# Moving Average, Exponential Smoothing, and Random Walk for Forecasting

I would like to confirm my understanding. Is it true that a (simple) exponential smoothing model with alpha (smoothing constant) = 1 is the same as MA(1), which is in turn the same as a random walk model? (i.e. using only the most recent observation as the forecast for all future periods)?

• Short answer: they are all different. Only the RW model uses the last observation as forecast. MA(1) and SES do something else. Check out the "Forecasting: Principles and Practice" textbook for some introductory material on these models. See also my comment under Carl's answer. – Richard Hardy Sep 3 '17 at 9:10

Is it true that a (simple) exponential smoothing model with alpha (smoothing constant) = 1 is the same as MA(1), which is in turn the same as a random walk model? (i.e. using only the most recent observation as the forecast for all future periods)?

No, it is not. Here are the forecasts by the three models:

• Simple exponential smoothing (SES; see section 7.1 of Hyndman & Athanasopoulos "Forecasting: Principles and Practice"): $$\hat x_{t+1|t} = \alpha x_t + \alpha(1-\alpha)x_{t-1} + \alpha(1-\alpha)^2 x_{t-2} + \alpha(1-\alpha)^3 x_{t-3} + \dots$$

• Moving average of order 1 (MA(1); see section 8.4 of the same textbook): $$\hat x_{t+1|t} = \mu + \alpha\hat\varepsilon_t$$ where $\varepsilon_t = x_t - \hat x_{t-1}$. By iterated substitution, $$\hat x_{t+1|t} = \mu(1-\alpha+\alpha^2-\alpha^3+\dots) + \alpha x_{t-1} - \alpha^2 x_{t-2} + \alpha^3 x_{t-3} - \dots$$

• Random walk (RW): $$\hat x_{t+1|t} = x_t.$$

As you can see, they are all quite different. It is only RW that uses the most recent observation as the forecast for all future periods. Meanwhile, both SES and MA(1) (implicitly) use a linear combination of all past observations to forecast the future.

When $\alpha=1$ for SES (but not for MA(1)), you get

• SES: $\hat x_{t+1|t} = x_t$.

Hence, there SES coincides with RW but MA(1) is different. Also, even if we take $\alpha=1$ for MA(1), it still does not coincide with SES or RW. However, if you replaced MA(1) with AR(1), then your conjecture would be correct.

Edit: If by MA(1) you mean moving average of one element rather than the moving average model of order 1 (which it is the standard notation for), then indeed that forecast will coincide with the RW and SES forecasts under $\alpha=1$.

A moving average can have a choice of a large set of decreasing weights into the past. One such is an exponential function, and exponential functions are memoryless in the sense that they have a constant rate of decreased weightings looking retrospectively. Thus, although exponential functions find use for moving averages, power functions and many other functions do as well.

{Random walk](https://en.wikipedia.org/wiki/Random_walk) is something entirely different. Morn gives this example in Wikipedia As you can see, these many random walks have increasing noise the further out they are extrapolated. Moving averages are smooths and have less noise than any single datum beyond the first. Moving averages are not extrapolations, they are estimates of location.

• Thanks so much @Carl for your explanation and for the knowledge. I guess I didn't clarify my question well enough. I am specifically looking for the application of these models in forecasting. My understanding is whether you use a (simple) MA or SES to generate forecasts, you will just generate a single forecast, based on the most recent observations, and use that forecast for all future periods (locally constant, i.e. non-trended, forecast). – Beth Sep 3 '17 at 5:53
• Then, both MA(1) and SES with smoothing constant of 1 rely completely on the most recent observation alone. Is this correct? Also, if you were to forecast based on a (simple, symmetric) random walk model, you would also assume the future is best described by the most recent observation. Is this correct? In other words, under these three forecasting methods, would you just use the most recent observation as your forecast for any future periods? (Of course, these specific models are not likely to do well in general.) – Beth Sep 3 '17 at 5:53
• Not correct. For example, a moving average of the numbers 1, 3, 2, 4, 6, 5 could be have 3, 2, 1 weighs, that is $$\frac{3*5+2*6+1*4}{3+2+1}=\frac{15+12+4}{6}=\frac{31}{6}=5\space \frac{1}{6}$$ – Carl Sep 3 '17 at 6:03
• Usually, regression models are used for forecasting. – Carl Sep 3 '17 at 7:08
• That is not a moving average model. The MA(q) model is defined as $$x_t=\mu+\varepsilon_t+\theta_1\varepsilon_{t-1}+\dots+\theta_q\varepsilon_{t-q}.$$ Check out Wikipedia for more details. Specifically, The moving-average model should not be confused with the moving average, a distinct concept despite some similarities. – Richard Hardy Sep 3 '17 at 8:58