the bit in red should be y - x for those integral terminals. The second integral should have x - y, which it does.