We have a U-statistic defined as $U=\frac{1}{N(N-1)}\sum_{n\neq m}p_{N}(Z_n,Z_m)$, where observations $\{Z_i\}_{i=1}^{N}$ are i.i.d. following distribution $F(z)$ with density $f(z)$.

$p_{N}(Z_n,Z_m)=\frac{1}{f(Z_n)}g(Z_m)\frac{1}{h^2}k'(\frac{Z_m-Z_n}{h})$, where $g(\cdot),f(\cdot)$ are three times differentiable, and $k(\cdot)$ is a standard symmetric second order kernel function with bounded support $[\underline{u},\overline{u}]$ and equals zero on the boundary of support, and $h\rightarrow 0$ is the bandwidth.


Do a change of variable of $\frac{Z_m-z}{h}=u$, we have $E[p_{N}(Zn,Z_m)|Z_m]=-g(Z_m)\frac{1}{h}\int k'(u)du=-g(Z_m)\frac{1}{h}[k(\overline{u})-k(\underline{u})]=0$

However, $E[p_{N}(Z_n,Z_m)|Z_n]=\frac{1}{f(Z_n)}\int g(z)\frac{1}{h^2}k'(\frac{z-Z_n}{h})f(z)dz$

Do a change of variable of $\frac{z-Z_n}{h}=u$, we have

$E[p_{N}(Z_n,Z_m)|Z_n]=\frac{1}{f(Z_n)}\int g(uh+Z_n)\frac{1}{h}k'(u)f(uh+Z_n)du$

Integration by parts gives $\frac{1}{f(Z_n)}\int g(uh+Z_n)\frac{1}{h}k'(u)f(uh+Z_n)du=0-\frac{1}{f(Z_n)}\int \frac{\partial [g(uh+Z_n)f(uh+Z_n)]}{\partial u}\frac{1}{h}k(u)du$

Taylor expansion then gives

$-\frac{1}{f(Z_n)}\int \frac{\partial [g(uh+Z_n)f(uh+Z_n)]}{\partial u}\frac{1}{h}k(u)du=-\frac{1}{f(Z_n)}[g'(Z_n)f(Z_n)+g(Z_n)f'(Z_n)]+O_p(h^2).$

By law of iterated expectations this imply $E[E[p_{N}(Z_n,Z_m)|Z_m]]=E[0]=0$, but $E[E[p_{N}(Z_n,Z_m)|Z_m]]=E[-\frac{1}{f(Z_n)}[g'(Z_n)f(Z_n)+g(Z_n)f'(Z_n)]+O_p(h^2)]\neq 0$.

However, by definition $E[E[p_{N}(Z_n,Z_m)|Z_m]]=E[E[p_{N}(Z_n,Z_m)|Z_n]]=E[p_{N}(Z_n,Z_m)]$. So where went wrong in the above derivation? How can we possibly fix it? Any comments or suggestions are welcome. Thanks!

  • $\begingroup$ @Xi'an Thanks! But I don't think symmetry is the problem here, as $E[m(X,Y)]=E[E[m(X,Y)|X]]=E[E[m(X,Y)|Y]]$ should hold for any (integrable) function $m(x,y)$ by law of iterated expectations, regardless of whether $m(x,y)$ is symmetry or not. Right? $\endgroup$ – T34driver Oct 2 '20 at 8:18
  • $\begingroup$ This is a correct point. I thus wonder at the impact of the Taylor approximation in the second derivation. $\endgroup$ – Xi'an Oct 2 '20 at 8:29
  • $\begingroup$ @Xi'an Thanks! That looks likely! $\endgroup$ – T34driver Oct 2 '20 at 22:57
  • $\begingroup$ @Xi'an Do you have any suggestions on how this might be fixed? $\endgroup$ – T34driver Oct 2 '20 at 23:06

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