# Forecasting with Holt's linear trend method

In Holt's linear trend method, given a time series $$y_1,\cdots,y_t$$, the forecasting equation at time $$t+h$$ based on data up to time $$t$$ is given by $$\hat{y}_{t+h|t}= \ell_t + h b_t \tag{1}$$ where $$\ell_t$$ is an estimate of the level of the series at time $$t$$ which satisfies $$\ell_{t}=\alpha y_t + (1-\alpha)(\ell_{t-1}+b_{t-1}) \tag{2}$$ and $$b_t$$ an estimate of the trend (slope) of the series at time $$t$$ which satisfies $$b_{t}=\beta \color{blue}{(\ell_t-\ell_{t-1})}+ (1-\beta)b_{t-1} \tag{3}$$

$$\alpha \in [0,1]$$ and $$\beta \in [0,1]$$ are the smoothing parameters.

The trend equation $$(3)$$ shows that $$b_t$$ is a weighted average of the estimated trend at time $$t$$ based on $$\ell_t-\ell_{t-1}$$ and the previous estimate of the trend, $$b_{t-1}$$. My question is, why not compute this average based on the real trend between $$t-1$$ and $$t$$? as follows:

$$b_{t}=\beta \color{red}{(y_t-y_{t-1})}+ (1-\beta)b_{t-1} \tag{3}$$

Suppose that the series $$y_t$$ actually satisfies a linear trend model, say $$y_t = a_0 + a_1 t + \epsilon_t$$ where $$\epsilon_t$$ is some i.i.d. random noise (mean 0, constant variance). Let's also assume that we are far enough along in applying whichever smoothing formulas we chose so that $$\ell_{\tau} \approx a_0 + a_1 \tau$$ and $$b_{\tau}\approx a_1.$$ According to the first Holt formula, \begin{align*} \ell_{t} & =\alpha(a_{0}+a_{1}t+\epsilon_{t})+(1-\alpha)(\ell_{t-1}+b_{t-1})\\ & \approx\alpha(a_{0}+a_{1}t+\epsilon_{t})+(1-\alpha)(a_{0}+a_{1}[t-1]+a_{1})\\ & =a_{0}+a_{1}t+\alpha\epsilon_{t}. \end{align*} Using Holt's second formula as stated, we get \begin{align*} b_{t} & =\beta(\ell_{t}-\ell_{t-1})+(1-\beta)b_{t-1}\\ & \approx\beta(a_{0}+a_{1}t+\alpha\epsilon_{t}-[a_{0}+a_{1}(t-1)])+(1-\beta)a_{1}\\ & =a_{1}+\beta\alpha\epsilon_{t}, \end{align*} whereas with your second formula we get \begin{align*} b_{t} & =\beta(y_{t}-y_{t-1})+(1-\beta)b_{t-1}\\ & \approx\beta(a_{1}+\epsilon_{t}-\epsilon_{t-1})+(1-\beta)a_{1}\\ & =a_{1}+\beta(\epsilon_{t}-\epsilon_{t-1}). \end{align*} Both estimates of slope are unbiased, but yours contains more noise.
Addendum: That was a bit hand-wavy even by my lax standards, so let me try to make it slightly more rigorous. Let $$\lambda_t$$ and $$\eta_t$$ be the errors in $$\ell_t$$ and $$b_t$$ respectively, i.e., $$\ell_t = a_0 + a_1 t +\lambda_t$$ and $$b_t = a_1 + \eta_t.$$ The Holt formula for $$b_t$$ reduces to $$b_t = a_1 + \beta(\lambda_t - \lambda_{t-1}) + (1-\beta)\eta_t$$and the proposed alternative reduces to $$b_t = a_1 + \beta(\epsilon_t - \epsilon_{t-1}) + (1-\beta)\eta_t.$$ Assuming that the exponential smoothing is actually smoothing things, we expect $$\lambda_t$$ to have lower variance than $$\epsilon_t.$$ I'm pretty sure you can prove that via an induction argument.