Finite-bandwidth Gaussian smoothing

Put $I={\int }_{\mathbb{R}}{e}^{-t^2}dt$. Tonelli [F5] and [F4] give $I^2=\pi$, so $I=\sqrt{\pi }$. Scaling by $s=t\mathrm{/}\sqrt{2}$ with [F6] yields $\int (2\pi {)}^{-1\mathrm{/}2}{e}^{-t^2\mathrm{/}2}dt=1$; repeated Tonelli gives $\int K_d=1$. Since $f_{\tau }(x)={\tau }^{-d}{K}_d(x\mathrm{/}\tau )$, [F6] with $x=\tau u$ gives $\int f_{\tau }=1$.

By [F1] and [F2], expectation of one KDE summand is integration of its Gaussian kernel against $f_{\tau }$. The summand is bounded for fixed $h$, hence integrable. The integrand is nonnegative; using [F6] with $T(u)=x-hu$ changes variables to $\int K_d(u)f_{\tau }(x-hu)du$. All summands have this expectation, so linearity [F3] gives $\mathbb{E}{\widehat{f}}_h(x)={\int }_{{\mathbb{R}}^d}{K}_d(u)f_{\tau }(x-hu)du\mathrm{.}$ The integral is finite because both Gaussian densities are bounded and integrable.

In the integrand of step 2.1, combine exponents using $\mathrm{\parallel }u{\mathrm{\parallel }}^2+{\tau }^{-2}\mathrm{\parallel }x-hu{\mathrm{\parallel }}^2=\frac{{\tau }^2+h^2}{{\tau }^2}{\parallel u-\frac{hx}{{\tau }^2+h^2}\parallel }^2+\frac{\mathrm{\parallel }x{\mathrm{\parallel }}^2}{{\tau }^2+h^2}\mathrm{.}$ The remaining translated Gaussian integral is $(2\pi {\tau }^2\mathrm{/}({\tau }^2+h^2){)}^{d\mathrm{/}2}$ by [F4], [F5], and affine scaling [F6]. Multiplying the normalizing constants gives the displayed formula.

At $x=0$, $\mathbb{E}{\widehat{f}}_h(0)=(2\pi ({\tau }^2+h^2){)}^{-d\mathrm{/}2}<(2\pi {\tau }^2{)}^{-d\mathrm{/}2}=f(0),$ because $h>0$. Therefore a finite positive bandwidth has nonzero bias in this example.