Conditional Expectation, discrete: extension?

Question

Original Thread

Given a discrete random variable $\mathbb{X}$ on $\Omega = \{1,2,3\}$ with the following pmf:

$f_{\mathbb{X}}(1) = P(\mathbb{X} = 1) = \frac{1}{3}$

$f_{\mathbb{X}}(2) = P(\mathbb{X} = 2) = \frac{1}{2}$

$f_{\mathbb{X}}(3) = P(\mathbb{X} = 3) = \frac{1}{6}$

Find the following value of the conditional expectation: $\text{E}[\mathbb{X}\space | \space\mathbb{X} \in [1,2]]$

That is, I am wondering about the conditional expectation across a closed continuous interval.

Does the answer change from the previous thread? My initial thought would be no since $\mathbb{X}$ is discrete.

Conditional Expectation Formula:$$\text{E}[\mathbb{X} | \mathbb{Y} = y] = \sum_x xf_{\mathbb{X}|\mathbb{Y}}(x,y)$$

Answer 1

$E(X|A)=\frac{E(X1_A)}{P(A)}$

Conditional_expectation_with_respect_to_an_event

Define $A=\{ 1\leq X \leq 2\}$ so

$$E(X|\{ 1\leq X \leq 2\})=\frac{E(X1_{\{ 1\leq X \leq 2\}})}{P(\{ 1\leq X \leq 2\})}$$

$$=\frac{\sum_{x=1}^{3} x 1_{x\in \{1,2\}} P(X=x)}{P(X\in \{1,2\})}$$

$$=\frac{\sum_{x=1}^{2} x p(X=x)}{\frac{1}{3}+\frac{1}{2}}$$

$$=\frac{1 p(X=1)+2 p(X=2)}{\frac{5}{6}}$$ $$=\frac{1 *\frac{1}{3}+2 *\frac{1}{2}}{\frac{5}{6}}=\frac{\frac{4}{3}}{\frac{5}{6}}=\frac{8}{5}$$

Answer 2

No, the answer is the same, because all the elements of the open interval $(1,2)$ are stated to have probability $0$ of occurring. Thus $$P(\mathbb{X}=1\bigcap\mathbb{X}=2)=P(\mathbb{X}\in[1,2])$$

As Masoud calculated in another answer, the conditional expectation you seek is $\frac85$, which is what it was in the other post.