0
$\begingroup$

Can somebody explain to me how I would interpret the result of the MRAE.

In my textbook the MRAE is defined as followed:

$$ MRAE= {1 \over n}\left(\sum_{t=1}^n \left|{e_t \over e_t^*}\right|\right) $$

with $e_t=\text{actual value}-\text{forecasted value}$ and $e_t^*$ being the benchmarking method where $e_t^*=e_{t-1}$.

Now consider these simple example: Example 1: | actual | forecast | error | |--------|----------|---------| | 10 | 5 | 5 | |8 | 4 | 4 | |20 | 10| 10|

this would result in $$ MRAE= {1 \over 2}\cdot\left({4 \over 5}+{10 \over 4}\right)=1,65 $$

Example 2: | actual | forecast | error | |--------|----------|---------| | 10 | 1 | 9 | |8 | 1 | 7 | |20 | 1| 19|

this would result in $$ MRAE= {1 \over 2}\cdot\left({7 \over 9}+{19 \over 7}\right)=1,75 $$

Now what does this tell me? Am I 65% (75% respectively) better than the benchmarking method? And if so, what does that tell me about the actual forecasting method?

Thanks for your help.

$\endgroup$
0
$\begingroup$

The MRAE is a particularly unhelpful way to measure forecast accuracy. For any reasonable forecasting method, the errors can be positive or negative, so the distribution of the errors will have non-zero density at 0. Therefore the ratio of the two errors in the MRAE will have an undefined mean and infinite variance. Consequently the MRAE cannot converge and is not a consistent estimator of anything. This is one of the problems I pointed out in Hyndman & Koehler (2006).

I suggest you do not use the MRAE, despite some textbooks still including it.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.