I have the following table of random values $x_1$ and $x_2$
I need to compute $\mathbb{E}(x_2\mid (x_1 - 1)^2 \ge 1)$. I have done it. However, there is some doubt.
$$ \begin{align} \mathbb{E}(x_2\mid (x_1 - 1)^2 \ge 1) &= \mathbb{E}(x_2\mid x_1 = -1\parallel x_1 = 0 \parallel x_1 = 2 ) =\\ & = \sum_j \mathbb{E}(x_2\mid x_1 = b_j ) P (x_1 = b_j ) = \\ &= \sum_j \left( \sum_i a_i P(x_2 = a_i \mid x_1 = b_j)\right) P (x_1 = b_j ) =\\ &= \sum_j \left( \sum_i a_i \frac{P(x_2 = a_i, x_1 = b_j)}{P (x_1 = b_j )}\right) P (x_1 = b_j ) = \\ &= \sum_j \sum_i a_i P(x_2 = a_i, x_1 = b_j) = \\ &= \sum_i a_i \sum_j P(x_2 = a_i, x_1 = b_j) \end{align} $$ where $\parallel$ is logic ''OR''
And that's all. However, I am almost sure that I need to get $P(x_1 = b_j)$ in the final expression and I don't like that this value just got reduced. Do I have mistake here?
Edit: I don't understand where I have problem with my soution. Is it correct: $$\mathbb{E} (A \mid B \parallel C \parallel D) = \mathbb{E} (A \mid B) P (B) + \dots $$ and Is it correct? $$P(A \mid B \parallel C \parallel D ) = P (A \mid B) + P (A \mid C) + \dots$$
I don't really understand what you are doing.
You should simply go for finding: $$\mathsf E(x_2\mid(x_1-1)^2\geq1)=$$$$-2\mathsf P(x_2=-2\mid(x_1-1)^2\geq1)-\mathsf P(x_2=-1\mid(x_1-1)^2\geq1)+2\mathsf P(x_2=2\mid(x_1-1)^2\geq1)$$
Here e.g.: $$\mathsf P(x_2=-2\mid(x_1-1)^2\geq1)=\mathsf P(x_2=-2\wedge (x_1-1)^2\geq1)/\mathsf P((x_1-1)^2\geq1)=$$$$(0.05+0.2+0.15)/(0.3+0.28+0.21)$$