# Forecast is simply equal to the lag of the original time series

I am currently dealing with the problem of short time series which often involves naive models as they already perform well enough. So I implemented an exponential smoothing that follows

$$F_t = \alpha y_{t-1} + (1-\alpha)f_{t-1}$$

where $$F_t$$ is the forecast value now based on the smoothed value of the true value of the previous time step $$y_{t-1}$$ and the previous forecast $$f_{t-1}$$ which was simply taken as the average of all previous true values (a.k.a. the expanding mean in Python).

I'm investigating the impact of the choice of alpha and generally, at least for my specific use case, $$\alpha \geq 0.9$$ performs well. But if you scrutinize the forecasted time series, it is simply a lagged version of the original time series. I did a visual inspection on this, and also calculated the correlation between the forecast for each alpha and a lag 1 of $$y$$.

My main question is, is a forecast which is simply a lag of the original time series said to be a good fit since the r2-score says so? I understand, this may be an overfit, but if you compare other values of alpha, they're already negative r2.

Here's a graph to visualize:

• R2 is never a good indicator for anything useful by itself IMO. I would judge based on a holdout accuracy or you can use the state space representation of ETS and get an information criterion. Just by looking at your data there is a trend so theoretically it makes sense that the best it could do is predict the last value. – Tylerr Apr 27 at 17:55
• @Tylerr i've only stated looking into TS so I'll need to read up on state space representations. Thanks for this! – HQ_nought Apr 29 at 1:28