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 $t-1$.

  • 3
    $\begingroup$ 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
  • 1
    $\begingroup$ yeah in priciple $E(y_{t+1} \mid y_{t})$ $\endgroup$ – Geoffrey Heideman Nov 17 '15 at 12:51
  • 1
    $\begingroup$ 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

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

| cite | improve this answer | |
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ $y_t$ 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
  • 1
    $\begingroup$ this comment of yours does not make sense. $\endgroup$ – Michael R. Chernick Mar 7 '17 at 17:56
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.