How to deal with expected value in the context of time series? For example, in this MA(2) model, 
$y_t = u_t + \phi u_{t-2}$
$u_t$ is identically, independently, normally distributed with a mean of 0 and a variance of $\sigma^2$. (Does variance matter here?)
I think $E[u_{t}|y_{t-2}] = 0$ because it is impossible to represent $y_{t-2}$ using an equation containing $u_{t}$.  
However, in the solution, it states that $E[u_{t-6}|y_{t-2}] = 0$ is also 0. But I think $y_{t-2}$ can be represented with an equation containing $u_{t-6}$ ($y_{t-2} = u_{t-2} + \phi (y_{t-4} - \phi u_{t-6})$). So why is $E[u_{t-6}|y_{t-2}] = 0$? Is it because "$u_t$ is identically, independently, normally distributed"? If yes, am I correct that any expected value of u conditional on y (no matter their index) is always 0 (including $E[u_t-2|y_{t}]$)?
Why is $E[y_{t-4}|y_{t-2}] = E[y_{t}|y_{t-2}]$? In the solution, it says "because $u_{t-2}|y_{t-2}$ and $u_{t-4}|y_{t-2}$ are independent conditionally". But how are the reasons given relevant? $y_{t}$ can not really be represented as equation containing $u_{t-2}$, $u_{t-4}$ and $y_{t-4}$.
Is it also analogous to autoregressive time series?
 A: From $y_t = u_t + \phi u_{t-2}$, we have
$$\left(\begin{matrix}u_{t-6}\\u_{t-4}\\u_{t-2}\\u_{t}\\y_{t-4}\\y_{t-2}\\y_{t} \end{matrix}\right) = \left(\begin{matrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\\\phi&1&0&0\\0 & \phi&1&0&\\0& 0&\phi&1\end{matrix}\right) \left(\begin{matrix}u_{t-6}\\u_{t-4}\\u_{t-2}\\u_{t}\end{matrix}\right)$$
or
$$Y=Au$$
$u$ is multi-normal distributed, so $Y$ also follows multinormal distribution which determined by the mean vector and variance-covariance matrix.
$$\mathrm{E}(Y)=0$$
$$\mathrm{Var}(Y) = A\mathrm{Var}(u)A' = \sigma^2\left(\begin{matrix} I_{4\times 4}& \Sigma_{12}\\\Sigma_{12}'&\Sigma_{22}\end{matrix}\right)$$
where $$\Sigma_{22}=\left(\begin{matrix}\phi +1 &\phi & 0\\ \phi & \phi +1 &\phi \\ 0  &\phi & \phi +1\end{matrix}\right) $$
$$\Sigma_{12} = \left(\begin{matrix}\phi & 0 & 0 \\1 &\phi & 0\\ 0 & 1 &\phi \\
0& 0 & \phi\end{matrix}\right) $$
Ignore the constant $\sigma^2$, because it appears in the numerator and denominator.
$\mathrm{E}(Y_{t-4}|Y_{t-2}) = E(Y_{t-4}) + \frac {Cov(Y_{t-4},Y_{t-2})}{Var(Y_{t-2})}(Y_{t-2}-E(Y_{t-2})) = 0 + \frac{\phi}{1+\phi}(Y_{t-2}-0) = \frac{\phi}{1+\phi}Y_{t-2}$
Replacing $Y_{t-4}$ by $Y_t$ the result is the same because $Cov(Y_{t-4},Y_{t-2}) = Cov(Y_{t},Y_{t-2})$.
For the rest, I think you can find the answers.
If not working with matrix, need following to get the variance and covariance based on $u_i$ is iid.
Let $Y_1 = a_1u_1+ ... + a_ku_k$ then $E(Y_1) = a_1E(u_1) +...+a_kE(u_k)$, $Var(Y_1) = a_1^2Var(u_1)+..+a_k^2Var(u_k)$
Let $Y_2 = b_1u_1+...+b_ku_k$ then $Cov(Y_1,Y_2)= a_1b_1Var(u_1)+...+a_kb_kVar(u_k)$
In your question about $Y_{t-2}$ and $Y_{t-4}$,
$\begin{align} Y_{t-2} = &1u_{t-2} + &\phi u_{t-4} + &0u_{t-6}\\
Y_{t-4} = &0u_{t-2} + &1 u_{t-4} + &\phi u_{t-6}
\end{align}$ 
So $Cov(Y_{t-2}, Y_{t-4}) = \phi Var(u_{t-4}) = \sigma^2 \phi$
