# Why isn't mean integrated square error (MISE) probability-weighted?

We often try to minimize the MISE of a KDE: $\text{E}_{\mathbb{P}^n}[\int (\hat{p}(x) - p(x))^2 dx]$. Why don't we instead try to minimize $\text{E}_{\mathbb{P}^n}[\int (\hat{p}(x) - p(x))^2 p(x) dx]$, so that we weight deviations from the true density more when they occur in high-probability regions?

The MISE is designed to give a measure of error of an estimator $\hat{p}$ from a true density $p$. For a fixed estimate, that distance can be obtained as the squared-error, which is the internal integral in the expectation. In the standard MISE we do not weight this integral, since we consider errors to be of the same importance regardless of where they occur in the density.
Since the MISE is an expectation, it is implicitly an integral over the estimator $\hat{p}$. If we explicitly show that $\hat{p}$ depends on the data vector $\mathbf{x}$ (which is different to the argument $x$) then we have:
\begin{equation} \begin{aligned} \text{MISE}(\hat{p}) &\equiv \mathbb{E} \Big[ \int (\hat{p}(x,\mathbf{X}) - p(x))^2 dx \Big] \\[6pt] &= \int \Big( \int (\hat{p}(x,\mathbf{x}) - p(x))^2 dx \Big) f(\mathbf{x}) d\mathbf{x} \\[6pt] &= \int \int (\hat{p}(x,\mathbf{x}) - p(x))^2 f(\mathbf{x}) dx d\mathbf{x}. \\[6pt] \end{aligned} \end{equation}
So as you can see, the MISE does involve an integral of the weighted squared-error, but the weighting is with respect to the density for the data that was used in the estimator $\hat{p}$.
• Not really, since the expectation is a weighting over the variable $\mathbf{x}$, not $x$. If you would like to estimate better in particular regions (and worse) in others, you could apply a separate weighting function on $x$ as you suggested in your post. It is really a matter of how you want to measure the distance between two distributions $\hat{p}$ and $p$. – Ben Jul 9 '18 at 23:33