The original question
Let $X$, $Y$ be random variables, $\Omega$ be the set of all possible outcomes, and $A$ be an event, such that $X \sim U_{[0,1]}$, $Y \sim U_{[0, 1-x]}$, $A = \{(x,y) \in \Omega : x,y < \frac{1}{2} \}$, and $\Omega = \{(x,y): x,y \geq 0, x+y \leq 1 \}$.
Find $\mathbb{E} (1-x-y\mid A)$.
My textbook's solution
They found $f(x,y)$ to be $\frac{1}{1-x}$ and $\mathbb{P}(A)$ to be $\frac{1}{2} \ln(2)$ by double integrating $f(x,y)$.
Then, they did $\mathbb{E} (1-x-y\mid A) = \frac{1}{\mathbb{P}(A)} \int_{A} (1-x-y) f(x,y)\, dx\, dy$ to find the solution.
My question
I'm not sure how they came up with the above integral to find the conditional expectation? I haven't seen this formula before. Any explanation of this would be appreciated!
A conditional $\mathbb{E}\big(h(z)\,\big|\,A\big)$ is computed just a like a normal expectation, but we disregard any outcome $z \notin A$.
Now let $f$ be the original density function. If we look just at the set of outcomes $A$, $f$ is not a density anymore, because we won't have $\int_A f(z)\,dz = 1$. So we must adjust $f$ by re-scaling it to ensure that the integral over $A$ is one. Instead of $f$, we must thus use the density $$ f_A(z) = \frac{1}{\int_A f(z)\,dz} f(z) = \frac{1}{\mathbb{P}(A)} f(z), $$ since this guarantees that $\int_A f_A(z)\,dz = \frac{\int_A f(z)\,dz}{\int_A f(z)\,dz} = 1$.
Using the modified density $f_A$, and the usual formula for the expectation $\mathbb{E}\, h(z)$, we then find for the conditional expectation $$ \mathbb{E}\big(h(z)\,\big|\,A\big) = \int_A h(z)f_A(z)\,dz = \frac{1}{\mathbb{P}(A)} \int_A h(z)f(z)\,dz. $$