Conditional Expectation has the minimum mean prediction error Consider $\{X_t\}$ as a general time series including random variables $X_t$. Assume that we have observed $X$'s until time t. The goal is to come up with a function of the observed $X$'s to predict a future X, let's say unobserved $X_{t+h}$. Condsider $h(\vec{X}) = h((..., X_{t-1}, X_t))$ as our proposed function. We define mean prediction error as:
$$E\{(X_{t+h} - h(\vec{x}))^2 | \vec{x}\}$$
Goal: I want to show that among all possible functions h, the function $E(X_{t+h}|\vec{x})$ has the minimum mean prediction error.
Thank you for your help
 A: You can adapt the following argument to get a full answer to your question. For random variables $X$ and $Y$, and a measurable function $f$, we have
$$
  \newcommand{\E}{\mathbb{E}}
  \E[(Y-f(X))^2] = \E[(Y-\E[Y\mid X]+\E[Y\mid X]-f(X))^2] 
$$
$$
  = \E[(Y-\E[Y\mid X])^2] - 2\,\E[(Y-\E[Y\mid X])(\E[Y\mid X]-f(X))] + \E[(\E[Y\mid X]-f(X))^2] \, .
$$
Consider the middle term of the last expression. By definition, $\E[Y\mid X]-f(X)$ is a function (call it $g$) of $X$. Hence, using the properties of the conditional expectation, we have
$$
  \E[(Y-\E[Y\mid X])(\E[Y\mid X]-f(X))] = \E[(Y-\E[Y\mid X])g(X)] 
$$
$$ 
  = \E[g(X)Y] - \E[g(X)\E[Y\mid X]]
$$
$$
  = \E[g(X)Y] - \E[\E[g(X)Y\mid X]]
$$
$$
  = \E[g(X)Y] - \E[g(X)Y] = 0 \, .
$$
Therefore,
$$
  \E[(Y-f(X))^2] = \E[(Y-\E[Y\mid X])^2] + \E[(\E[Y\mid X]-f(X))^2] \, ,
$$
implying that
$$
  \E[(Y-f(X))^2] \geq \E[(Y-\E[Y\mid X])^2] \, ,
$$
with equality if we choose $f(X)=\E[Y\mid X]$ a.s. $\newcommand{\cF}{\mathscr{F}}$
Your proof follows mutatis mutandis from this one. Define $\cF_t=\sigma(X_{t-k}:k\geq 0)$. Let $Z$ be $\cF_t$-measurable. Use what we just did to prove that
$$
  \E[(X_{t+k}-Z)^2\mid\cF_t] \geq \E[(X_{t+k}-\E[X_{t+k}\mid\cF_t])^2\mid\cF_t]
$$
holds a.s. The result follows.
