I am performing daily and weekly forecasts for 28 days and 4 weeks respectively.

Once I have used the same model to obtain the respective forecasts an root mean square errors (RMSE), I will like to make a conclusion based on just these results, as to whether the model is more suited for daily or weekly forecasting.

Denote the daily forecasts RMSE to be $RMSE_d$ and weekly forecasts RMSE to be $RMSE_w$. I then arrived at the following formulas and proposition:

$$RMSE_d = \sqrt{\frac{1}{n_i} \sum_j (y_{ij} - \hat{y}_{ij})^2}$$

$$RMSE_w = \sqrt{\frac{1}{n} \sum_i (y_i - \hat{y}_i )^2}$$

where $n = \sum_i n_i , y_i = \sum_j y_{ij}$, for weeks $i = 1, \dots , n$ and days $j=1,\dots,7$.

Then for comparing both metrics, I suppose that $y_i \approx 7 y_{ij}$ for each $i$. Am I right to conclude that $RMSE_w \approx 7^2 RMSE_d$?

I can't seem to find any related online literature nor on this site, but I apologize if there has indeed been a post like this before..

Some insights will be deeply appreciated!

  • $\begingroup$ I think you are slightly off the right track. A cumulative standard deviation of residuals for a n-day forecast = sqrt(n) * RMSE, so try taking a square root instead of power, and I guess you should see very similar values for both models. $\endgroup$ – Alexey Burnakov Sep 12 at 13:14
  • $\begingroup$ @AlexeyBurnakov Thanks for your feedback. Just to clarify, do you mean something like $RMSE_w \approx 7 \times RMSE_d$ or $RMSE_w \approx \sqrt{7} RMSE_d$, instead of $RMSE_w \approx 7^2 RMSE_d$? $\endgroup$ – Stoner 2 days ago
  • $\begingroup$ I mean the second formula, which is correct. $\endgroup$ – Alexey Burnakov 2 days ago

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