For $X, Y$ random variables, given $E[X] = \mu$ ; $\operatorname{Var}(X) = \sigma^2$; $Y\mid X \sim \text{Unif}(X,1)$: Find $E[Y]$ and $\operatorname{Var}(Y)$.
(1) To find E[Y], I used the law of conditional expectation:
From Y|X ~ U(X,1) --> E[Y|X] = X+1/2 Thus: E[Y] = E[E[Y|X]] = E[X+1/2] = E[X/2] + E[1/2] = 0.5*(µ+1)
(2) To find Var(Y), I'm not sure how to evaluate the Var(Y|X) expression in:
Var(Y) = E[Var(Y|X)] + Var(E[Y|X])
I know that: Var(Y|X) = E[(Y - E[Y|X])² | X], but have no idea how to evaluate that.
Help is appreciated!
The law of total variance $$\operatorname{Var}[Y] = \operatorname{E}[\operatorname{Var}[Y \mid X]] + \operatorname{Var}[\operatorname{E}[Y \mid X]]$$ can be evaluated just like you did for the law of total expectation, by first computing the inner conditional terms.
Clearly, you already know $$\operatorname{E}[Y \mid X] = \frac{X+1}{2}.$$ But you also know that for a uniform distribution on $[X,1]$, that $$\operatorname{Var}[Y \mid X] = \frac{(1 - X)^2}{12}.$$ So now we have to compute $$\operatorname{Var}[\operatorname{E}[Y \mid X]] = \operatorname{Var}\left[\frac{X+1}{2}\right] = \frac{1}{4}\operatorname{Var}[X+1] = \frac{1}{4}\operatorname{Var}[X],$$ and $$\operatorname{E}[\operatorname{Var}[Y \mid X]] = \operatorname{E}\left[\frac{(1-X)^2}{12}\right] = \frac{1}{12}\operatorname{E}[(1-X)^2].$$ Here's a cute trick to calculate this expectation: note $$(1-X)^2 = (X-1)^2 = ((X-\mu)+(\mu-1))^2 = (X-\mu)^2 + 2(X-\mu)(\mu-1) + (\mu-1)^2$$ so that $$\operatorname{E}[(1-X)^2] = \operatorname{Var}[X] + 2(\mu-1)\operatorname{E}[X-\mu] + \operatorname{E}[(\mu-1)^2].$$ But $\operatorname{E}[X-\mu] = 0$ because $\mu = \operatorname{E}[X]$, and $(\mu-1)^2$ is constant, so we just have $$\operatorname{E}[(1-X)^2] = \operatorname{Var}[X] + (\mu-1)^2.$$