# Thread: Silly mistake somewhere in integration?

1. ## Silly mistake somewhere in integration?

Hi, I have the following problem.

Suppose that the random variables X and Y vary in accordance with the joint pdf $f_{X, Y}(x, y) = c(x + y), 0 < x < y <1$
Find c.

As I see it then we can say $F_{X, Y}(x, y) = \int_{0}^1 \int_{0}^y \! c(x + y) \, \mathrm{d}x \, \mathrm{d}y = 1$

Then separating the variables $F_{X, Y}(x, y) = \int_{0}^1 cy \int_{0}^y \! cx \, \mathrm{d}x \, \mathrm{d}y = 1$

So $F_{X, Y}(x, y) = \int_{0}^1 \! cy [{c\over2} y^2] \, \mathrm{d}y = \int_{0}^1 \! {c^2\over2} y^3 \, \mathrm{d}y = 1$

So, ${c^2 \over 8} = 1$

so $c = \sqrt{8}$

But according to the solution manual $c = 2$

Can anyone point out where I have gone wrong?
$F_{X, Y}(x, y) = \int_{0}^1 cy^2$ $\mathrm{d}y + \int_{0}^1 \int_{0}^y \! cx \, \mathrm{d}x \, \mathrm{d}y = 1$