# Why $E[E[p_{N}(Z_n,Z_m)|Z_m]]\neq E[E[p_{N}(Z_n,Z_m)|Z_n]]$ here? What went wrong?

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.

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

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!

• @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? – T34driver Oct 2 '20 at 8:18
• This is a correct point. I thus wonder at the impact of the Taylor approximation in the second derivation. – Xi'an Oct 2 '20 at 8:29
• @Xi'an Thanks! That looks likely! – T34driver Oct 2 '20 at 22:57
• @Xi'an Do you have any suggestions on how this might be fixed? – T34driver Oct 2 '20 at 23:06