This problem is found in p. 77 of Wand & Jones' (1995) book. If you are familiar with nonparametric estimation you may skip this introduction.

Suppose we want to minimize the integrated squared bias (ISB) w.r.t. the bandwidth $h$, defined as $$ISB(h)=\int_{\mathbb{R}} \left[(K_h * f)(x) - f(x)\right]^2 dx,$$ where $*$ denotes the convolution operator, $K_h(x)=h^{-1}K(x/h)$ with $K$ being a symmetric kernel function and $f$ a probability density function. Here $f$ is unknown, so it is reasonable to plug in some estimator. In particular, let $n^{-1}\sum_{i=1}^n\hat{f}_{L,-i}(x;g)$ be this estimator, with $\hat{f}_{L,-i}(x;g)=(n-1)^{-1}\sum_{j \neq i}^n L_g(x-X_j)$, where $L$ and $g$ are a symmetric kernel function and a bandwidth, possibly different from $K$ and $h$, respectively. This estimator is called "leave-one-out".


Show the following equality \begin{align} \widetilde{ISB}(h)& {}= n^{-1}\sum_{i=1}^{n}\int_{\mathbb{R}} \left[(K_h * \hat{f}_{L,-i})(x;g) - \hat{f}_{L,-i}(x;g)\right]^2 dx\\ & {} =\{n(n-1)\}^{-1}\sum_{i=1}^n\sum_{j \neq i}^n\{K_h*K_h*L_g*L_g\\ & {}-2K_h*L_g*L_g+L_g*L_g\}(X_i-X_j). \end{align}

Could you please advise me on how to proceed?


I started calculating each of the following terms in the integral in turn $$\int \left[(K_h * \hat{f}_{L,-i})(x;g)^2 -2(K_h * \hat{f}_{L,-i})(x;g)\hat{f}_{L,-i}(x;g)+ \hat{f}_{L,-i}(x;g)^2\right] dx.$$

The first term above is the most demanding and is given by

\begin{align} & {} \int (K_h * \hat{f}_{L,-i})(x;g)^2dx \\ & {} =(n-1)^{-2}\sum_{j \neq i}\sum_{t \neq i}\int\left[\int K_h(x-y)L_g(y-X_j)dy\right]\left[\int K_h(x-z)L_g(z-X_t)dz\right]dx\\ & {} =(n-1)^{-2}\sum_{j \neq i}\sum_{t \neq i}\int L_g(y-X_j)\int L_g(z-X_t) \int K_h(x-z)K_h(x-y)dx dz dy\\ & {} =(n-1)^{-2}\sum_{j \neq i}\sum_{t \neq i}\int L_g(y-X_j)\int L_g(z-X_t) (K_h*K_h)(y-z) dz dy\\ & {} =(n-1)^{-2}\sum_{j \neq i}\sum_{t \neq i}\int L_g(y-X_j)(K_h*K_h*L_g)(y-X_t) dy\\ & {} =(n-1)^{-2}\sum_{j \neq i}\sum_{t \neq i} (K_h*K_h*L_g*L_g)(X_j-X_t)\\ \end{align}

assuming the absolute value of the integrals are finite and using the Fubini's theorem, integration by substitution and the symmetry of $K_h$ and $L_g$. In the same fashion, it is now straightforward to obtain the remaing terms which are given by \begin{align} \int (K_h * \hat{f}_{L,-i})(x;g)\hat{f}_{L,-i}(x;g)dx & {} =(n-1)^{-2}\sum_{j \neq i}\sum_{t \neq i} (K_h*L_g*L_g)(X_j-X_t)\\ \int \hat{f}_{L,-i}(x;g)^2 dx & {} =(n-1)^{-2}\sum_{j \neq i}\sum_{t \neq i} (L_g*L_g)(X_j-X_t) \end{align}

Hence, combining all these results yields \begin{align} \widetilde{ISB}(h)= & {}n^{-1}(n-1)^{-2}\sum_{i=1}^{n}\sum_{j \neq i}\sum_{t \neq i} (K_h*K_h*L_g*L_g \\ & {} -2K_h*L_g*L_g+L_g*L_g)(X_j-X_t). \end{align}

Comparing the indices of my result and the expression of the question, looks like that there is something I'm missing with the cases where $t=j$.

Could someone give me advices or point out mistakes?

Thanks in advance!


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