A kernel is a density

Topic: Hierarchical clustering and kernel density estimation · more questions on this topic
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Show that \widehat f(\mathbf{x};H) integrates to 1 whenever K does, using the change of variables \mathbf{u}=H^{-1/2}\mathbf{x} in each term.

Worked answer (click to show)

The bandwidth stretches the kernel, so the factor |H|^{-1/2} cancels the Jacobian. For each sample point,

\int K_H(\mathbf{x}-\mathbf{x}_i)\,d\mathbf{x} =|H|^{-1/2}\int K\bigl(H^{-1/2}(\mathbf{x}-\mathbf{x}_i)\bigr)\,d\mathbf{x}.

Put \mathbf{u}=H^{-1/2}(\mathbf{x}-\mathbf{x}_i). Then d\mathbf{x}=|H|^{1/2}\,d\mathbf{u}, because the Jacobian determinant of \mathbf{x}=H^{1/2}\mathbf{u}+\mathbf{x}_i is |H|^{1/2}. The integral becomes \int K(\mathbf{u})\,d\mathbf{u}=1. Each of the n terms integrates to 1, and the factor 1/n gives total mass 1.