I am currently reading the book "Forecasting with Exponential Smoothing: The State Space Approach". On the bottom of page 14, it says that “In the special case where $\alpha=\beta^*$, Holt’s method is equivalent to Brown’s double exponential smoothing”. enter image description here

I also found a similar statement in the Duke university’s online material: https://people.duke.edu/~rnau/411avg.htm

I tried to work it out by myself, but I can’t prove the equivalency, can anyone help me please?

I started from Brown’s double 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} = \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 :=\beta S^{(1)}_t + (1-\beta)S^{(2)}_{t-1} = \beta\sum_{j=0}^\infty (1-\beta)^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}

Thus, we have \begin{align} S^{(1)}_t &= L_t - \frac{1-\alpha}{\alpha} T_t\\ S^{(2)}_t &= L_t - \frac{1-\alpha}{\alpha}T_t - \frac{1-\beta}{\beta}T_t\\ Y_t &= S^{(1)}_{t-1} + \frac{1}{\alpha}T_{t-1}+e_t \end{align} By substituting above expression of $S_t^{(1)}$ and $S_t^{(2)}$ into the exponential smoothing function about $S_t^{(2)}$ we have \begin{align} T_t &= \beta\left(S^{(1)}_t - S^{(1)}_{t-1}\right) + (1-\beta) T_{t-1}\\ &=\beta\left(L_t - \frac{1-\alpha}{\alpha} T_t - L_{t-1} + \frac{1-\alpha}{\alpha} T_{t-1}\right) + (1-\beta) T_{t-1}\\ &=\beta\left(L_t - L_{t-1} \right) + (1-\beta) T_{t-1} - \beta\frac{1-\alpha}{\alpha} \underbrace{(T_t-T_{t-1})}_{\alpha\beta e_t}\\ & =\beta\left(L_t - L_{t-1} \right) + (1-\beta) T_{t-1} - \beta^2(1-\alpha)e_t \end{align} Similarly, \begin{align} \underbrace{L_t - \frac{1-\alpha}{\alpha} T_t}_{S^{(1)}_t} & = \alpha Y_t + (1-\alpha) \underbrace{\left(L_{t-1} - \frac{1-\alpha}{\alpha} T_{t-1}\right)}_{S^{(1)}_{t-1}}\\ L_t & = \alpha Y_t + (1-\alpha)(L_{t-1}+T_{t-1}) + \frac{1-\alpha}{\alpha}(T_t-T_{t-1}) \\ & = \alpha Y_t + (1-\alpha)(L_{t-1}+T_{t-1}) + \beta(1-\alpha)e_t \end{align} Therefore, I can’t derive the Holt’s exponential smoothing from Brown’s double exponential smoothing, due to the extra error term $e_t$.


2 Answers 2


OK.... I think I have been misleading by the notation of smoothing constants in the Brown's double exponential smoothing and Holt's exponential smoothing.

Recall from my question that \begin{align} T_t &= \beta\left(S^{(1)}_t - S^{(1)}_{t-1}\right) + (1-\beta) T_{t-1}\\ &=\beta\left(L_t - \frac{1-\alpha}{\alpha} T_t - L_{t-1} + \frac{1-\alpha}{\alpha} T_{t-1}\right) + (1-\beta) T_{t-1}\\ &=\beta\left(L_t - L_{t-1} \right) + (1-\beta) T_{t-1} - \beta\frac{1-\alpha}{\alpha}(T_t-T_{t-1})\\ \alpha T_t + \beta (1-\alpha) T_t& =\alpha\beta\left(L_t - L_{t-1} \right) + \alpha(1-\beta) T_{t-1} + \beta (1-\alpha)T_{t-1} \\ T_t &= b\left(L_t - L_{t-1} \right) + (1-b) T_{t-1} \end{align} where \begin{equation} b=\frac{\alpha\beta}{\alpha+\beta(1-\alpha)}. \end{equation} Similarly, \begin{align} \underbrace{L_t - \frac{1-\alpha}{\alpha} T_t}_{S^{(1)}_t} & = \alpha Y_t + (1-\alpha) \underbrace{\left(L_{t-1} - \frac{1-\alpha}{\alpha} T_{t-1}\right)}_{S^{(1)}_{t-1}}\\ L_t & = \alpha Y_t + (1-\alpha)(L_{t-1}+T_{t-1}) + \frac{1-\alpha}{\alpha}(T_t-T_{t-1}) \end{align} because \begin{equation} T_t-T_{t-1} = \frac{\alpha\beta}{\alpha+\beta(1-\alpha)}[L_t - (L_{t-1}+T_{t-1})] \end{equation} we have \begin{align} \alpha L_t - \frac{\alpha\beta(1-\alpha)}{\alpha+\beta(1-\alpha)}L_t & = \alpha^2 Y_t + \alpha(1-\alpha)(L_{t-1}+T_{t-1}) - \frac{\alpha\beta(1-\alpha)}{\alpha+\beta(1-\alpha)}(L_{t-1}+T_{t-1})\\ \frac{\alpha^2}{\alpha+\beta(1-\alpha)} L_t&=\alpha^2 Y_t + \frac{\alpha^2(1-\alpha)(1-\beta)}{\alpha+\beta(1-\alpha)}(L_{t-1}+T_{t-1})\\ L_t & = a Y_t + (1-a)(L_{t-1}+T_{t-1}) \end{align} where \begin{equation} a=\alpha+\beta(1-\alpha). \end{equation}

We could verify the equivalency from ARIMA point of view, the Holt’s exponential smoothing can be written as ARIMA(0,2,2) such that \begin{equation} (1-\mathcal{L})^2 Y_t=e_t+(a + ab-2)\mathcal{L}e_t+(1-a) \mathcal{L}^2 e_t. \end{equation} By subsitituting $a$ and $b$ into it yields \begin{equation} (1-\mathcal{L})^2 Y_t=e_t+(\alpha + \beta-2)\mathcal{L}e_t+(1-\alpha)(1-\beta) \mathcal{L}^2 e_t. \end{equation} by letting $\beta = \alpha$, we have \begin{equation} (1-\mathcal{L})^2 Y_t=e_t+(2\alpha-2)\mathcal{L}e_t+(1-\alpha)^2 \mathcal{L}^2 e_t, \end{equation} which is the ARIMA representation of the Brown's double exponential smoothing.


Hi: I've done this once but I can't find it. Here are some suggestions. ( I'm not telling what to do but just giving you the best route ).

  1. I would forget trying to go the arima(0,2,2) route because that will be more confusing in my opinion.

  2. Reading your explanation was somewhat clear but when you were trying to equate brown and holt ( like I said, this is the way to go ), you introduced an error term when there really is not one. The underlying model being estimated is $y_{t} = a + b \times t$. I think the error term must have come about when you were working on the other one ? So, I got somewhat confused by this and decided to find something more along what I once did. Below is a link to a nice, atleast from a framework description point of view, comparison of holt and brown.

The key page is page 57. Note though that that, in order to prove what you want to prove, all you have to do is show that (7), when written out, is equivalent to (2).

The difference in the way the two models look is due to that, in brown's case, he is exponentially smoothing the level using $S^{1}_{t}$ and the (level + slope = ACTUAL RESULTING PREDICTION) using $S^2_{t}$. Conversely, in holt's, he keeps things seperate so that the level, $l_{t}$, and the slope, $b_{t}$, are smoothed separately. So, the trick is to add them together in holt and get a relation for exponential smoothing of $l_{t} + b_{t}$. That can get a little ugly algebraically why is wish I had my scrap but I can tell you from experience that 2) and 7) can be equated. Keep in mind that my notation for level and slope might not be in exact sync with the paper's notation at the link below but that was my intention.


So, this is not an answer but I hope what I said helps you some. Oh, the approach where you use the ARIMA(0,2,2) equivalence can lead down dark roads because holt and brown are optimal for PREDICTION of the arima(0,2,2) so the error terms do go away. But this did not seem like the route you took so I got confused there. Good luck.

  • $\begingroup$ Now I see your update. You did exactly what I said to avoid but you got it, so congrats and it shows that you should never take my advice. But we had different underlying models so that's why. As you showed, it's definitely key to have what is what notationally correct. Also, I learned something so thanks. $\endgroup$
    – mlofton
    Nov 17, 2023 at 10:50
  • $\begingroup$ It seems like you're into this type of material so, if you don't have it and like to waste $ on books, I recommend "Smoothing, Forecasting and Prediction of Discrete Time Series" by Robert Brown. $\endgroup$
    – mlofton
    Nov 17, 2023 at 10:57
  • $\begingroup$ thanks a lot! I got that book from Robert Brown, really nice one! $\endgroup$
    – Stephen Ge
    Nov 18, 2023 at 2:42

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