Question about Conditional Expectation from One Variable

Question

I'm having a bit of trouble understanding how to solve this question in my textbook.

It says to allow X to have the following distribution:

Suppose you are then told that $X \geq 2$. Record a prediction of a future value of X that uses this information.

I know that I need the expectation, but I was struggling a bit to find $\text{E}(X|X \geq 2)$. $$\text{E}(X|X \geq 2) = \sum_x{x\Pr(X=x|X \geq 2)}$$ and $$\Pr(X=x|X \geq 2) = \frac{\Pr(X=x \cap X\geq2)}{\Pr(X \geq 2)} \\ \Pr(X=x \cap X\geq2) = \frac{1}{2} \\ \Pr(X \geq 2) = \frac{1}{2}$$ so $$\text{E}(X|X \geq 2)= 8$$ which doesn't seem right at all.

edit: I'm really not sure how I got 8 lol!

Answer 1

The probability that $X=x$ and $X\geq 2$ is not a flat $1/2$.   It depends on what value of $x$ we have.

Clearly $\mathsf P(X=1\cap X\geq 2)=0$ but for all $x\geq 2$ there $\mathsf P(X=x\cap X\geq 2)=\mathsf P(X=x)$

$$\begin{array}{|c|c|}\hline x & 1 & 2 & 3 & 4 \\\hline\mathsf P(X=x\cap X\geq 2) & 0 & 1/4 & 1/8 & 1/8 \\\hline \end{array}$$

$$\mathsf E(X\mid X\geq 2) =\dfrac{2\cdot 1/4+3\cdot 1/8+4\cdot 1/8}{1/2}=\dfrac{11}{4}$$

Answer 2

The summation binds $x$ to a (series of) particular value(s). In your first line that ends with 1/2, you're using the union of all values $\geq 2$, but that's wrong for the probability expression in the sum. You have to fill in 2, 3, 4 for $x$, one by one for both $x$'s in the sum.

(I'll let you work out the rest.)

[Not sure how you got 8, but it doesn't matter!]