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Following Eqs. (3.10a) and (3.10b) from (Hyndman et al., 2008) I obtained a simulated series $y_t=l_{t-1}+\varepsilon_t$ and level $l_t=l_{t-1}+\alpha\,\varepsilon_t$, $t=1.2,\ldots,40$, see data below in case of $l_0=10$ and $\varepsilon_t\sim N(0,2)$:

$y_t={}${8.78512, 10.0658, 10.9666, 10.9225, 5.77473, 7.20857, 11.3946, 8.52732, 10.7475, 9.65533, 9.69253, 8.3939, 6.13765, 9.68515, 10.6551, 11.7868, 11.1522, 9.70109, 11.9185, 8.83022, 12.56, 9.16055, 10.3244, 11.2713, 6.99303, 11.644, 9.85491, 11.1847, 7.37897, 11.0723, 9.76387, 11.244, 9.19405, 8.65215, 9.05387, 9.0939, 11.1757, 5.75803, 11.8584, 10.1116}

$l_t = {}${9.87851, 9.89724, 10.0042, 10.096, 9.66388, 9.41835, 9.61597, 9.50711, 9.63115, 9.63357, 9.63947, 9.51491, 9.17718, 9.22798, 9.37069, 9.6123, 9.76629, 9.75977, 9.97565, 9.8611, 10.131, 10.0339, 10.063, 10.1838, 9.86475, 10.0427, 10.0239, 10.14, 9.86388, 9.98472, 9.96264, 10.0908, 10.0011, 9.86621, 9.78497, 9.71587, 9.86185, 9.45146, 9.69215, 9.7341}

Sorry for this, I'm not allowed to post images. Anyway it is easy do produce graphs using these data.

Then I applied the single exponential smoothing model $\hat{y}_{t+1} = \hat{y}_t+\alpha\,(y_t-\hat{y}_t)$ and simultaneously found optimal values for both $\alpha$ and $\hat{y}_1$ using least squares. They were $\alpha\approx-0.321$ and $\hat{y}_1\approx9.43$. I'm aware that the following should hold: $0<\alpha<2$. Nevertheless it works somehow, see results:

$\hat{y}_t = {}${9.42631, 9.63197, 9.49281, 9.02011, 8.40994, 9.25516, 9.91159, 9.43593, 9.72735, 9.40013, 9.31828, 9.19825, 9.45623, 10.5206, 10.7886, 10.8315, 10.525, 10.3239, 10.5236, 10.0762, 10.4759, 9.80742, 10.0149, 9.91562, 9.48079, 10.2787, 9.84081, 9.83629, 9.4038, 10.0532, 9.72639, 9.71437, 9.22375, 9.23328, 9.41967, 9.53699, 9.67911, 9.1991, 10.3028, 9.80387, 9.70518}

In fact I tried the same procedure many times. Quite often I got negative $\alpha$. Why does it work this way?

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The restriction $0<\alpha<2$ is to enforce the forecastability constraint (equivalent to the invertibility constraint for ARIMA modelling). Just as it is possible to estimate a non-invertible ARIMA model and use it, it is possible to estimate a non-forecastable ETS model and use it. The constraint is to ensure that observations long ago have negligible influence on forecasts. The ets() function from the forecast package in R will impose the constraint automatically.

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Rob, thank you very much. I hoped you should react to this question and I am glad you answered it. In fact I still do not understand if it's a common practice to get $\alpha$ and $\hat{y}_1$ simultaneously as two arguments of a minimized function (I mean squared errors). If one estimates only $\alpha$ then its optimal value is inside (0;2) interval for standard initial $\hat{y}_1$ (obtained for example, as intercept of linear regression) –  Gregory Fridman Oct 17 '11 at 16:04
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