This looks confused? I don't understand what you're saying in the second paragraph...
Comment 1: "Best" forecast depends on what you mean by "best."
Let $Y$ be a random variable and $\mathcal{F}$ be the information set. The "best" forecast depends on what the loss function is. If you're minimizing the expectation of squared loss:
\begin{equation}
\begin{array}{*2{>{\displaystyle}r}}
\mbox{minimize (over $x$)} & \mathbb{E}[(Y - x)^2 \mid \mathcal{F}]
\end{array}
\end{equation}
You have the solution that $x$ is the conditional expectation of $Y$ given information set $\mathcal{F}$
$$x^* = \mathbb{E}[Y \mid \mathcal{F}]$$
Of course you can have other loss functions. Consider minimizing the expected absolute error:
\begin{equation}
\begin{array}{*2{>{\displaystyle}r}}
\mbox{minimize (over $x$)} & \mathbb{E}[|Y - x| \mid \mathcal{F}]
\end{array}
\end{equation}
The solution here is that $x$ is the median of $Y$. Let $F^{-1}_Y$ be the quantile function for $Y$ conditional on information set $\mathcal{F}$.
$$ x^* = F_Y^{-1}(.5 \mid \mathcal{F})$$
Comment 2: The importance of stationarity
Let $\{ Y_t\}$ be a stochastic process. $Y_1$, $Y_2$, $Y_3$ etc... are all random variables.
Speaking with imprecise language:
- Stationarity says that the unconditional distribution of $Y_1$ is the same as $Y_2$ is the same as $Y_3$ is the same as $Y_{1000}$ etc...
- Ergodicity says that process doesn't get stuck somewhere.
Stationarity says that you can talk about a time invariant expectation $\mathbb{E}[Y]$. With ergodicity, a time-series mean $\frac{1}{T} \sum_{t=1}^T Y_t$ will estimate that time invariant expectation (by an ergodic theorem). With stationarity and ergdocity, averages over time converge to averages over space.
With a non-stationary series, that's not true! Example. Let $\{X_t\}$ be a stochastic process. Let $X_1$ be result of a die roll. Let $X_2$ be winning total of the Golden State Warriors vs. the LA Lakers. Let $X_3$ be the number of votes cast for BRexit. Let $X_4$ be the return on Apple stock December 10th. If I found a way to keep doing this, $X$ would be a non-stationary process. Talking about the expectation $\mathbb{E}[X]$ is non-sensical. There is no time-invariant expectation. And taking the average over time of $X$ does nothing useful at all.
(Note: Often times people have in mind a random walk when talking about a non-stationary process. In the case of a random walk, the first differences are stationary.)