A basic question on conditional expectation and variance

Question

How calculate variance of $X$ from the following statements :

$$ E[X\mid Y=1] = 2 + E[X]$$ $$ E[X\mid Y=2] = 3 + E[X]$$ $$ E[X\mid Y=3] =0$$ $$ P(Y=1)=0.5, P(Y=2)=0.3 ~~\text{and}~~ P(Y=3) =0.2 $$

Answer 1

I think you need to give more information. You can easily find $$ \mathbb{E}[X] = \mathbb{E}[\mathbb{E}[X\mid Y]] = 0.5(2 + \mathbb{E}[X]) + 0.3(3 + \mathbb{E}[X]) = \frac{19}{10} + \frac{4}{5}\mathbb{E}[X] $$ so that $\mathbb{E}[X] = \frac{19}{2}$, but to find the variance you will need to know $\mathbb{E}[X^2]$ or at least $\mathbb{E}[X^2\mid Y]$. Otherwise, there may be two different random variables with the properties you list yet having different variances. For example, the condition you gave is the same as $$ \mathbb{E}[X \mid Y ] = -2Y^2 + 7Y - 3 + \frac{19}{2} \chi\{ Y \neq 3 \}, $$ so let $$ X_1 = -2Y^2 + 7Y - 3 + \frac{19}{2} \chi\{ Y \neq 3\} $$ and let $$ X_2 = -2 N(Y^2,10) + 7U(0,2Y) - 3 + \frac{19}{2}\chi\{ Y \neq 3\}, $$ where $N(Y^2,10)$ is a normal random variable with mean $Y^2$ and variance $10$, and $U(0,2Y)$ is a uniform random variable on the interval $(0,2Y)$. Both satisfy the given conditions, yet obviously have different variances.