# What's the intuitive purpose of RMSE (root mean square error) compared to MAE (mean average error)?

When I want to find out what the average difference / error there is between two datasets, such as a predicted output vs. observed output of any system (i.e.: I predict output to be 100V, how does the actual measured output compare?) , intuitively I would do:

\begin{align} \frac{\sum_{i = 1}^{i = n} {|{P(i)-O(i)}|}}{n} \end{align}

where $$P$$ is a predicted value, $$O$$ is the observed value, for every instance $$i$$ up to $$n$$ instances. This would technically be defined as the MAE or mean average error.

However, I have seen another way to compare observed and predicted values using the RMSE, or root mean square error, defined as,

\begin{align} \sqrt{\frac{\sum_{i = 1}^{i = n} {\{{P(i)-O(i)}\}^2}}{n}} \end{align}

For someone who just wants a good idea of the average differences between two sets of data (in the case of a predicted output vs. observed output scenario), which method would be more useful, RMSE or MAE?

• They should both be fine. RMS is more common since in many cases squaring/square root operations are easier to deal with than absolute value calculations. Think about derivatives, etc. – Michael Biro Apr 3 at 0:15
• Also, if you are trying to find an regression model for a data set, defining you error by a RMS is going to penalize higher errors much more in comparison to small errors. – D.B. Apr 3 at 0:20
• That property of RMS can also be a disadvantage when your data set includes outliers. I think the main appeal of RMS for regression is it is much easier to calculate. (The "predicted" value is part of the output of that process.) – David K Apr 3 at 2:03
• @DavidK Any example of how a predicted value can be a part of the output of a process? – plu Apr 3 at 23:48
• A regression model is a estimate of the underlying trend of the data, which is a kind of prediction. You might get a plot with a line (for linear regression) that predicts what $y$ value you would get at each $x$ value if the data followed the model perfectly--but usually the observed $y$ values are not exactly on the line, so now at each of those points you can measure a difference between the predicted value and the observed value. – David K Apr 4 at 1:39