RMSE(Root mean square error) value as a model comparison method Most of times RMSE is used to compare models where:
$RMSE=\sqrt{(\sum_i (y_{i,pred}-y_{i,obs})^2)/n}$ 
or namely
$RMSE=(\sum_i (y_{i,pred}-y_{i,obs})^2/n)^{(1/2)}$
For some compared models, the gap between $y_{i,pred}$ and $y_{i,obs}$ can be large for one/some specific $i$, because of randomness.
And, that gap may dominate $RMSE$. 
Even a model is better than other in real, because of even one big gap, $RMSE$ may choose the worse model.
For such situations, is it realistic to use below criteria:
$RMSE=(\sum_i (y_{i,pred}-y_{i,obs})^2/n)^{(1/p)}$
where $0<p<1$.
For example, if $p$ is chosen as 0.5, the effect of large deviations will be smaller. 
I want to learn


*

*Is this method a realistic method?

*If yes, what should be the optimum value of $p$. Is there any method to chose optimum $p$?

*What are the any other methods to eliminate effect of large deviations.


I will be very glad for any help or source recommendation.
Thanks a lot.
 A: First, you should not be taking square roots the way you are.$^{\dagger}$ Either take the square root of everything, including the $/n$ denominator, to give RMSE, or don’t take a square root at all to give MSE. Your first equation, MSE, and RMSE are all basically the same in the you’ll rank model performance the same way no matter which of the three you use (assuming the same data set), but you’ll cause confusion with what you gave. That equation is not standard.
Second, what you might be looking for is absolute error:
$$MAE=\dfrac{\sum_{i=1}^N \big\vert y_i -\hat{y}_i \big\vert}{N}$$
This tends to lessen the impact of extremely bad predictions, making your model somewhat more willing to make a couple of awful predictions in exchange for a better fit of most of the points. This also results in a somewhat more straightforward interpretation of the average error in that MAE is the average amount by which a prediction misses. 
What you propose about taking the $p^{th}$ root of the sum of squared errors does not make sense. Check it out on a couple of models on the same dataset. No matter which $p$ you choose, you’ll rank your models the same way.
What could make sense is the following:
$$\bigg(\sum_{i=1}^N \big\vert y_i -\hat{y}_i \big\vert^p\bigg)^{1/p}$$
This is related to $L^p$ norms in linear algebra and functional analysis.
When $p=1$, this is absolute loss. When $p=2$, this is square loss. 
As $p$ gets smaller, the model is more and more willing to make a few awful predictions in order to get a better fit of most of the points. As $p$ gets larger, the model obsessed over not making those awful errors, perhaps at the expense of poorly fitting most of the points. When $p=\infty$, all the model wants to do is minimize the maximum error.
$p=1$ and $p=2$ are popular because they can be interpreted as giving models predicting the conditional median and mean, respectively.
$^{\dagger}$There were some errors in the original post that seem to have been corrected.
