I've been given two formulas:
Discrete Case: $$\mathbb{E}(X|Y=y)=\sum_{x\in S_{X}} xp_{X|Y}(x|y)$$
and
Continuous Case: $$\mathbb{E}(X|Y=y) = \int_{x \in S_{X}}xf_{X|Y}(x|y)$$
where $S_X$ is the domain for variation for $X$ in both cases.
Just based on this, is it possible to deduce the expectation of $\mathbb{E}(X|A)$ for a continuous random variable $X$ and an event $A$?
I have not been given any other definitions and I need to find this for a question.
I was thinking of instead using an indicator random variable $$\mathbb{1}_A = \begin{cases} 1 \quad \text{If A occurs} \\ 0 \quad \text{otherwise}\end{cases}$$
and then instead trying to find the expectation $\mathbb{E}(X|\mathbb{1}_{A}=1)$. But I'm not sure how to find this since $X$ is continuous but $\mathbb{1}_{A}$ is discrete.
Specifically, I'm trying to find $E(X_1| X_1\le\frac{1}{2}, X_2 \le \frac{1}{2})$ and $E(X_2| X_1\le\frac{1}{2}, X_2 \le \frac{1}{2})$ where $X_1 \sim U(0,1)$ and $X_2|X_1 \sim U(0,X_1)$. I believe I need to know the first the definition I have asked about above.
I am trying just to answer to the final problem avoiding too many general reasonings.
We have $X_1,X_2$ r.v. with joint p.d.f. $p(x_1,x_2)$. If we have $I \subset \mathbf{R}^2$, $A=\{ (X_1,X_2) \in I\}$ and $f(x_1,x_2)$ a scalar function than:
, where the first step is just by definition of conditional measure and the second is simply the computation of the integral using the joint density.
In our case we first find $I$. Since $X_2 \le X_1$ we have that when $X_1\le1/2$ also $X_2\le1/2$, so that $A=\{X_1 \le 1/2\}$ and $I=\{(x_1,x_2)|x_1 \le 1/2\}$.
Now for the exercise:
where $1_A$ are indicator functions.
We have than $p(A)=1/2$ and for example, applying [1] and [2] to $f(x_1,x_2)=x_2$, I find:
$E[X_2|A]=2\int_0^{1/2}dx_1\int_0^{1}dx_2 1_{[0,1]}(x_1)1_{[0,x_1]}(x_2)x_2/x_1=1/8$