10.5 Appendix
Proof
(Proof of Theorem 1) Let us take into account a random sample \(\left( X_1,\ldots ,X_n\right) \) which comes from a stationary process and the smoothed moving blocks bootstrap version of the kernel density estimator, \(\hat{f}^*_h(x)\). The bootstrap version of the mean integrated squared error is given by:
$$\begin{aligned} MISE^*(h)= B^*(h) + V^*(h), \end{aligned}$$
(10.6)
where
$$\begin{aligned} B^*(h)= & {} \int \left[ \mathbb {E}^*\left( \hat{f}_h^*(x)\right) - \hat{f}_g(x)\right] ^2 dx\text {, and }\\ V^*(h)= & {} \int \text {Var}^*\left( \hat{f}_h^*(x)\right) dx.\\ \end{aligned}$$
Now, straight forward calculations lead to
$$\begin{aligned} B^*(h)= & {} \int \left[ \mathbb {E}^*\left( \dfrac{1}{n}\sum \limits _{i=1}^n K_h(x-X_i^*) \right) - \hat{f}_g(x)\right] ^2 dx \\= & {} \int \left[ \dfrac{1}{n}\sum \limits _{i=1}^n \int K_h(x-y)\hat{f}_g^{(i)}(y)dy - \hat{f}_g(x)\right] ^2 dx, \end{aligned}$$
where
$$\hat{f}_g^{(i)}(y)= \dfrac{1}{n-b+1} \sum \limits _{j=t_i}^{n-b+t_i} K_g(y-X_j), $$
considering \(t_i=[(i-1)\text {mod }b]+1\).
Let us now assume that n is an integer multiple of b:
$$\begin{aligned}&\int \left[ \dfrac{1}{n}\sum \limits _{i=1}^n \int K_h(x-y)\hat{f}_g^{(i)}(y)dy - \hat{f}_g(x)\right] ^2 dx\\= & {} \int \left[ \dfrac{1}{n}\sum \limits _{i=1}^b \dfrac{n}{b} \left( K_h *\hat{f}_g^{(i)}\right) (x) - \hat{f}_g(x)\right] ^2 dx\\= & {} \int \left[ \dfrac{1}{b}\sum \limits _{i=1}^b \left( K_h *\hat{f}_g^{(i)}\right) (x) - \hat{f}_g(x)\right] ^2 dx\\= & {} \int \left[ \dfrac{1}{b}\sum \limits _{i=1}^b \left( \dfrac{1}{n-b+1} \sum \limits _{j=t_i}^{n-b+t_i} K_h*K_g(\cdot -X_j) \right) (x) - \hat{f}_g(x)\right] ^2 dx\\= & {} \int \left[ \dfrac{1}{b}\sum \limits _{i=1}^b \left( \dfrac{1}{n-b+1} \sum \limits _{j=t_i}^{n-b+t_i} \int K_h(x-y)K_g(y-X_j)dy \right) - \hat{f}_g(x)\right] ^2 dx\\= & {} \int \left[ \dfrac{1}{b}\sum \limits _{i=1}^b \left( \dfrac{1}{n-b+1} \sum \limits _{j=t_i}^{n-b+t_i} \int K_h(x-u-X_j)K_g(u)du \right) - \hat{f}_g(x)\right] ^2 dx\\= & {} \int \left[ \dfrac{1}{b}\sum \limits _{i=1}^b \left( \dfrac{1}{n-b+1} \sum \limits _{j=t_i}^{n-b+t_i} K_h *K_g (x-X_j) \right) - \hat{f}_g(x)\right] ^2 dx\\= & {} \int \left[ \dfrac{1}{b(n-b+1)}\sum \limits _{i=1}^b \sum \limits _{j=t_i}^{n-b+t_i} K_h *K_g (x-X_j) - \hat{f}_g(x)\right] ^2 dx.\\ \end{aligned}$$
Furthermore, if \(b<n\)
$$\begin{aligned}&\dfrac{1}{b(n-b+1)}\sum \limits _{i=1}^b \sum \limits _{j=t_i}^{n-b+t_i} K_h *K_g (x-X_j)\\= & {} \dfrac{1}{n-b+1} \sum \limits _{j=b}^{n-b+1} K_h *K_g (x-X_j)+\dfrac{1}{b(n-b+1)} \sum \limits _{j=1}^{b-1}j(K_h*K_g) (x-X_j)\\&+\dfrac{1}{b(n-b+1)} \sum \limits _{j=n-b+2}^{n} (n-j+1) (K_h *K_g) (x-X_j)\\= & {} \sum \limits _{j=1}^{n}a_j (K_h*K_g)(x-X_j),\\ \end{aligned}$$
where \(a_j\) (10.3).
If \(b=n\),
$$\begin{aligned} \dfrac{1}{b(n-b+1)}\sum \limits _{i=1}^b \sum \limits _{j=t_i}^{n-b+t_i} K_h *K_g (x-X_j)= & {} \dfrac{1}{n}\sum \limits _{j=1}^{n} K_h*K_g(x-X_j)\\= & {} \sum \limits _{j=1}^{n} a_j \left( K_h*K_g\right) (x-X_j), \end{aligned}$$
considering \(a_j=\dfrac{1}{n}\), if \(b=n\).
Hence, carrying on with the calculations of the integrated bootstrap bias (including several changes of variable and using the symmetry of K) results in:
$$\begin{aligned} B^*(h)= & {} \int \left[ \sum \limits _{j=1}^{n}a_j \left( K_h *K_g \right) (x-X_j) - \hat{f}_g(x)\right] ^2 dx\\= & {} \int \left[ \sum \limits _{j=1}^{n}a_j \left( K_h *K_g \right) (x-X_j) - \dfrac{1}{n} \sum \limits _{j=1}^{n} K_g(x-X_j)\right] ^2 dx\\= & {} \sum \limits _{j=1}^{n}\sum \limits _{k=1}^{n}\int \left[ a_j \left( K_h *K_g \right) (x-X_j) - \dfrac{1}{n} K_g(x-X_j)\right] \\&\times \left[ a_k \left( K_h *K_g \right) (x-X_k) - \dfrac{1}{n} K_g(x-X_k)\right] dx\\= & {} \sum \limits _{j=1}^{n}\sum \limits _{k=1}^{n}\int \left[ a_j a_k \left( K_h *K_g \right) (x-X_j) \left( K_h*K_g\right) (x-X_k) \right. \\&- \left. \dfrac{2a_j}{n} \left( K_h *K_g\right) (x-X_j) K_g(x-X_k) + \dfrac{1}{n^2} K_g(x-X_j) K_g(x-X_k) \right] dx\\= & {} \sum \limits _{j=1}^{n}\sum \limits _{k=1}^{n}a_j a_k\int \left[ \left( K_h *K_g \right) (x-X_j) \left( K_h*K_g\right) (x-X_k)\right] dx \\&- \dfrac{2}{n}\sum \limits _{j=1}^{n}a_j \sum \limits _{k=1}^{n} \int \left[ \left( K_h *K_g\right) (x-X_j) K_g(x-X_k) \right] dx\\&+ \dfrac{1}{n^2}\sum \limits _{j=1}^{n}\sum \limits _{k=1}^{n}\int \left[ K_g(x-X_j) K_g(x-X_k) \right] dx\\= & {} \sum \limits _{j=1}^{n}\sum \limits _{k=1}^{n}a_j a_k\int \left[ \left( K_h *K_g \right) (-v) \left( K_h*K_g\right) (X_j-X_k-v)\right] dv \\&- \dfrac{2}{n} \sum \limits _{j=1}^{n}a_j \sum \limits _{k=1}^{n} \int \left[ \left( K_h *K_g\right) (-v) K_g(X_j-X_k-v) \right] dv\\&+ \dfrac{1}{n^2}\sum \limits _{j=1}^{n}\sum \limits _{k=1}^{n}\int \left[ K_g(-v) K_g(X_j-X_k-v) \right] dv\\= & {} \sum \limits _{j=1}^{n}\sum \limits _{k=1}^{n}a_j a_k\left[ \left( K_h *K_g \right) *\left( K_h*K_g\right) \right] (X_j-X_k) \\&- \dfrac{2}{n}\sum \limits _{j=1}^{n}a_j \sum \limits _{k=1}^{n}\left[ \left( K_h *K_g\right) *K_g\right] (X_j-X_k) + \dfrac{1}{n^2}\sum \limits _{j=1}^{n}\sum \limits _{k=1}^{n}\left[ K_g*K_g\right] (X_j-X_k).\\ \end{aligned}$$
Thus,
$$\begin{aligned} B^*(h)= & {} \sum \limits _{j=1}^{n} a_j \sum \limits _{k=1}^{n}a_k\left[ \left( K_h *K_g \right) *\left( K_h*K_g\right) \right] (X_j-X_k) \\ \nonumber&- \dfrac{2}{n}\sum \limits _{j=1}^{n} a_j \sum \limits _{k=1}^{n} \left[ \left( K_h *K_g\right) *K_g\right] (X_j-X_k) + \dfrac{1}{n^2}\sum \limits _{j=1}^{n}\sum \limits _{k=1}^{n}\left[ K_g*K_g\right] (X_j-X_k) . \end{aligned}$$
(10.7)
We now focus on the integrated bootstrap variance, which needs a deeper insight:
$$\begin{aligned} V^{*}\left( h\right)= & {} \int {Var}^{*}\left( n^{-1}\sum _{i=1}^nK_h\left( \nonumber x-X_i^{*}\right) \right) dx \\ \nonumber= & {} n^{-2}\int \sum \limits _{i=1}^{n}{Var}^{*}\left( K_h\left( x-X_i^{*}\right) \right) dx\\ \nonumber&+\,n^{-2}\sum \limits _{\begin{array}{c} i,j=1 \\ i\ne j \end{array}} ^n\int Cov^{*}\left( K_h\left( x-X_i^{*}\right) ,K_h\left( x-X_j^{*}\right) \right) dx\\\nonumber= & {} n^{-2}\sum \limits _{i=1}^{n}\int \mathbb {E}^{*}\left( K_h\left( x-X_i^{*}\right) ^2\right) dx\\ \nonumber&-\,n^{-2}\sum \limits _{i=1}^{n}\int \left[ \mathbb {E}^{*}\left( K_h\left( x-X_i^{*}\right) \right) \right] ^2 dx \\&+\,n^{-2}\sum \limits _{\begin{array}{c} i,j=1 \\ i\ne j \end{array}} ^n\int Cov^{*}\left( K_h\left( x-X_i^{*}\right) ,K_h\left( x-X_j^{*}\right) \right) dx.\\ \nonumber \end{aligned}$$
(10.8)
The first term in (10.8), after some changes of variable, is given by:
$$\begin{aligned}&n^{-2}\sum \limits _{i=1}^{n}\int \mathbb {E}^{*}\left( K_h\left( x-X_i^{*}\right) \nonumber ^2\right) dx = n^{-2}\sum \limits _{i=1}^{n}\int \left[ \int K_h (x-y)^2 \hat{f}_g^{(i)}(y)dy\right] dx\\ \nonumber= & {} n^{-2}\sum \limits _{i=1}^{n}\int \left[ \int K_h(x-y)^2\nonumber \left[ \dfrac{1}{n-b+1}\sum \limits _{j=t_i}^{n-b+t_i}K_g(y-X_j)\right] dy\right] dx\\ \nonumber= & {} \dfrac{1}{n^2(n-b+1)}\sum \limits _{i=1}^{n}\sum \limits _{j=t_i}^{n-b+t_i}\int K_g(y-X_j) \left[ \int K_h(x-y)^2 dx \right] dy\\\nonumber= & {} \dfrac{1}{n^2(n-b+1)}\sum \limits _{i=1}^{n}\sum \limits _{j=t_i}^{n-b+t_i}\int K_g(y-X_j) \left[ \dfrac{1}{h}\int K\left( z\right) ^2 dz \right] dy\\\nonumber= & {} \dfrac{R(K)}{n^2(n-b+1) h}\sum \limits _{i=1}^{n}\sum \limits _{j=t_i}^{n-b+t_i}\int K_g(y-X_j) dy\\\nonumber= & {} \dfrac{R(K)}{n^2(n-b+1) h}\sum \limits _{i=1}^{n}\sum \limits _{j=t_i}^{n-b+t_i}\int K\left( u\right) du= \dfrac{R(K)}{n h}.\\ \end{aligned}$$
(10.9)
Focusing now on the second term, including several changes of variable and using the symmetry of K:
$$\begin{aligned}&n^{-2}\sum \limits _{i=1}^{n}\int \left[ \mathbb {E}^{*}\left( K_h\left( x-X_i^{*}\right) \right) \right] ^2 dx = n^{-2}\sum \limits _{i=1}^n \int \left[ \int K_h(x-y) \hat{f}_g^{(i)}(y)dy\right] ^2 dx\\= & {} n^{-1}b^{-1} \sum \limits _{i=1}^{b} \int \left[ \left( K_h*\hat{f}_g^{(i)}\right) (x)\right] ^2 dx\\= & {} n^{-1}b^{-1}\sum \limits _{i=1}^{b} \int \left[ \sum \limits _{j=t_i}^{n-b+t_i}\dfrac{1}{n-b+1} (K_h*K_g)(x-X_j)\right] \\&\times \left[ \sum \limits _{k=t_i}^{n-b+t_i}\dfrac{1}{n-b+1} (K_h*K_g)(x-X_k)\right] dx\\= & {} \dfrac{1}{nb(n-b+1)^2}\sum \limits _{i=1}^{b} \sum \limits _{j=t_i}^{n-b+t_i}\sum \limits _{k=t_i}^{n-b+t_i} \int (K_h*K_g)(x-X_j) (K_h*K_g)(x-X_k)dx\\= & {} \dfrac{1}{nb(n-b+1)^2}\sum \limits _{i=1}^{b} \sum \limits _{j=t_i}^{n-b+t_i}\sum \limits _{k=t_i}^{n-b+t_i} \int (K_h*K_g)(v) (K_h*K_g)(X_j-X_k-v)dv\\= & {} \dfrac{1}{nb(n-b+1)^2}\sum \limits _{i=1}^{b} \sum \limits _{j=i}^{n-b+i}\sum \limits _{k=i}^{n-b+i} \left[ (K_h*K_g)*(K_h*K_g)\right] (X_j-X_k). \end{aligned}$$
Let us consider the function \(\psi \) defined in Theorem 1. Whenever \(b<n\), we have:
$$\begin{aligned}&n^{-2}\sum \limits _{i=1}^{n}\int \left[ \mathbb {E}^*\left( K_h(x-X_j^*)\right) \right] ^2 dx\nonumber \\ \nonumber= & {} \dfrac{1}{nb(n-b+1)^2}\sum \limits _{i=1}^{b} \sum \limits _{j=i}^{n-b+i}\sum \limits _{k=i}^{n-b+i} \left[ (K_h*K_g)*(K_h*K_g)\right] (X_j-X_k)\\\nonumber= & {} \dfrac{1}{nb(n-b+1)^2}\sum \limits _{i=1}^{b} \sum \limits _{j=i}^{n-b+i}\sum \limits _{k=i}^{n-b+i} \psi (X_j-X_k)\\\nonumber= & {} \dfrac{1}{nb(n-b+1)^2}\left[ \sum \limits _{i=1}^{b} \sum \limits _{j=i}^{b-1} \sum \limits _{k=i}^{b-1} \psi (X_j-X_k)\right. \\\nonumber&\left. + \sum \limits _{i=1}^{b} \sum \limits _{j=i}^{b-1} \sum \limits _{k=b}^{n-b+1} \psi (X_j-X_k) + \sum \limits _{i=1}^{b} \sum \limits _{j=i}^{b-1} \sum \limits _{k=n-b+2}^{n-b+i} \psi (X_j-X_k)\right. \\ \nonumber&\left. + \sum \limits _{i=1}^{b} \sum \limits _{j=b}^{n-b+1} \sum \limits _{k=i}^{b-1} \psi (X_j-X_k)+ \sum \limits _{i=1}^{b} \sum \limits _{j=b}^{n-b+1} \sum \limits _{k=b}^{n-b+1} \psi (X_j-X_k)\right. \\ \nonumber&\left. + \sum \limits _{i=1}^{b} \sum \limits _{j=b}^{n-b+1} \sum \limits _{k=n-b+2}^{n-b+i} \psi (X_j-X_k)+ \sum \limits _{i=1}^{b} \sum \limits _{j=n-b+2}^{n-b+i} \sum \limits _{k=i}^{b-1} \psi (X_j-X_k)\right. \\\nonumber&\left. + \sum \limits _{i=1}^{b} \sum \limits _{j=n-b+2}^{n-b+i} \sum \limits _{k=b}^{n-b+1} \psi (X_j-X_k) + \sum \limits _{i=1}^{b} \sum \limits _{j=n-b+2}^{n-b+i} \sum \limits _{k=n-b+2}^{n-b+i} \psi (X_j-X_k)\right] \\ \nonumber= & {} \dfrac{1}{nb(n-b+1)^2}\left[ \sum \limits _{j=1}^{b-1} \sum \limits _{k=1}^{b-1}\nonumber \sum \limits _{i=1}^{\min \{j,k\}} \psi (X_j-X_k)\right. \\\nonumber&\left. + \sum \limits _{j=1}^{b-1} \sum \limits _{k=b}^{n-b+1}\sum \limits _{i=1}^{j} \psi (X_j-X_k) + \sum \limits _{j=1}^{b-1}\sum \limits _{k=n-b+2}^{n} \sum \limits _{i=\max \{(k+b-n),1\}}^{j} \psi (X_j-X_k)\right. \\ \nonumber&\left. + \sum \limits _{j=b}^{n-b+1} \sum \limits _{k=1}^{b-1}\sum \limits _{i=1}^{k} \psi (X_j-X_k)+ \sum \limits _{j=b}^{n-b+1} \sum \limits _{k=b}^{n-b+1} \sum \limits _{i=1}^{b} \psi (X_j-X_k)\right. \\\nonumber&+ \sum \limits _{j=b}^{n-b+1} \sum \limits _{k=n-b+2}^{n}\sum \limits _{i=\max \{(k-n+b),1\}}^{b} \psi (X_j-X_k) \\ \nonumber&+ \sum \limits _{j=n-b+2}^{n} \sum \limits _{k=1}^{b-1} \sum \limits _{i=\max \{(j+b-n),1\}}^{k} \psi (X_j-X_k)\\ \nonumber&\left. + \sum \limits _{j=n-b+2}^{n} \sum \limits _{k=b}^{n-b+1}\sum \limits _{i=\max \{(j-n+b),1\}}^{b} \psi (X_j-X_k)\right. \\\nonumber&\left. + \sum \limits _{j=n-b+2}^{n} \sum \limits _{k=n-b+2}^{n} \nonumber \sum \limits _{i=\max \{(j-n+b),(k-n+b)\}}^{b} \psi (X_j-X_k)\right] \\ \nonumber= & {} \dfrac{1}{nb(n-b+1)^2}\left[ \sum \limits _{j=1}^{b-1} \sum \limits _{k=1}^{b-1} \min \{j,k\} \psi (X_j-X_k)\right. \\\nonumber&+ \sum \limits _{j=1}^{b-1}j \sum \limits _{k=b}^{n-b+1} \psi (X_j-X_k)\\\nonumber&+ \sum \limits _{j=1}^{b-1}\sum \limits _{k=n-b+2}^{n} \min \{(n-b+j-k+1),j\} \psi (X_j-X_k) \\ \nonumber&\left. + \sum \limits _{j=b}^{n-b+1} \sum \limits _{k=1}^{b-1}k \psi (X_j-X_k)+ b \sum \limits _{j=b}^{n-b+1} \sum \limits _{k=b}^{n-b+1} \psi (X_j-X_k)\right. \\\nonumber&\left. + \sum \limits _{j=b}^{n-b+1} \sum \limits _{k=n-b+2}^{n} \min \{(n-k+1),b\} \psi (X_j-X_k)\right. \\\nonumber&\left. + \sum \limits _{j=n-b+2}^{n} \sum \limits _{k=1}^{b-1} \min \{(n-b+k-j+1),k\} \psi (X_j-X_k)\right. \\\nonumber&\left. + \sum \limits _{j=n-b+2}^{n} \min \{(n-j+1),b\} \nonumber \sum \limits _{k=b}^{n-b+1} \psi (X_j-X_k)\right. \\ \nonumber&\left. + \sum \limits _{j=n-b+2}^{n} \sum \limits _{k=n-b+2}^{n} \min \{(n-j+1),(n-k+1)\} \psi (X_j-X_k)\right] \\\nonumber= & {} \dfrac{1}{nb(n-b+1)^2}\left[ \sum \limits _{j=1}^{b-1} \sum \limits _{k=1}^{b-1}\nonumber \min \{j,k\} \psi (X_j-X_k)\right. \\ \nonumber&+ \sum \limits _{j=1}^{b-1}j \sum \limits _{k=b}^{n-b+1} \psi (X_j-X_k) \\ \nonumber&+ \sum \limits _{j=1}^{b-1}\sum \limits _{k=n-b+2}^{n} \min \{(n-b+j-k+1),j\} \psi (X_j-X_k) \\ \nonumber&\left. + \sum \limits _{j=b}^{n-b+1} \sum \limits _{k=1}^{b-1}k \psi (X_j-X_k)+ b \sum \limits _{j=b}^{n-b+1} \sum \limits _{k=b}^{n-b+1} \psi (X_j-X_k)\right. \\\nonumber&\left. + \sum \limits _{j=b}^{n-b+1} \sum \limits _{k=n-b+2}^{n} \min \{(n-k+1),b\} \psi (X_j-X_k)\right. \\ \nonumber&\left. + \sum \limits _{j=n-b+2}^{n} \sum \limits _{k=1}^{b-1} \min \{(n-b+k-j+1),k\} \psi (X_j-X_k)\right. \\ \nonumber&\left. + \sum \limits _{j=n-b+2}^{n} \min \{(n-j+1),b\} \sum \limits _{k=b}^{n-b+1} \psi (X_j-X_k)\right. \\&\left. + \sum \limits _{j=n-b+2}^{n} \sum \limits _{k=n-b+2}^{n} \left( n+1-\max \{j,k\}\right) \psi (X_j-X_k)\right] . \end{aligned}$$
(10.10)
On the other hand, if \(b=n\):
$$\begin{aligned}&\dfrac{1}{nb(n-b+1)^2}\sum \limits _{i=1}^b\sum \limits _{j=i}^{n-b+i}\sum \limits _{k=i}^{n-b+i}\psi \left( X_j-X_k\right) = \dfrac{1}{n^2}\sum \limits _{i=1}^n \psi (X_i-X_i)= \dfrac{\psi (0)}{n}. \end{aligned}$$
Finally, we investigate the covariance term further. It is now necessary to take into account the following notation, naming the n / b blocks as follows:
$$J_r=\left\{ (r-1)b+1,(r-1)b+2,\ldots ,rb\right\} , r=1,2,\ldots ,n/b.$$
Thus, \(X_i^*\) and \(X_j^*\) turn out to be independent (in the bootstrap universe) whenever it does not exist \(r \in \{1,2,\ldots ,n/b\}\) which satisfies \(i,j\in J_r\). In that case, \(X_i^{*}\) and \(X_j^{*}\) do not belong to the same bootstrap block, implying:
$$Cov^*\left( \left. K_h(x-X_i^*),K_h(x-X_j^*)\right. \right) =0.$$
On the other hand, if there exists \(r \in \{1,2,\ldots ,n/b\}\) satisfying \(i,j\in J_r\), then the bootstrap distribution of the pair \((X_i^*,X_j^*)\) is exactly identical to that of the pair \((X_{t_i}^*,X_{t_j}^*)\), where \(t_i =[(i-1)\text {mod }b]+1\). Let us consider \(r \in \{1,2,\ldots ,n/b\}\) satisfying \(i,j \in J_r\), then \(X_i^{*}\) e \(X_j^{*}\) belong to the same bootstrap block. As a consequence,
$$Cov^*\left( \left. K_h(x-X_i^*),K_h(x-X_j^*)\right. \right) = Cov^*\left( K_h(x-X_{t_i}^*), K_h(x-X_{t_j}^*)\right) .$$
Thus:
$$Cov^*\left( K_h(x-X_i^*), K_h(x-X_j^*)\right) = \left\{ \begin{array}{lll} Cov^*\left( K_h(x-X_{t_i}^*), K_h(x-X_{t_j}^*)\right) , &{} \text {if}\,\, \exists r/ i,j \in J_r\\ \\ 0, &{} \text {otherwise} \end{array}. \right. $$
Notice that: \(\mathbb {E}^*\left[ \left. K_h\left( x-X_i^*\right) \right. \right] =\left( K_h*\hat{f}_g^{(i)}\right) , \text { and}\) \(\mathbb {E}^*\left[ \left. K_h\left( x-X_j^*\right) \right. \right] =\left( K_h*\hat{f}_g^{(j)}\right) . \) Now, consider \(k,\ell \in \{1,2,\ldots ,b\}\) satisfying \(k < \ell \). Carrying on with the calculations of the covariance term and using:
$$\mathbb {P}^*\left( \left( X_k^{*(d)},X_{\ell }^{*(d)}\right) =\left( X_j,X_{j+\ell -k}\right) \right) =\dfrac{1}{n-b+1}, j=k,k+1,\ldots ,n-b+k,$$
leads to:
$$\begin{aligned}&\dfrac{1}{n^2} \sum \limits _{\begin{array}{c} i,j=1 \\ i \ne j \end{array}}^{n} Cov^{*}\left( K_{h}\left( x-X_{i}^{*}\right) ,K_{h}\left( x-X_{j}^{*}\right) \right) \\ \!= & {} \! \dfrac{1}{n^2} \dfrac{n}{b} \sum \limits _{\begin{array}{c} k,\ell =1 \\ k\ne \ell \end{array}}^{b} Cov^{*}\left( K_{h}\left( x-X_{k}^{*}\right) ,K_{h}\left( x-X_{\ell }^{*}\right) \right) \\ \!= & {} \! \dfrac{2}{nb} \sum \limits _{\begin{array}{c} k,\ell =1 \\ k< \ell \end{array}}^{b} Cov^{*}\left( K_{h}\left( x-X_{k}^{*}\right) ,K_{h}\left( x-X_{\ell }^{*}\right) \right) \\= & {} \dfrac{2}{nb} \sum \limits _{\begin{array}{c} k,\ell =1 \\ k< \ell \end{array}}^{b} \left[ \mathbb {E}^{*}\left( K_{h}\left( x-X_{k}^{*}\right) K_{h}\left( x-X_{\ell }^{*}\right) \right) -\mathbb {E}^{*}\left( K_{h}\left( x-X_{k}^{*}\right) \right) \mathbb {E}^{*}\left( K_{h}\left( x-X_{\ell }^{*}\right) \right) \right] \\= & {} \dfrac{2}{nb} \sum \limits _{\begin{array}{c} k,\ell =1 \\ k< \ell \end{array}}^{b} \left[ \mathbb {E}^{*}\left[ \mathbb {E}^{*}\left( \left. K_{h}\left( x-X_{k}^{*}\right) K_{h}\left( x-X_{\ell }^{*}\right) \right| _{U_{k}^{*},U_{\ell }^{*}}\right) \right] - \left( K_h *\hat{f}_g^{(k)}\right) \left( K_h *\hat{f}_g^{(\ell )}\right) \right] \\= & {} \dfrac{2}{nb} \sum \limits _{\begin{array}{c} k,\ell =1 \\ k< \ell \end{array}}^{b} \left[ \mathbb {E}^{*}\left[ \mathbb {E}^{*}\left( \left. K_{h}\left( x-X_{k}^{*(d) }-gU_k^*\right) K_{h}\left( x-X_{\ell }^{*(d)}-gU_{\ell }^*\right) \right| _{U_{k}^{*},U_{\ell }^{*}}\right) \right] \right. \\&\left. - \left( K_h *\hat{f}_g^{(k)}(x)\right) \left( K_h *\hat{f}_g^{(\ell )}(x)\right) \right] \\= & {} \dfrac{2}{nb} \sum \limits _{\begin{array}{c} k,\ell =1 \\ k< \ell \end{array}}^{b} \left[ \dfrac{1}{n-b+1}\sum \limits _{j=k}^{n-b+k} \mathbb {E}^{*}\left[ K_{h}\left( x-X_{j}-gU_k^*\right) K_{h}\left( x-X_{j+\ell -k}-gU_{\ell }^*\right) \right] \right. \\&\left. - \left( K_h *\hat{f}_g^{(k)}(x)\right) \left( K_h *\hat{f}_g^{(\ell )}(x)\right) \right] \\= & {} \dfrac{2}{nb} \sum \limits _{\begin{array}{c} k,\ell =1 \\ k< \ell \end{array}}^{b}\left[ \dfrac{1}{n-b+1}\sum \limits _{j=k}^{n-b+k} \int \int K_h(x-X_j-gu) K_h(x-X_{j+\ell -k}-gv) K(u) K(v) du dv\right. \\&\left. - \left( K_h *\hat{f}_g^{(k)}(x)\right) \left( K_h *\hat{f}_g^{(\ell )}(x)\right) \right] \\= & {} \dfrac{2}{nb} \sum \limits _{\begin{array}{c} k,\ell =1 \\ k< \ell \end{array}}^{b}\left[ \dfrac{1}{n-b+1}\sum \limits _{j=k}^{n-b+k} \int \int K_h(x-X_j-s) K_h(x-X_{j+\ell -k}-t) K_g(s) K_g(t) ds dt\right. \\&\left. - \left( K_h *\hat{f}_g^{(k)}(x)\right) \left( K_h *\hat{f}_g^{(\ell )}(x)\right) \right] \\= & {} \dfrac{2}{nb} \sum \limits _{\begin{array}{c} k,\ell =1 \\ k< \ell \end{array}}^{b}\left[ \dfrac{1}{n-b+1}\sum \limits _{j=k}^{n-b+k} (K_h*K_g)(x-X_j) (K_h *K_g)(x-X_{j+\ell -k})\right. \\&\left. - \left( K_h *\hat{f}_g^{(k)}(x)\right) \left( K_h *\hat{f}_g^{(\ell )}(x)\right) \right] \\= & {} \dfrac{2}{nb} \sum \limits _{\begin{array}{c} k,\ell =1 \\ k< \ell \end{array}}^{b}\left[ \dfrac{1}{n-b+1}\sum \limits _{j=k}^{n-b+k} (K_h*K_g)(x-X_j) (K_h *K_g)(x-X_{j+\ell -k})\right. \\&\left. - \left( \dfrac{1}{n-b+1}\sum \limits _{i=k}^{n-b+k}\int K_h(x-y) K_g(y-X_i)dy\right) \right. \\&\left. \times \left( \dfrac{1}{n-b+1}\sum \limits _{j=\ell }^{n-b+\ell }\int K_h(x-y) K_g(y-X_j)dy \right) \right] \\= & {} \dfrac{2}{nb} \sum \limits _{\begin{array}{c} k,\ell =1 \\ k< \ell \end{array}}^{b}\left[ \dfrac{1}{n-b+1}\sum \limits _{j=k}^{n-b+k} (K_h*K_g)(x-X_j) (K_h *K_g)(x-X_{j+\ell -k})\right. \\&\left. - \left( \dfrac{1}{n-b+1}\sum \limits _{i=k}^{n-b+k}\int K_h(x-X_i-u) K_g(u)du\right) \right. \\&\left. \times \left( \dfrac{1}{n-b+1}\sum \limits _{j=\ell }^{n-b+\ell }\int K_h(x-X_j-u) K_g(u)du \right) \right] \\= & {} \dfrac{2}{nb} \sum \limits _{\begin{array}{c} k,\ell =1 \\ k < \ell \end{array}}^{b}\left[ \dfrac{1}{n-b+1}\sum \limits _{j=k}^{n-b+k} (K_h*K_g)(x-X_j) (K_h *K_g)(x-X_{j+\ell -k})\right. \\&\left. - \dfrac{1}{(n-b+1)^2}\sum \limits _{i=k}^{n-b+k}\sum \limits _{j=\ell }^{n-b+\ell } (K_h*K_g)(x-X_i) (K_h*K_g)(x-X_j) \right] . \\ \end{aligned}$$
The integral with respect to x is now computed (using some changes of variable and the symmetry of the kernel K):
$$\begin{aligned}&\int \dfrac{1}{n^2} \sum \limits _{\begin{array}{c} i,j=1 \\ i \ne j \end{array}}^{n} Cov^*\left( K_h\left( x-X_i^*\right) , K_h\left( x-X_j^*\right) \right) dx\\= & {} \int \left[ \dfrac{2}{nb} \sum \limits _{\begin{array}{c} k,\ell =1 \\ k< \ell \end{array}}^{b}\left[ \dfrac{1}{n-b+1}\sum \limits _{j=k}^{n-b+k} (K_h*K_g)(x-X_j) (K_h *K_g)(x-X_{j+\ell -k})\right. \right. \\&\left. \left. - \dfrac{1}{(n-b+1)^2}\sum \limits _{i=k}^{n-b+k}\sum \limits _{j=\ell }^{n-b+\ell } (K_h*K_g)(x-X_i) (K_h*K_g)(x-X_j) \right] \right] dx\\= & {} \dfrac{2}{nb} \left[ \sum \limits _{\begin{array}{c} k,\ell =1 \\ k< \ell \end{array}}^{b}\left[ \dfrac{1}{n-b+1}\sum \limits _{j=k}^{n-b+k} \int (K_h*K_g)(X_{j+\ell -k}-X_j-u) (K_h *K_g)(u)du\right. \right. \\&\left. \left. - \dfrac{1}{(n-b+1)^2}\sum \limits _{i=k}^{n-b+k}\sum \limits _{j=\ell }^{n-b+\ell } \int (K_h*K_g)(u) (K_h*K_g)(X_i-X_j-u)du \right] \right] \\= & {} \dfrac{2}{nb(n-b+1)} \sum \limits _{\begin{array}{c} k,\ell =1 \\ k< \ell \end{array}}^{b} \sum \limits _{j=k}^{n-b+k}\left[ (K_h*K_g)*(K_h*K_g)\right] (X_{j+\ell -k}-X_j)\\&- \dfrac{2}{nb(n-b+1)^2}\sum \limits _{\begin{array}{c} k,\ell =1 \\ k < \ell \end{array}}^{b} \sum \limits _{i=k}^{n-b+k}\sum \limits _{j=\ell }^{n-b+\ell } \left[ (K_h*K_g)*(K_h*K_g)\right] (X_{i}-X_j). \end{aligned}$$
Notice that, whenever \(b<n\):
$$\begin{aligned}&\sum \limits _{\begin{array}{c} k,\ell =1 \\ k < \ell \end{array}}^{b} \sum \limits _{j=k}^{n-b+k}\left[ (K_h*K_g)*(K_h*K_g)\right] (X_{j+\ell -k}-X_j)\\= & {} \sum \limits _{k=1}^{b-1} \sum \limits _{j=k}^{n-b+k}\sum \limits _{\ell =k+1}^{b} \left[ (K_h*K_g)*(K_h*K_g)\right] (X_{j+\ell -k}-X_j)\\= & {} \sum \limits _{k=1}^{b-1} \sum \limits _{j=k}^{n-b+k}\sum \limits _{s=1}^{b-k} \left[ (K_h*K_g)*(K_h*K_g)\right] (X_{j+s}-X_j)\\= & {} \sum \limits _{s=1}^{b-1} \sum \limits _{j=1}^{n-s}\sum \limits _{k=\max \{1,j+b-n\}}^{\min \{j,b-s\}} \left[ (K_h*K_g)*(K_h*K_g)\right] (X_{j+s}-X_j)\\= & {} \sum \limits _{s=1}^{b-1} \sum \limits _{j=1}^{n-s} \left( \min \{j,b-s\}-\max \{1,j+b-n\}+1\right) \left[ (K_h*K_g)*(K_h*K_g)\right] (X_{j+s}-X_j). \end{aligned}$$
Now, using the function \(\psi \), and considering \(b<n\), we have:
$$\begin{aligned}&n^{-2}\sum \limits _{\begin{array}{c} i,j=1 \\ i\ne j \end{array}}^{n}\int Cov^*\nonumber \left( K_h(x-X_i^*),K_h(x-X_j^*)\right) dx \nonumber \\ \nonumber= & {} \sum \limits _{\begin{array}{c} k,\ell =1 \\ k< \ell \end{array}}^{b} \sum \limits _{i=k}^{n-b+k}\sum \limits _{j=\ell }^{n-b+\ell } \psi (X_i-X_j)\\ \nonumber= & {} \sum \limits _{\begin{array}{c} k,\ell =1 \\ k< \ell \end{array}}^{b} \sum \limits _{i=k}^{b-2}\sum \limits _{j=\ell }^{b-1} \psi (X_i-X_j) + \sum \limits _{\begin{array}{c} k,\ell =1 \\ k< \ell \end{array}}^{b} \sum \limits _{i=k}^{b-2}\sum \limits _{j=b}^{n-b+2} \psi (X_i-X_j)\\ \nonumber&+ \sum \limits _{\begin{array}{c} k,\ell =1 \\ k< \ell \end{array}}^{b} \sum \limits _{i=k}^{b-2}\sum \limits _{j=n-b+3}^{n-b+\ell } \psi (X_i-X_j)\\ \nonumber&+ \sum \limits _{\begin{array}{c} k,\ell =1 \\ k< \ell \end{array}}^{b} \nonumber \sum \limits _{i=b-1}^{n-b+1}\sum \limits _{j=n-b+3}^{n-b+\ell } \psi (X_i-X_j) + \sum \limits _{\begin{array}{c} k,\ell =1 \\ k< \ell \end{array}}^{b} \sum \limits _{i=n-b+2}^{n-b+k}\sum \limits _{j=\ell }^{b-1} \psi (X_i-X_j)\\ \nonumber&+ \sum \limits _{\begin{array}{c} k,\ell =1 \\ k< \ell \end{array}}^{b} \sum \limits _{i=n-b+2}^{n-b+k}\sum \limits _{j=b}^{n-b+2}\psi (X_i-X_j) +\sum \limits _{\begin{array}{c} k,\ell =1 \\ k< \ell \end{array}}^{b} \sum \limits _{i=n-b+2}^{n-b+k}\sum \limits _{j=n-b+3}^{n-b+\ell } \psi (X_i-X_j)\\ \nonumber= & {} \sum \limits _{\begin{array}{c} k,\ell =1 \\ k< \ell \end{array}}^{b} \sum \limits _{i=k}^{b-2}\sum \limits _{j=\ell }^{b-1} \psi (X_i-X_j) + \sum \limits _{k=1}^{b-1}\sum \limits _{\ell =k+1}^{b} \sum \limits _{i=k}^{b-2}\sum \limits _{j=b}^{n-b+2} \psi (X_i-X_j)\\ \nonumber&+ \sum \limits _{\begin{array}{c} k,\ell =1 \\ k< \ell \end{array}}^{b} \sum \limits _{i=k}^{b-2}\sum \limits _{j=n-b+3}^{n-b+\ell } \psi (X_i-X_j)\\ \nonumber&+ \sum \limits _{\ell =2}^{b}\sum \limits _{k=1}^{\ell -1} \nonumber \sum \limits _{i=b-1}^{n-b+1}\sum \limits _{j=\ell }^{b-1}\psi (X_i-X_j) + \sum \limits _{i=b-1}^{n-b+1}\sum \limits _{j=b}^{n-b+2}\sum \limits _{\begin{array}{c} k,\ell =1 \\ k< \ell \end{array}}^{b} \psi (X_i-X_j)\\ \nonumber&+ \sum \limits _{\ell =2}^{b} \sum \limits _{k=1}^{\ell -1}\sum \limits _{i=b-1}^{n-b+1} \sum \limits _{j=n-b+3}^{n-b+\ell }\psi (X_i-X_j) + \sum \limits _{\begin{array}{c} k,\ell =1 \\ k< \ell \end{array}}^{b} \sum \limits _{i=n-b+2}^{n-b+k}\sum \limits _{j=\ell }^{b-1} \psi (X_i-X_j)\\ \nonumber&+ \sum \limits _{k=1}^{b-1}\sum \limits _{\ell =k+1}^{b}\nonumber \sum \limits _{i=n-b+2}^{n-b+k}\sum \limits _{j=b}^{n-b+2}\psi (X_i-X_j) \\ \nonumber&+\sum \limits _{\begin{array}{c} k,\ell =1 \\ k< \ell \end{array}}^{b} \sum \limits _{i=n-b+2}^{n-b+k}\sum \limits _{j=n-b+3}^{n-b+\ell } \psi (X_i-X_j)\\ \nonumber= & {} \sum \limits _{\begin{array}{c} k,\ell =1 \\ k< \ell \end{array}}^{b} \sum \limits _{i=k}^{b-2}\sum \limits _{j=\ell }^{b-1} \psi (X_i-X_j) + \sum \limits _{k=1}^{b-1}(b-k) \sum \limits _{i=k}^{b-2}\sum \limits _{j=b}^{n-b+2} \psi (X_i-X_j)\\ \nonumber&+ \sum \limits _{\begin{array}{c} k,\ell =1 \\ k< \ell \end{array}}^{b} \sum \limits _{i=k}^{b-2}\sum \limits _{j=n-b+3}^{n-b+\ell } \psi (X_i-X_j)\\ \nonumber&+ \sum \limits _{\ell =2}^{b}(\ell -1) \sum \limits _{i=b-1}^{n-b+1}\sum \limits _{j=\ell }^{b-1}\psi (X_i-X_j) + \dfrac{b(b-1)}{2}\sum \limits _{i=b-1}^{n-b+1}\sum \limits _{j=b}^{n-b+2} \psi (X_i-X_j)\\ \nonumber&+ \sum \limits _{\ell =2}^{b}(\ell -1) \sum \limits _{i=b-1}^{n-b+1} \sum \limits _{j=n-b+3}^{n-b+\ell }\psi (X_i-X_j) + \sum \limits _{\begin{array}{c} k,\ell =1 \\ k< \ell \end{array}}^{b} \sum \limits _{i=n-b+2}^{n-b+k}\sum \limits _{j=\ell }^{b-1} \psi (X_i-X_j)\\ \nonumber&+ \sum \limits _{k=1}^{b-1}(b-k) \sum \limits _{i=n-b+2}^{n-b+k}\sum \limits _{j=b}^{n-b+2}\psi (X_i-X_j) \\&+\sum \limits _{\begin{array}{c} k,\ell =1 \\ k < \ell \end{array}}^{b} \sum \limits _{i=n-b+2}^{n-b+k}\sum \limits _{j=n-b+3}^{n-b+\ell } \psi (X_i-X_j). \end{aligned}$$
(10.11)
On the other hand, if \(b=n\) and using the symmetry of the kernel K, we obtain:
$$\begin{aligned}&\dfrac{2}{nb(n-b+1)} \sum \limits _{\begin{array}{c} k,\ell =1 \\ k< \ell \end{array}}^{b} \nonumber \sum \limits _{j=k}^{n-b+k}\left[ (K_h*K_g)*(K_h*K_g)\right] (X_{j+\ell -k}-X_j) \nonumber \\ \nonumber&- \dfrac{2}{nb(n-b+1)^2}\sum \limits _{\begin{array}{c} k,\ell =1 \\ k < \ell \end{array}}^{b} \sum \limits _{i=k}^{n-b+k}\sum \limits _{j=\ell }^{n-b+\ell } \left[ (K_h*K_g)*(K_h*K_g)\right] (X_{i}-X_j)\\= & {} \dfrac{2}{n^2} \sum \limits _{k=1}^{n-1}\sum \limits _{\ell =k+1}^{n} \psi \left( X_\ell -X_k\right) - \dfrac{2}{n^2} \sum \limits _{k=1}^{n-1}\sum \limits _{\ell =k+1}^{n} \psi \left( X_k-X_\ell \right) =0. \end{aligned}$$
(10.12)
Using (10.9), (10.10) and (10.11) in (10.8), and this and (10.7) in (10.6) gives the statement of Theorem 1 for \(b<n\). The case \(b=n\) is even simpler using (10.12).
