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I am learning the double (Brown) exponential smoothing. Comparing to the simple exponential smoothing, the Brown exponential smoothing smooths the output sequence $Y_t$ twice, such that \begin{align} S^{(1)}_t &= \alpha Y_t + (1-\alpha)S^{(1)}_{t-1}\\ S^{(2)}_t &= \alpha S^{(1)}_t + (1-\alpha)S^{(2)}_{t-1} \end{align} I tried to derive its equivalent smoothing equations via level and trend functions \begin{align} L_t & = 2S^{(1)}_t - S^{(2)}_t\\ T_t &= \alpha \left(S^{(1)}_t - S^{(1)}_{t-1}\right) + (1-\alpha)T_{t-1} \end{align} where $Y_{t+1} = L_t + T_t + e_{t+1} = S^{(1)}_t + \frac{1}{\alpha} T_t + e_{t+1}$.

But I failed, can anyone help me to derive smoothing equations from the double exponential smoothing? many thanks!

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  • $\begingroup$ Those aren't the level and trend functions corresponding to Brown's exponential smoothing; check the Wikipedia page en.wikipedia.org/wiki/Exponential_smoothing. $\endgroup$
    – jbowman
    Commented Nov 13, 2023 at 3:09
  • $\begingroup$ Thank you, I corrected it. $\endgroup$
    – Stephen Ge
    Commented Nov 14, 2023 at 10:37

2 Answers 2

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I think one of advantages of using exponential smoothing functions \begin{equation} S^{(1)}_t :=\alpha Y_t + (1-\alpha)S^{(1)}_{t-1} = \alpha \sum_{i=0}^\infty (1-\alpha)^i Y_{t-i}, \end{equation} and \begin{equation} T_t = \alpha\left(S^{(1)}_t - S^{(1)}_{t-1}\right) + (1-\alpha) T_{t-1}, \end{equation} is that it makes more easier/convenient to prove the equivalency between Brown's double exponential smoothing and ARIMA(0,2,2) model.

Because \begin{equation} Y_t = L_{t-1} + T_{t-1} + e_t = S^{(1)}_{t-1} + \frac{1}{\alpha}T_{t-1} + e_t, \end{equation} we have \begin{align} S^{(1)}_t &= \alpha \left(S^{(1)}_{t-1} + \frac{1}{\alpha}T_{t-1} + e_t\right) + S^{(1)}_{t-1} - \alpha S^{(1)}_{t-1}\\ (1-\mathcal{L})S^{(1)}_t & = S^{(1)}_t - S^{(1)}_{t-1} =T_{t-1} + \alpha e_t \end{align} and \begin{align} T_t &= \alpha \left(T_{t-1} + \alpha e_t\right) + T_{t-1} - \alpha T_{t-1}\\ (1-\mathcal{L})T_t & = \alpha^2 e_t. \end{align} Thus, we have \begin{align} (1-\mathcal{L})Y_t &= (1-\mathcal{L})\left(S^{(1)}_{t-1} + \frac{1}{\alpha}T_{t-1} + e_t\right)\\ & = T_{t-2} + 2\alpha\mathcal{L}e_t + (1-\mathcal{L})e_t, \end{align} and \begin{align} (1-\mathcal{L})^2Y_t &= (Y_t-Y_{t-1})- (Y_{t-1}-Y_{t-2}) \\ &= (1-\mathcal{L})T_{t-2}+2\alpha(1-\mathcal{L})\mathcal{L}e_t + (1-\mathcal{L})^2e_t \\ & = \left[1-(1-\alpha)\mathcal{L}\right]^2e_t, \end{align} which is an ARIMA(0,2,2) model.

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OK, I think I proved it. Here is the answer:

Let us define first smoothing function $S^{(1)}_t$ such that \begin{equation} S^{(1)}_t :=\alpha Y_t + (1-\alpha)S^{(1)}_{t-1} = \alpha \sum_{i=0}^\infty (1-\alpha)^i Y_{t-i} = \mathbb{E}\{Y_t\}, \end{equation} where $Y_t$ are observed outputs.

If we exponentially smooth the $S^{(1)}_t$ again, we yield second smoothing function $S^{(2)}_t$, such that \begin{equation} S^{(2)}_t :=\alpha S^{(1)}_t + (1-\alpha)S^{(2)}_{t-1} = \alpha \sum_{j=0}^\infty (1-\alpha)^j S^{(1)}_{t-j}= \mathbb{E}\left\{S^{(1)}_t\right\}. \end{equation} Now if outputs $Y_t$ can be discribed using following linear function: \begin{equation} Y_{t\pm k} = L_t \pm kT_t + e_{t\pm k} \end{equation} and \begin{equation} \mathbb{E}\{Y_{t\pm k}\} = L_t \pm kT_t, \end{equation} where $L_t$ and $T_t$ are called level and trend function at time $t$, respectively.

Then, we could derive the relationship between smoothing functions and level/trend function \begin{equation} S^{(1)}_t = \alpha \sum_{i=0}^\infty (1-\alpha)^iY_{t - i} = \alpha \sum_{i=0}^\infty (1-\alpha)^i (L_t - iT_t) + \underbrace{\alpha \sum_{i=0}^\infty (1-\alpha)^i e_{t-i}}_{\mathbb{E}\{e_t\}=0} \end{equation} you should convince yourself that \begin{equation} \alpha \sum_{i=0}^\infty (1-\alpha)^i \times i = \frac{1-\alpha}{\alpha} \quad \text{and} \quad \alpha \sum_{i=0}^\infty (1-\alpha)^i = \sum_i w_i = 100\% \end{equation} hence \begin{equation} S^{(1)}_t = L_t - \frac{1-\alpha}{\alpha} T_t \end{equation} Similarly, we have \begin{equation} S^{(1)}_{t-j} = \alpha \sum_{i=0}^\infty (1-\alpha)^i (L_t - (i+j)T_t) = L_t - jT_t - \frac{1-\alpha}{\alpha} T_t. \end{equation} For the second smoothing function $S^{(2)}_t$, we have \begin{align} S^{(2)}_t &= \alpha \sum_{j=0}^\infty (1-\alpha)^jS^{(1)}_{t-j}\\ &= \alpha \sum_{j=0}^\infty (1-\alpha)^j \left(L_t - jT_t - \frac{1-\alpha}{\alpha} T_t\right) = L_t - 2\frac{1-\alpha}{\alpha}T_t. \end{align} Finally, it yields \begin{equation} \left\{\begin{matrix}L_t = 2S^{(1)}_t - S^{(2)}_t \\ T_t = \frac{\alpha}{1-\alpha}\left(S^{(1)}_t - S^{(2)}_t\right)\end{matrix}\right. \end{equation} By substituting \begin{equation} S^{(2)}_t = S^{(1)}_t - \frac{1-\alpha}{\alpha}T_t \end{equation} into the second smoothing equation about $S^{(2)}_t$, we found that trend function $T_t$ also follows exponential decays \begin{equation} T_t = \alpha\left(S^{(1)}_t - S^{(1)}_{t-1}\right) + (1-\alpha) T_{t-1}, \end{equation} This is exact the second smoothing function that Robert R. Brown was introduced in his original paper "Exponential Smoothing for Predicting Demand". BTW, in his original paper, Robert R. Brown derived this second smoothing function about $T_t$ using the linear least squares.

Therefore, the forcasted outputs can be written as \begin{equation} \mathbb{E}\{Y_{t + k}\} = L_t + kT_t = S^{(1)}_t + \frac{1-\alpha}{\alpha} T_t + kT_t = S^{(1)}_t + \frac{1}{\alpha}T_t + (k-1)T_t. \end{equation}

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