I have two jointly distributed variables, X and Y. Their joint density function is x + y whenever 0 <= x, y <= 1.
I'm asked to find P(X + Y < 1). What confuses me here is that I have this huge range outside of 0 <= x, y <= 1 where the "probability" is 0, but the sum of x and y is (for the positive side) always larger than 1. How do I go about accounting for this, and solving this probability in general?
$\mathbb P (X + Y < 1) = \mathbb P( (X,Y) \in A) = \mu_{(X,Y)}(A)$, where:
$\mu_{(X,Y)}(A) = \int_A f_{(X,Y)}(x,y)d\lambda_2(x,y)$, $ A = \{ (x,y) \in \mathbb R^2 : x+y < 1\}$, and $f_{(X,Y)}(x,y) = (x+y)\chi_{[0,1]^2}(x,y)$
Since $x \in [0,1]$ almost surely, then $1-x \ge 0$ almost surely, so we get the bound of $y$ on the set $A$: $y \in (0,1-x)$
So we just need to integrate, $x$ over whole interval $(0,1)$ and $y$ on the interval $(0,1-x)$.
$\mu_{(X,Y)}(A) = \int_0^1 \int_0^{1-x} (x+y) dydx = \int_0^1 |(xy + \frac{y^2}{2})|^{y=1-x}_{y=0} dx = \int_0^1 x(1-x) + \frac{(1-x)^2}{2} dx =$
$= | \frac{x^2}{2} - \frac{x^3}{3} - \frac{(1-x)^3}{6}|^{x=1}_{x=0} = \frac{1}{2} - \frac{1}{3} + \frac{1}{6} = \frac{3-2+1}{6} = \frac{2}{6}= \frac{1}{3}$