Let X and Y be random variables with a joint probability density function given by: f(x,y) = (x+y)/1000 for 0 <= x < 10; 0 <= y < 10.
Find P{X+Y<4}.
So to do this I've got my double integral set up with the outer integral being taken from 0 - 10 in terms of dy and the inner my lower bound is 0 however I'm not sure whether my upper bound should be 4-X or 10-X, and I'm taking this integral in terms of dx. I think it should be 10-X however I'm not 100% sure.
Any help is appreciated!
Your joint PDF is given by
$$ f(x,y) = \frac{x+y}{1000}~\mathbf 1_{(x,y)\in[0;10]^2} $$
And the probability in question is
$$ P(x+y < 4) = P(x < 4-y) $$
Using what is above, you can set up the integral
$$\begin{align} P(x<4-y) & = \int_0^{10}\int_0^{4-y} f(x,y) \mathrm{d}x \mathrm{d}y \\[1ex] & = \int_0^{4}\int_0^{4-y} \left( \frac{x+y}{1000} \right) \mathrm{d}x \mathrm{d}y \end{align}$$