Bias of a kernel density estimator

For a symmetric bandwidth-$h$ averaging kernel and a density with bounded first derivative, the interior pointwise bias is at most a constant times $h\Vert f^{\prime }{\Vert }_{\infty }$. Near a support boundary, an uncorrected symmetric kernel can instead have nonvanishing bias.

For a nonnegative symmetric unit-integral kernel for density estimation and $f\in H^2(\mathbb{R})$, the integral Taylor remainder and Minkowski inequality give $\Vert K_h\ast f-f{\Vert }_2\le h^2{\mu }_2(K)\Vert f^{\prime \prime }{\Vert }_2/2$. Translation preserves the L2 norm. For a signed kernel the same proof requires the absolute moment $\int u^2|K(u)|du$.

An uncorrected nonnegative symmetric kernel density estimator can leak an order-one amount across a jump in a probability density function. If its transition region has width proportional to the smoothing bandwidth $h$, the integrated mean squared error includes squared bias of an estimator of order $h$, rather than the order $h^4$ familiar for twice-smooth interior densities. For the unit-width uniform window and the exponential distribution of rate one, the integrated squared bias of an estimator is $h/12+o(h)$.

For a density with finitely many jumps ${\Delta }_r$ at separated points and the unit-width box kernel, take $h$ smaller than all consecutive jump spacings. The smoothing bias is supported in disjoint length-$h$ neighbourhoods of the jumps. At each jump its two sides form triangles of height $|{\Delta }_r|/2$, and exact integration givesCombined with the integrated variance of a kernel density estimator, this gives expected $L^2$ error $O(n^{-1/4})$ at $h\asymp n^{-1/2}$.

Let a nonnegative kernel for density estimation $K$ have integral one, zero first moment and finite second moment ${\mu }_2(K)$. If a probability density function $f$ has bounded second derivative, its kernel density estimator has pointwise bias at most $h^2{\mu }_2(K)\Vert f^{\prime \prime }{\Vert }_{\infty }/2$. Apply the Taylor theorem with Lagrange remainder to $f(x-hu)$ and integrate: the first-order term vanishes and the remainder is bounded by $h^2{u}^2\Vert f^{\prime \prime }{\Vert }_{\infty }/2$. For $K=1_{[-1/2,1/2]}$, ${\mu }_2(K)=1/12$.