Conditional Expectation, discrete?

Question

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: $\text{E}[\mathbb{X}\space | \space\mathbb{X} \in \{1,2\}]$

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

so does:

$$\text{E}[\mathbb{X}\space | \space\mathbb{X} \in \{1,2\}] = 1*\frac{1}{3} + 2*\frac{1}{2} = \frac{4}{3}$$ since $\mathbb{X}$ is conditioned on a subset of itself?

Thanks!

Update:

Additional Related Question

Answer 1

What you did is not correct. Note that the sums of the probabilities you considered is not $1$.

When we are dealing with conditional expectation, we deal with a new probability measure on the space. In this case, it's the probability $$P(\cdot \mid X \in \{1,2\}).$$

Therefore, the conditional expectation is $$E(X \mid X\in\{1,2\}) = 1 \times P(X=1 \mid X\in \{1,2\})+2\times P(X=2 \mid X\in \{1,2\}) + 3 \times P(X=3 \mid X\in \{1,2\})=1\times \frac{1/3}{5/6}+2 \times \frac{1/2}{5/6}+3 \times 0 = \frac{8}{5}.$$

I hope this helps you.

Answer 2

You want to find first the distribution of the random variable $Z$ which is "$X$ given $X \in \{1,2\}.$" Notice $P(Z = 1) = \dfrac{\frac{1}{3}}{\frac{1}{3} + \frac{1}{2}} = \dfrac{2}{5},$ and $P(Z = 2) = 1 - P(Z = 1) = \dfrac{3}{5}.$ Thus, $E(Z)=\dfrac{2}{5} + 2\cdot \dfrac{3}{5}=\dfrac{8}{5}.$