Is Brown’s double exponential smoothing a special case of Holt’s exponential smoothing

exponential-smoothingtime series

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”.

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 $e_t$.

Best Answer

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.

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From your model, $s_t$ is treated as an (unknown) constant; hence, it affects the mean of $y_t$, which affects the calculation of the sample ACF as you have shown in the graph. Normally when you calculate the ACF/PACF/EACF, you calculate it for the residuals since all of the residuals will come from the same family. Clearly, ${\tt acf(y - rep(rnorm(d), n/d), type = "corr")}$ will give a well-behaved ACF.

However if we must estimate $s_t$ for $t=1,\cdots,p$, then it is clear that this can be written as a linear model with \begin{eqnarray*} y_t = \sum_{i=0}^{p-1} \beta_i \mbox{I}\left[t\bmod p = i\right] + \varepsilon_t, \end{eqnarray*} where $\beta_0$ represents $s_p$.

Instead of working with $\boldsymbol{y} = \left(y_1, \cdots, y_n\right)^{\prime}$, let us define $\boldsymbol{y}_i = \left(y_{i1}, \cdots, y_{ip}\right)^{\prime}$ and $\boldsymbol{y} = \left(\boldsymbol{y}_1^{\prime}, \cdots, \boldsymbol{y}_{n/p}^{\prime}\right)^{\prime}$. Clearly the design matrix $\boldsymbol{X} = \boldsymbol{j}_{n/p} \otimes \boldsymbol{I}_p$; hence, $\left(\boldsymbol{X}^{\prime} \boldsymbol{X} \right)^{-1} = \frac{p}{n}\boldsymbol{I}_p$, and the MLE of $\boldsymbol{\beta} = \left(\beta_0, \cdots, \beta_{p-1}\right)^{\prime}$ is \begin{eqnarray*} \widehat{\boldsymbol{\beta}} = \left(\boldsymbol{X}^{\prime} \boldsymbol{X} \right)^{-1} \boldsymbol{X}^{\prime} \boldsymbol{y} = \frac{p}{n}\sum_{i=1}^{n/p}\boldsymbol{y}_i \end{eqnarray*}

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set.seed(1)
d <- 7
n <- 100 * d
z <- rep(rnorm(d), n/d)
y <- z + rnorm(n)
acf(y, type = "correlation")
X = kronecker(rep(1,n/d),diag(1,d))
res = (diag(1,n) - X%*%solve(t(X)%*%X)%*%t(X))%*%y
acf(res, type = "correlation")

It is interesting to note that this set-up is identical the cell-means model formulation of a one-way ANOVA

Why does the level update equation change in triple exponential smoothing (but the trend equation does not)

In Holt-Winters exponential smoothing for additive model, the time series $y_t$ is forecast as $$ F_{t+k} = l_t + h T_t + s_{t - p + 1 + (h-1) \text{mod} p} ,$$ where $l_t$ is the level of the time series at time $t$, $T_t$ is the trend and $s$ represents the seasonal component.

$T_{t-1}$ is the trend at time $t-1$.
Defining the seasonal term in terms of $y_t$ and the level $l_t$ works because of an iterative definition, where $l_t$ itself is defined in terms of $y_t$ and past seasonal component.
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The equations in the link in the answer contains mistakes. Please see the correct equations in other places, e.g., in the R manual for the function.

Best Answer

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