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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 $t-1$.

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    $\begingroup$ Don't you actually want to find $E(y_{t+1}|y_1, y_2, ..., y_t)$? $\endgroup$ – Glen_b Nov 17 '15 at 11:31
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    $\begingroup$ yeah in priciple $E(y_{t+1} \mid y_{t})$ $\endgroup$ – Geoffrey Heideman Nov 17 '15 at 12:51
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    $\begingroup$ Yes, that follows, but the point was, you'll need to use the conditioning. $\endgroup$ – Glen_b Nov 17 '15 at 12:54
  • $\begingroup$ You answered your own question in the last equation if you replace t-1 with t! $\endgroup$ – Michael R. Chernick Mar 7 '17 at 17:51
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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

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    $\begingroup$ This is a Markov Chain. However a random walk is pure white noise with no correlation. So a random walk predicts the next step to be the same state as the current state. $\endgroup$ – Michael R. Chernick Mar 6 '17 at 15:30
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    $\begingroup$ $y_t$ is NOT a random walk. For a random walk, $E(y_{t+1}|past)=y_t$ (martingale). But a random walk with noise added is no longer a random walk. Exponential smoothing with the right $\lambda$ is the best compromise between removing the noise (averaging) and following the walk. $\endgroup$ – Benoit Sanchez Mar 7 '17 at 17:53
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    $\begingroup$ this comment of yours does not make sense. $\endgroup$ – Michael R. Chernick Mar 7 '17 at 17:56
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    $\begingroup$ Ok. Let's make it clearer. $(z_t)$ is a Markov chain but $(y_t)$ is not a Markov Chain. $\endgroup$ – Benoit Sanchez Mar 8 '17 at 10:59
  • $\begingroup$ I guess normality is irrelevant in your argument, or is it? $\endgroup$ – Richard Hardy Mar 11 '17 at 20:05

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