A finite-history conditional mean can be nonlinear

A common shift $s$ changes phase $\omega$ to $(\omega +s)<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>3$, a permutation of the equally weighted outcomes, so every shifted finite coordinate vector has the original law. One-coordinate laws give a common mean, pair laws at each lag give covariance independent of $t$, and boundedness gives finite moments; hence the process is weakly stationary by [F3].

The values of $(X_1,X_2)$ at $\omega =0,1,2$ are $(-1,0),(0,1),(1,-1)$; hence both means are zero, $\mathrm{Var}(X_1)=2\mathrm{/}3$, and $\mathrm{Cov}(X_2,X_1)=-1\mathrm{/}3$. By [F4], the unique best affine forecast from this one-variable history is $(-1\mathrm{/}3)\mathrm{/}(2\mathrm{/}3)X_1=-X_1\mathrm{/}2$.

Since $X_1:\mathrm{\Omega }\to \{-1,0,1\}$ is bijective, $\sigma (X_1)$ is the full event sigma-algebra and $X_2$ is measurable with respect to it. The conditional-expectation definition and known-variable property give $\mathbb{E}[X_2\mid \sigma (X_1)]=X_2$, with values $(0,1,-1)$ at $\omega =0,1,2$. Every outcome has positive probability; if $X_2=a+b{X}_1$ almost surely, the outcomes with $X_1=0$ and $X_1=-1$ force $a=b=1$, which would give $X_2=2$ when $X_1=1$, contrary to its value $-1$. Thus the conditional mean is not affine in $X_1$, while step 2.1 gives the best affine forecast $-X_1\mathrm{/}2$.