Best Predictor of MA(1) I have read a statement in a lecture note that for an MA(1) model 
$X_t = \theta \epsilon_{t-1} + \epsilon_t$ with $|\theta| < 1$, where $\epsilon_t$ are white noise variates:
We can forecast only one step ahead as $X_{t+2}$ is independent of observations up to time $t$. Hence the best predictor of $X_{t+2}$ is zero.
I doubt about the claim of independence. Is it just that $X_{t+2}$ is uncorrelated with other $X_t$? And also isn't it should be best linear predictor?
 A: In an $\text{MA}(1)$ model of that form we have auto-covariance given by:
$$\gamma(k) \equiv \mathbb{Cov}(X_t, X_{t+k}) = \mathbb{E}(X_t X_{t+k}) = \begin{cases} 
(1+\theta^2) \sigma^2 & & \text{for } |k|=0, \\[6pt]
\theta \sigma^2 & & \text{for } |k|=1, \\[6pt]
0 & & \text{for } |k|>1. \\[6pt]
\end{cases}$$
This gives the auto-correlation function:
$$\rho(k) \equiv \mathbb{Corr}(X_t, X_{t+k}) = \frac{\gamma(k)}{\gamma(0)} = \begin{cases} 
1 & & \text{for } |k|=0, \\[6pt]
\frac{\theta}{1+\theta^2} & & \text{for } |k|=1, \\[6pt]
0 & & \text{for } |k|>1. \\[6pt]
\end{cases}$$
You have not specified an error distribution for your process, but if you use the normal distribution for your error term then the observable values in the process are jointly normal and so uncorrelated values are indeed independent.  In that case it is correct to state that $X_t \text{ } \bot \text{ } X_{t+k}$ for $k>1$.  In this case the best predictor of $X_{t+2}$ given data up to $t$ is indeed zero.  (Incidentally, nothing in these results requires $|\theta| < 1$; the results are the same even if this restriction does not hold.)
