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The mean integrated square error of a density estimate $\hat p$ of a true density $p$ is $$\begin{equation} \begin{aligned} \text{MISE}(\hat{p}) &\equiv \mathbb{E} \Big[ \int (\hat{p}(x,\mathbf{X}) - p(x))^2 dx \Big] \\[6pt] &= \int \Big( \int (\hat{p}(x,\mathbf{x}) - p(x))^2 dx \Big) f(\mathbf{x}) d\mathbf{x} \\[6pt] &= \int \int (\hat{p}(x,\mathbf{x}) - p(x))^2 f(\mathbf{x}) dx d\mathbf{x}. \\[6pt] \end{aligned} \end{equation}$$

I wanted to compute this so I looked up some source code and found the following code in the deamer package for R:

#calculation of the MISE
mise = function(density, obj){
if(class(obj)=="deamer"){
 supp <- obj$supp
 est <- obj$f
 } else { stop("argument 'obj' must be of class 'deamer'")} 

if(class(density)=="function"){
     dens <- density(supp)
     } else {
     stop("argument 'density' should be a function (see ?mise)")
   }

mise <- ((max(supp)-min(supp))/length(supp))*sum((dens-est)^2)
return(mise)
}

This does not look correct to me. It seems they are just calculating the ISE $\int (\hat{p}(x,\mathbf{x}) - p(x))^2 dx$ and not the MISE since I don't see how they are computing the expectation?

So have I misinterpreted something, how can the above code produce the MISE when it doesn't perform an integration with respect to the sample?

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