Hi, I have the following problem.

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

Find c.

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

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

so $\displaystyle c = \sqrt{8}$

But according to the solution manual $\displaystyle c = 2$

Can anyone point out where I have gone wrong?

Thanks in advance. MD.