Let $X$ and $Y$ be random losses with joint density function $$f(x, y) = 2x \;\; 0<x, y<1$$ An insurance policy is written to cover the loss $X+Y$. The policy has a deductible of $1$. Calculate the expected payment under the policy.
My attempt:
Let $Z$ denote the payment under this policy. $$Z=\begin{cases} 0 \; \;\;\; \text{if}\; 0<x+y<1, 0<x, y<1\\ X+Y-1 \; \text{if}\; 1\le x+y\le 2, 0<x, y<1 \end{cases}$$
$E[Z] = 0+ E[X+Y-1\mid 1-x<y<1] \cdot P[1-x<y<1]$.
Note that $E[X+Y-1\mid 1-x<y<1] = \int_0^1\int_{1-x}^1 (x+y-1)\cdot 2x \; \; \text{dy dx} = \frac{1}{4}$ and $P[1-x<y<1] = \int_0^1\int_{1-x}^1 2x \; \; \text{dy dx} = \frac{2}{3}$.
This is not the correct solution. The correct solution only computed $E[X+Y-1\mid 1-x<y<1]$ using the same formula as the one above but did not consider $P[1-x<y<1]$. Why is that approach correct (and my approach incorrect)?
Your calculation of the conditional expectation is actually the complete calculation because in fact, $$\operatorname{E}[X+Y-1 \mid 1-X < Y < 1] = \frac{\int_{x=0}^1 \int_{y=1-x}^1 (x+y-1) 2x \, dy \, dx}{\Pr[1 - X < Y < 1]}.$$ This is because in order for the conditional expectation to be conditioned, it must be divided by the probability of observing the event $1 - X < Y < 1$.
Another way to see this is to write $$Z = \max(0, X+Y-1)$$ hence $$\begin{align} \operatorname{E}[Z] &= \int_{x=0}^1 \int_{y=0}^1 \max(0, x+y-1) 2x \, dy \, dx \\ &= \int_{x=0}^1 \int_{y=0}^{1-x} 0 \, dy \, dx + \int_{x=0}^1 \int_{y=1-x}^1 (x+y-1) 2x \, dy \, dx. \end{align}$$