Why do you take the sqrt of 1/n for RMSE? Updated question:
Why do we use RMSE:
$$RMSE = \sqrt{\frac{1}{n}\Sigma_{i=1}^{n}{\Big(\hat{y}_i -y_i\Big)^2}}$$
Why is it not MRSE:
$$MRSE = \frac{1}{n}\sqrt{\Sigma_{i=1}^{n}{\Big(\hat{y}_i -y_i\Big)^2}}$$
I understand that other methods (e.g., MAE and MAPE) can be used as a metric for error. My question is specifically about why we use RMSE over MRSE.
Original:
Why is the equation for RMSE:
$$RMSE = \sqrt{\frac{1}{n}\Sigma_{i=1}^{n}{\Big(\hat{y}_i -y_i\Big)^2}}$$
Why is it not:
$$RMSE = \frac{1}{n}\sqrt{\Sigma_{i=1}^{n}{\Big(\hat{y}_i -y_i\Big)^2}}$$
What is the reason for taking the square root of 1/n?
 A: The goal is to have an unbiased estimator for the error your model makes on average. Let's call that $\bar \epsilon$. Now let's see what's the relationship of the two estimators you asked about with the $\bar \epsilon$:
$\hat y_{i} - y_{i} = \epsilon_{i}$
$\frac{1}{n}\sum_{i=1}^{n}(\hat y_{i} - y_{i})^2 = \frac{1}{n}\sum_{i=1}^{n}(\epsilon_{i})^2 = \bar \epsilon^2$
thus
$ \sqrt{\frac{1}{n}\sum_{i=1}^{n}\left(\hat{y}_{i}-y_{i}\right)^{2}} \approx \bar \epsilon$
which is what we aimed for. Now let's see what the other estimator will give you:
$\frac{1}{n} \sqrt{\sum_{i=1}^{n}\left(\hat{y}_{i}-y_{i}\right)^{2}} = $
$\frac{1}{n} \sqrt{n \times \frac{1}{n}\sum_{i=1}^{n}\left(\hat{y}_{i}-y_{i}\right)^{2}} = $
$\frac{\sqrt{n}}{n} \times \bar \epsilon$
As you can see the second estimator has a bias of $\frac{\sqrt{n}}{n}$ in estimating the average error you aimed for. For example, if for a data generating process of $f(x)  = 0$ you always predict 2 then you would want the estimator to give you $\bar \epsilon = 2$ which is given by the first estimator while the second estimator (assuming n = 10) will give you ($2 \times \frac{\sqrt 10}{10}$).
A: While Demetri's answer gives a very good derivation or RMSE, it doesn't really explain why not the other method you suggest.  I think you can get a little more insight by observing that MRSE is not a valid name for your suggested measure.  Look closely and the steps are

*

*Square the residuals

*Add them up

*Square root

*Divide by the number of samples

A "mean" needs to have the sum and the divide consecutive.  So the MRSE would actually be:
$$ MRSE = \frac{1}{n} \sum \sqrt{(\hat{y}_i - y_i)^2} = \frac{1}{n}\sum |\hat{y}_i - y_i| = MAE$$
So, RMSE is the square-root of a mean - it is then just transformed (by square root) for convenience.  The MAE is itself a mean.  What you have created, isn't a mean - you are not adding things up and dividing by the number there are, you are adding things up, then square rooting, then dividing by the number there are.  In fact  the construct before the 1/n is a Euclidean distance - the total distance that the sample is from the predicted y-vector.  As pointed out by Amin's answer, this error naturally grows as sqrt of the size of the y-vector, so by dividing by n your error will systematically get smaller the larger the sample.
A: Interesting question. Let's break this down into:  Why squared error, why mean squared error, and then why root mean squared error.  I think that should answer your question.

Why squared error (SE)

Squared error happens to be a proper scoring rule, which is a really desirable property for your loss function to have (feel free to read up on proper scoring rules by searching this site).  However, the squared error can grow simply by just adding more data.  So if I have two data sets (maybe one from yesterday and one from today), and they are of different sizes, I could be fooled into thinking my model is doing poorly simply because I had more data today than yesterday.  Which leads me to...

Why mean squared error (MSE)

Taking the mean of the squared eliminates this problem of different data sizes.  By taking the average loss, we retain the nice properties of the proper scoring rule, but now can compare the loss of a model on different data sets of possibly different sizes.  But the interpretation of MSE is kind of hard.  If $y$ is measured in dollars, what is a dollar squared?  Which leads me too...

Why root mean squared error (RMSE)

MSE has weird units, but if we took the square root of MSE the result would be on the scale of $y$.  This makes interpretation a little easier.
In summation:

*

*SE is a proper scoring rule.  We like that

*To prevent misleading inflation of the error due to sample sizes, we take the average of SE, or MSE

*MSE is hard to interpret, so instead we take the square root of MSE to get RMSE and have the error units on the same scale as the outcome.

A: I think both RMSE and MRSE could be potentially used for the purpose of creating a metric related to residuals per data point. The difference, and why the RMSE is commonly used and MRSE is not probably lies in interpretation of the terms and their related metrics.
If we roll back RMSE to SE at every step of the way we get a term that is commonly used and interpretable. We square RMSE and get MSE (variance), we square root it and get SE. However, rolling back MRSE does not give as nice a terms. We multiply MRSE by sample size and get RSE, which is not commonly used to evaluate anything. This may be the reason why one is used and the other isn't.
