eriv\])). Therefore, Taylor expansion gives $$\label{taylor}
g(u)=\sum_{k=0}^3g^{(i)}(0)\frac{u^i}{ i! }+\frac{u^4}{ 4!}E_\tau g^{(4)}(\tau u)$$ where $\tau$ is a random variable with density $\lambda(x)=4(1-x)^3$, $0\le x\le 1,$ that does not depend on $t$, $w$ or $u$, and $E_\tau$ denotes expectation with respect to this variable. Equation (\[taylor\]) can be easily verified by integration by parts in $\int_0^14(1-t)^3g^{(4)}(tu)dt$. Next note that $$\label{firstt}
\int_{-B}^Bg_{t,w}(0)dw=\int_{-B}^Bf^{3/2}(t)K(wf^{1/2}(t))dw=\int_{-Bf^{1/2}(t)}^{Bf^{1/2}(t)}f(t)K(v)dv=f(t)$$ since the support of $K$ is contained in $[-Bf^{1/2}(t),Bf^{1/2}(t)]$ by the hypothesis on $B$ and since $K$ integrates to 1. Further, since $s'$ contains a $w$ factor, there are functions $c_i(f,t)$, $i=1,2$, such that $$\int_{-B}^Bwg_{t,w}'(0)dw=\int_{-Bf^{1/2}(t)}^{Bf^{1/2}(t)}(c_1(f,t)vK(v)+c_2(f,t)v^2K'(v))dv=0$$ because $K$ is even and $K'$ is odd. Similarly (that is, using only the symmetry properties of $K$ and its derivatives), we also get $\int_{-B}^B w^3g_{t,w}'''(0)dw=0.$ That these two integrals vanish is obvious and not surprising; what is remarkable is that also $\int_{-B}^Bw^2g_{t,w}''(0)dw=0$, and this fact is the main reason for the bias reduction achieved by Abramson’s (1982) ‘inverse square root rule’. We sketch an argument for completeness. Note first that, from the expression for $g''$ in (\[deriv\]), integrating by parts, $$\begin{aligned}
\int_{-B}^Bw^2[r(s')^2K''(s)](0)dw&=&{1\over 4}f^{1/2}(t)(f')^2(t)\int_{-B}^Bw^4K''(wf^{1/2}(t))dw\\
&=&-(f')^2(t)\int_{-B}^Bw^3K'(wf^{1/2}(t))dw.\end{aligned}$$ Collecting terms in $K'$, this gives $$\begin{aligned}
\int_{-B}^Bw^2g_{t,w}''(0)dw&=&\left[{3\over 4}f^{-1/2}(t)(f')^2(t)+{3\over 2}f^{1/2}(t)f''(t)\right]\int_{-B}^Bw^2K(wf^{1/2}(t))dw\\
&&~~~~~~~~+\left[{1\over 4}(f')^2(t)+{1\over 2}f(t)f''(t)\right]\int_{-B}^Bw^3K'(wf^{1/2}(t))dw,\end{aligned}$$ and, integrating by parts the second integral, we get zero. \[See Novak (1999) for a proof that, if one replaces $f^{1/