Difference between best predictor and best linear predictor Due to lack of examples, I am having a hard time understanding the difference between a best predictor and a best linear predictor. 
I think that $E[Y|X]$ is the best predictor for Y in terms of X, and if it happens to be linear, then it would also be the best linear predictor, (but please correct me if I'm wrong.)
So consider $Y=sin(X)+\epsilon$, where $X \sim U(0,2\pi)$ and $\epsilon \sim N(0, \sigma ^2)$.
Then, I think that the best predictor would be $E[Y|X]=E[sin(X)+\epsilon|X]=E[sin(X)|X]+E[\epsilon|X]=E[sin(X)]=0$. (However I am not sure on the last step, I used $E[sin(X)]=\int_0^{2\pi} \frac{sin(x)}{2\pi}$.)
Since $E[Y|X]$ is not linear, then we would have to find the best linear predictor. I think to find the best linear predictor we would use $Y=E[Y]+\frac{ Cov(X,Y)}{Var(X)}(X-E[X])$. But I wanted to make sure before I started solving for the parameters.
I got that $E[X]=\pi$, $E[Y]=0$, but I cannot find Cov(X,Y). If $E[Y|X]$ truly does equal 0 then I think Cov(X,Y) will also equal 0.
So I get that both the best predictor and linear predictor equal 0, and I don't think that is correct.
It is difficult for me to learn without being given examples so I am sorry in advanced for my many mistakes, thank you for your help.
 A: The best predictor is 
$$
\begin{eqnarray}
E(Y|X) &=& E(sin(X) + \epsilon|X) \\
&=& E(sin(X)|X) + E(\epsilon|X) \\
&=& sin(X) + 0 \\
&=& sin(X)
\end{eqnarray}
$$
Notice that $sin(X)$ is a constant because $X$ is given in $E(sin(X)|X)$, i.e., $E(sin(X)) \neq E(sin(X)|X)$
Your rationale in the second part is correct. Only the calculation of $Cov(X, Y)$ should be reviewed:
$$
\begin{eqnarray}
Cov(X, Y) &=& Cov(X, sin(X) + \epsilon) \\
&=& Cov(X, sin(X)) + Cov(X, \epsilon) \\
&=& Cov(X, sin(X)) \\
&=& E([X - E(X)][sin(X) - E(sin(X))]) \\
&=& E([X - \pi][sin(X) - 0]) \\
&=& E([X - \pi]sin(X)) \\
&=& \int_0^{2\pi} [x - \pi]sin(X)\frac{1}{2\pi}dx \\
&=& \int_0^{2\pi} x sin(x)\frac{1}{2\pi}dx - \int_0^\frac{1}{2}sin(x)\frac{1}{2\pi}dx \\
&=& \int_0^{2\pi} x sin(x)\frac{1}{2\pi}dx \\
&=& \frac{1}{2\pi}\left(- x cos(x)|_0^{2\pi} + \int_0^{2\pi}cos(x)dx\right) \\
&=& \frac{1}{2\pi}\left(-2\pi + cos(x)|_0^{2\pi}\right) \\
&=& \frac{1}{2\pi}\left(-2\pi + 0\right) \\
&=& -1
\end{eqnarray}
$$
Then, your best linear predictor is
\begin{eqnarray}
g(Y|X) &=& E(Y) + \frac{Cov(X, Y)}{Var(X)}(X - E(X)) \\
&=& 0 - \frac{3}{\pi^2}(X - \pi) \\
&=& -\frac{3}{\pi^2}(X - \pi)
\end{eqnarray}
