What's the forecast of a Random Walk with Noise model? I have a RW with noise model defined as:
$$ y_{t} = z_{t} + v_{t}$$
where $ z_{t} = z_{t-1} + e_{t}$.
$v_{t}$ and $e_{t}$ are mutually independent with expectation $0$ and variance $\sigma_{v}^{2}$ and $\sigma_{e}^{2}$, respectively. 
My question is: what is the 1-step ahead optimal forecast of $y_{t+1}$? If I express it in terms of expectations, I get the forecast to be:
$$E(y_{t+1}) = z_{t-1} $$
But this is not correct, as it's apparently related to the simple exponential smoothing function, i.e. it is a weighted mix of  $y_{t}$ and the forecast of $y_{t}$ made in $t-1$. 
 A: You must first clarify the conditioning.
If you can remember a single value $y_t$ then $E(y_{t+1}|y_t)=y_t$. This is straightforward.
But when you write $E(y_{t+1})$ at a certain time $t$, you implicitly mean the conditional expectation given all the past observations. Logically, what you want to know is :
$E(y_{t+1}|y_1,y_2,...,y_t)$
If you assume $e_t$ and $v_t$ to be all normally distributed variables and independent, it is possible to compute this conditional expectation exactly as the orthogonal projection of $y_{t+1}$ on the linear span of $y_1,y_2,...,y_t$. The result is not simple even though it relies on basic methods in an euclidean space.
There is a special case however, when you assume t is large enough to neglect the effects of your series being limited in the past (starting at $t=1$). If the series is unlimited in the past, then you can prove :
$E(y_{t+1}|y_t,y_{t-1},y_{t-2},...)=\lambda\sum_{u=0}^\infty(1-\lambda)^u y_{t-u}$
for a certain value of $\lambda$ depending on $\frac{\sigma_e}{\sigma_v}$
You recognize exponential smoothing. A proof can be found in this article as well as the value of $\lambda$: http://www.tandfonline.com/doi/abs/10.1080/01621459.1960.10482064
