Derive the estimator for the integrated squared bias $\int \left(\operatorname{E}\hat{f} - f\right)^2$

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".

QUESTION

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?

MY ATTEMPT

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?