[proofplan] We expand the ordinary kernel density estimator and the derivative plug-in estimator by [Taylor's theorem](/theorems/827), using compact support to keep every evaluation inside the neighbourhood where $f$ is $C^4$. The leading $h_n^2f''(x)$ bias cancels in expectation, leaving only $O(h_n^4)+O(h_n^2b_n^2)$, which is negligible relative to the standard-error scale. We then write the bias-corrected estimator as a triangular-array average, verify the nondegenerate variance and Lindeberg hypotheses of the triangular-array Lindeberg [central limit theorem](/theorems/521) explicitly, and finish by Slutsky's theorem and the [continuity theorem](/theorems/1145) for convergence in distribution. [/proofplan]

[step:Expand the two deterministic biases and cancel the leading term]Let $U\subset\mathbb R$ be an open neighbourhood of $x$ on which $f\in C^4(U)$. Choose $\delta>0$ with $[x-\delta,x+\delta]\subset U$. Since $f^{(4)}$ is continuous, define \begin{align*} M_4:=\sup\{|f^{(4)}(y)|:y\in[x-\delta,x+\delta]\}<\infty. \end{align*} Let $A_K>0$ and $A_L>0$ satisfy $\operatorname{supp}K\subset[-A_K,A_K]$ and $\operatorname{supp}L\subset[-A_L,A_L]$. For all sufficiently large $n$, $x-h_nu\in(x-\delta,x+\delta)$ whenever $u\in\operatorname{supp}K$. By the Taylor theorem with Lagrange remainder applied to the map $t\mapsto f(x+t)$ at $t=0$, for each such $u$ there is a point $\xi_{n,u}$ between $x$ and $x-h_nu$ such that \begin{align*} f(x-h_nu)=f(x)-h_nu f'(x)+\frac{h_n^2u^2}{2}f''(x)-\frac{h_n^3u^3}{6}f^{(3)}(x)+\frac{h_n^4u^4}{24}f^{(4)}(\xi_{n,u}). \end{align*} Using the substitution $u=(x-y)/h_n$, for which $y=x-h_nu$ and $d\mathcal L^1(y)=h_n\,d\mathcal L^1(u)$, gives \begin{align*} \mathbb E[\hat f_{n,h_n}(x)]=\int_{\mathbb R}K(u)f(x-h_nu)\,d\mathcal L^1(u). \end{align*} The symmetry of $K$ makes $uK(u)$ and $u^3K(u)$ odd, hence their Lebesgue integrals over $\mathbb R$ vanish. Therefore \begin{align*} \mathbb E[\hat f_{n,h_n}(x)]-f(x)=\frac{h_n^2\mu_2(K)}{2}f''(x)+O(h_n^4), \end{align*} where the $O(h_n^4)$ constant is bounded by \begin{align*} \frac{M_4}{24}\int_{\mathbb R}|u|^4|K(u)|\,d\mathcal L^1(u). \end{align*} The last integral is finite because $K$ is bounded and compactly supported. Similarly, for all sufficiently large $n$, $x-b_nu\in(x-\delta,x+\delta)$ whenever $u\in\operatorname{supp}L$. The same Taylor theorem gives \begin{align*} f(x-b_nu)=f(x)-b_nu f'(x)+\frac{b_n^2u^2}{2}f''(x)-\frac{b_n^3u^3}{6}f^{(3)}(x)+\frac{b_n^4u^4}{24}f^{(4)}(\zeta_{n,u}), \end{align*} where $\zeta_{n,u}$ lies between $x$ and $x-b_nu$. The substitution $u=(x-y)/b_n$ gives \begin{align*} \mathbb E[\widehat{f''}_{n,b_n}(x)]=\frac{1}{b_n^2}\int_{\mathbb R}L(u)f(x-b_nu)\,d\mathcal L^1(u). \end{align*} The moment conditions on $L$ kill the constant and linear terms, symmetry kills the cubic term, and the normalization $\int_{\mathbb R}u^2L(u)\,d\mathcal L^1(u)=2$ leaves the quadratic term equal to $f''(x)$. Since $L$ is bounded and compactly supported, \begin{align*} \int_{\mathbb R}|u|^4|L(u)|\,d\mathcal L^1(u)<\infty, \end{align*} and the fourth-order remainder is $O(b_n^2)$. Hence \begin{align*} \mathbb E[\widehat{f''}_{n,b_n}(x)]=f''(x)+O(b_n^2). \end{align*} Combining the two expansions in the definition of $\hat f_{n,\mathrm{bc}}(x)$ gives \begin{align*} \mathbb E[\hat f_{n,\mathrm{bc}}(x)]-f(x)=O(h_n^4)+O(h_n^2b_n^2). \end{align*}[/step]

[guided]The two bias calculations use the same principle: because the kernels are compactly supported, the points at which $f$ is evaluated remain in the neighbourhood where $f$ has four continuous derivatives. Let $U\subset\mathbb R$ be an open neighbourhood of $x$ on which $f\in C^4(U)$, choose $\delta>0$ with $[x-\delta,x+\delta]\subset U$, and define \begin{align*} M_4:=\sup\{|f^{(4)}(y)|:y\in[x-\delta,x+\delta]\}<\infty. \end{align*} Choose $A_K>0$ and $A_L>0$ such that $\operatorname{supp}K\subset[-A_K,A_K]$ and $\operatorname{supp}L\subset[-A_L,A_L]$. For the ordinary kernel estimator, the affine change of variables $u=(x-y)/h_n$ has inverse $y=x-h_nu$ and transforms measure by $d\mathcal L^1(y)=h_n\,d\mathcal L^1(u)$. Thus \begin{align*} \mathbb E[\hat f_{n,h_n}(x)]=\int_{\mathbb R}K(u)f(x-h_nu)\,d\mathcal L^1(u). \end{align*} For all sufficiently large $n$, every $u\in\operatorname{supp}K$ satisfies $x-h_nu\in(x-\delta,x+\delta)$. Therefore the Taylor theorem with Lagrange remainder applies to $t\mapsto f(x+t)$ at $t=0$ and gives \begin{align*} f(x-h_nu)=f(x)-h_nu f'(x)+\frac{h_n^2u^2}{2}f''(x)-\frac{h_n^3u^3}{6}f^{(3)}(x)+\frac{h_n^4u^4}{24}f^{(4)}(\xi_{n,u}). \end{align*} The theorem's hypotheses are satisfied because $f$ has four continuous derivatives on the interval containing all points between $x$ and $x-h_nu$. Integrating the expansion against $K(u)\,d\mathcal L^1(u)$, the constant term gives $f(x)$ because $\int K\,d\mathcal L^1=1$, the linear and cubic terms vanish because $K$ is symmetric, and the quadratic term gives $h_n^2\mu_2(K)f''(x)/2$. The remainder is bounded in absolute value by \begin{align*} \frac{M_4h_n^4}{24}\int_{\mathbb R}|u|^4|K(u)|\,d\mathcal L^1(u), \end{align*} which is finite because $K$ is bounded and compactly supported. Hence \begin{align*} \mathbb E[\hat f_{n,h_n}(x)]-f(x)=\frac{h_n^2\mu_2(K)}{2}f''(x)+O(h_n^4). \end{align*} For the derivative estimator, the affine substitution $u=(x-y)/b_n$ gives $y=x-b_nu$ and $d\mathcal L^1(y)=b_n\,d\mathcal L^1(u)$, so \begin{align*} \mathbb E[\widehat{f''}_{n,b_n}(x)]=\frac{1}{b_n^2}\int_{\mathbb R}L(u)f(x-b_nu)\,d\mathcal L^1(u). \end{align*} For all sufficiently large $n$, the support condition on $L$ ensures $x-b_nu\in(x-\delta,x+\delta)$ for all $u\in\operatorname{supp}L$. Taylor's theorem gives \begin{align*} f(x-b_nu)=f(x)-b_nu f'(x)+\frac{b_n^2u^2}{2}f''(x)-\frac{b_n^3u^3}{6}f^{(3)}(x)+\frac{b_n^4u^4}{24}f^{(4)}(\zeta_{n,u}). \end{align*} After multiplication by $b_n^{-2}$ and integration against $L(u)\,d\mathcal L^1(u)$, the terms involving $1$ and $u$ vanish by the stated moment assumptions on $L$, and the cubic term vanishes because $u^3L(u)$ is odd. The quadratic term is \begin{align*} \frac{1}{b_n^2}\int_{\mathbb R}L(u)\frac{b_n^2u^2}{2}f''(x)\,d\mathcal L^1(u)=\frac{f''(x)}{2}\int_{\mathbb R}u^2L(u)\,d\mathcal L^1(u)=f''(x). \end{align*} The fourth-order remainder is bounded by \begin{align*} \frac{b_n^2M_4}{24}\int_{\mathbb R}|u|^4|L(u)|\,d\mathcal L^1(u), \end{align*} and the integral is finite because $L$ is bounded and compactly supported. Hence \begin{align*} \mathbb E[\widehat{f''}_{n,b_n}(x)]=f''(x)+O(b_n^2). \end{align*} Subtracting $h_n^2\mu_2(K)\widehat{f''}_{n,b_n}(x)/2$ therefore cancels the leading $h_n^2\mu_2(K)f''(x)/2$ bias in expectation and leaves \begin{align*} \mathbb E[\hat f_{n,\mathrm{bc}}(x)]-f(x)=O(h_n^4)+O(h_n^2b_n^2). \end{align*}[/guided]

[step:Identify the deterministic variance scale] Define \begin{align*} R(K):=\int_{\mathbb R}K(u)^2\,d\mathcal L^1(u), \end{align*} and \begin{align*} R(L):=\int_{\mathbb R}L(u)^2\,d\mathcal L^1(u). \end{align*} Let \begin{align*} A_{n,1}:=\frac{1}{h_n}K\left(\frac{x-X_1}{h_n}\right) \end{align*} and \begin{align*} B_{n,1}:=\frac{h_n^2\mu_2(K)}{2b_n^3}L\left(\frac{x-X_1}{b_n}\right). \end{align*} The same change of variables as above gives \begin{align*} \mathbb E[A_{n,1}^2]=\frac{1}{h_n}\int_{\mathbb R}K(u)^2f(x-h_nu)\,d\mathcal L^1(u). \end{align*} Since $K^2$ is integrable and $f(x-h_nu)\to f(x)$ for each $u\in\operatorname{supp}K$, dominated convergence on the compact support of $K$ gives \begin{align*} \operatorname{Var}(A_{n,1})=\frac{f(x)R(K)}{h_n}+o\left(\frac{1}{h_n}\right). \end{align*} Similarly, local boundedness of $f$ on $(x-\delta,x+\delta)$ and compact support of $L$ give \begin{align*} \operatorname{Var}(B_{n,1})=O\left(\frac{h_n^4}{b_n^5}\right). \end{align*} By the [Cauchy-Schwarz inequality](/theorems/432) applied to the covariance [inner product](/page/Inner%20Product) in $L^2(\Omega,\mathcal F,\mathbb P)$, \begin{align*} |\operatorname{Cov}(A_{n,1},B_{n,1})|\le \operatorname{Var}(A_{n,1})^{1/2}\operatorname{Var}(B_{n,1})^{1/2}=O\left(\frac{h_n^{3/2}}{b_n^{5/2}}\right). \end{align*} The derivative variance term is relatively $O(h_n^5/b_n^5)$ compared with $1/h_n$, and the covariance term is relatively $O(h_n^{5/2}/b_n^{5/2})$ compared with $1/h_n$. Both tend to $0$ because $h_n^2\sqrt{h_n}/b_n^{5/2}\to0$. Therefore \begin{align*} \operatorname{Var}(A_{n,1}-B_{n,1})=\frac{f(x)R(K)}{h_n}+o\left(\frac{1}{h_n}\right). \end{align*} Since $f(x)>0$ and $R(K)>0$, it follows that \begin{align*} \sigma_{\mathrm{bc},n}(x)^2=\frac{\operatorname{Var}(A_{n,1}-B_{n,1})}{n}=\frac{f(x)R(K)}{nh_n}+o\left(\frac{1}{nh_n}\right). \end{align*} Consequently $\sigma_{\mathrm{bc},n}(x)>0$ for all sufficiently large $n$ and $\sigma_{\mathrm{bc},n}(x)\asymp(nh_n)^{-1/2}$. [/step]

[step:Show that the corrected deterministic bias is negligible] From the bias expansion and the variance scale, \begin{align*} \frac{\mathbb E[\hat f_{n,\mathrm{bc}}(x)]-f(x)}{\sigma_{\mathrm{bc},n}(x)}=O(\sqrt{nh_n}\,h_n^4)+O(\sqrt{nh_n}\,h_n^2b_n^2). \end{align*} The two displayed bandwidth assumptions imply \begin{align*} \frac{\mathbb E[\hat f_{n,\mathrm{bc}}(x)]-f(x)}{\sigma_{\mathrm{bc},n}(x)}\to0. \end{align*} [/step]

[step:Verify Lindeberg's condition for the centred triangular array]For $1\le i\le n$, define \begin{align*} Y_{n,i}:=\frac{1}{h_n}K\left(\frac{x-X_i}{h_n}\right)-\frac{h_n^2\mu_2(K)}{2b_n^3}L\left(\frac{x-X_i}{b_n}\right). \end{align*} Then $Y_{n,1},\dots,Y_{n,n}$ are independent and identically distributed, and \begin{align*} \hat f_{n,\mathrm{bc}}(x)=\frac{1}{n}\sum_{i=1}^nY_{n,i}. \end{align*} Set \begin{align*} s_n^2:=\sum_{i=1}^n\operatorname{Var}(Y_{n,i})=n\operatorname{Var}(Y_{n,1}). \end{align*} The previous variance computation gives \begin{align*} s_n^2=\frac{nf(x)R(K)}{h_n}+o\left(\frac{n}{h_n}\right), \end{align*} so $s_n^2>0$ for all sufficiently large $n$ and $s_n\asymp\sqrt{n/h_n}$. Let \begin{align*} M_n:=\frac{\|K\|_\infty}{h_n}+\frac{h_n^2|\mu_2(K)|\|L\|_\infty}{2b_n^3}. \end{align*} Since $|Y_{n,i}|\le M_n$ and $|\mathbb E[Y_{n,i}]|\le M_n$, we have \begin{align*} |Y_{n,i}-\mathbb E[Y_{n,i}]|\le2M_n. \end{align*} Dividing by $s_n$, the first envelope contribution is $O((nh_n)^{-1/2})$, which tends to $0$ because $nh_n\to\infty$. The second contribution is \begin{align*} O\left(\frac{h_n^{5/2}}{\sqrt n\,b_n^3}\right)=O\left(\frac{h_n^{5/2}}{b_n^{5/2}}\frac{1}{\sqrt{nb_n}}\right). \end{align*} The factor $h_n^{5/2}/b_n^{5/2}$ tends to $0$ by $h_n^2\sqrt{h_n}/b_n^{5/2}\to0$, and $nb_n\to\infty$ follows from $nb_n^5\to\infty$ and $b_n\to0$. Hence \begin{align*} \frac{2M_n}{s_n}\to0. \end{align*} Therefore, for every $\varepsilon>0$, the event \begin{align*} \{|Y_{n,i}-\mathbb E[Y_{n,i}]|>\varepsilon s_n\} \end{align*} is empty for all $1\le i\le n$ and all sufficiently large $n$. Thus the Lindeberg expression uses $\mathbb 1_A$ to denote the indicator function of an event $A$, and is \begin{align*} \frac{1}{s_n^2}\sum_{i=1}^n\mathbb E\left[(Y_{n,i}-\mathbb E[Y_{n,i}])^2\mathbb 1_{\{|Y_{n,i}-\mathbb E[Y_{n,i}]|>\varepsilon s_n\}}\right] \end{align*} tends to $0$.[/step]

[guided]We now verify the exact hypotheses of the triangular-array [central limit theorem](/theorems/1848) rather than merely citing it. The version needed is the Lindeberg-Feller Central Limit Theorem for triangular arrays: if, for each $n$, the row variables are independent, centred, square-integrable real random variables, if their total variance $s_n^2$ is positive, and if the Lindeberg condition holds for every $\varepsilon>0$, then the row sum divided by $s_n$ converges in distribution to $\mathcal N(0,1)$. Define \begin{align*} Y_{n,i}:=\frac{1}{h_n}K\left(\frac{x-X_i}{h_n}\right)-\frac{h_n^2\mu_2(K)}{2b_n^3}L\left(\frac{x-X_i}{b_n}\right). \end{align*} Because $X_1,\dots,X_n$ are independent and identically distributed and $Y_{n,i}$ is a measurable function of $X_i$, the variables $Y_{n,1},\dots,Y_{n,n}$ are independent and identically distributed. They are square-integrable because $K$ and $L$ are bounded. Define the centred variables $Y_{n,i}-\mathbb E[Y_{n,i}]$ and their total row variance \begin{align*} s_n^2:=\sum_{i=1}^n\operatorname{Var}(Y_{n,i})=n\operatorname{Var}(Y_{n,1}). \end{align*} Write $Y_{n,1}=A_{n,1}-B_{n,1}$, where \begin{align*} A_{n,1}:=\frac{1}{h_n}K\left(\frac{x-X_1}{h_n}\right) \end{align*} and \begin{align*} B_{n,1}:=\frac{h_n^2\mu_2(K)}{2b_n^3}L\left(\frac{x-X_1}{b_n}\right). \end{align*} The same substitution $u=(x-y)/h_n$ used for the bias calculation gives \begin{align*} \operatorname{Var}(A_{n,1})=\frac{f(x)R(K)}{h_n}+o\left(\frac{1}{h_n}\right). \end{align*} The compact support and boundedness of $L$ and the local boundedness of $f$ give \begin{align*} \operatorname{Var}(B_{n,1})=O\left(\frac{h_n^4}{b_n^5}\right). \end{align*} Cauchy-Schwarz gives \begin{align*} |\operatorname{Cov}(A_{n,1},B_{n,1})|\le \operatorname{Var}(A_{n,1})^{1/2}\operatorname{Var}(B_{n,1})^{1/2}=O\left(\frac{h_n^{3/2}}{b_n^{5/2}}\right). \end{align*} Relative to the leading $1/h_n$ variance, these correction terms are $O(h_n^5/b_n^5)$ and $O(h_n^{5/2}/b_n^{5/2})$, both tending to $0$ by $h_n^2\sqrt{h_n}/b_n^{5/2}\to0$. Therefore \begin{align*} s_n^2=\frac{nf(x)R(K)}{h_n}+o\left(\frac{n}{h_n}\right). \end{align*} Since $f(x)>0$ and $R(K)>0$, this proves that $s_n^2>0$ for all sufficiently large $n$ and that $s_n\asymp\sqrt{n/h_n}$. This is the nondegenerate variance hypothesis in the Lindeberg-Feller theorem. It remains to prove Lindeberg's condition. Define the deterministic envelope \begin{align*} M_n:=\frac{\|K\|_\infty}{h_n}+\frac{h_n^2|\mu_2(K)|\|L\|_\infty}{2b_n^3}. \end{align*} The boundedness of $K$ and $L$ gives $|Y_{n,i}|\le M_n$. Taking expectations and using monotonicity of expectation gives $|\mathbb E[Y_{n,i}]|\le\mathbb E[|Y_{n,i}|]\le M_n$, and therefore \begin{align*} |Y_{n,i}-\mathbb E[Y_{n,i}]|\le2M_n. \end{align*} After division by $s_n\asymp\sqrt{n/h_n}$, the ordinary kernel part contributes \begin{align*} O\left(\frac{1/h_n}{\sqrt{n/h_n}}\right)=O((nh_n)^{-1/2}), \end{align*} which tends to $0$ because $nh_n\to\infty$. The derivative-kernel part contributes \begin{align*} O\left(\frac{h_n^2/b_n^3}{\sqrt{n/h_n}}\right)=O\left(\frac{h_n^{5/2}}{\sqrt n\,b_n^3}\right)=O\left(\frac{h_n^{5/2}}{b_n^{5/2}}\frac{1}{\sqrt{nb_n}}\right). \end{align*} The first factor tends to $0$ by the bandwidth assumption $h_n^2\sqrt{h_n}/b_n^{5/2}\to0$. The second factor tends to $0$ because $nb_n\to\infty$, and this follows from $nb_n^5\to\infty$ together with $b_n\to0$. Hence \begin{align*} \frac{2M_n}{s_n}\to0. \end{align*} Now fix $\varepsilon>0$. For all sufficiently large $n$, $2M_n\le\varepsilon s_n$, so \begin{align*} \{|Y_{n,i}-\mathbb E[Y_{n,i}]|>\varepsilon s_n\}=\varnothing \end{align*} for every $1\le i\le n$. Thus the Lindeberg sum uses $\mathbb 1_A$ for the indicator function of an event $A$, and equals \begin{align*} \frac{1}{s_n^2}\sum_{i=1}^n\mathbb E\left[(Y_{n,i}-\mathbb E[Y_{n,i}])^2\mathbb 1_{\{|Y_{n,i}-\mathbb E[Y_{n,i}]|>\varepsilon s_n\}}\right] \end{align*} is eventually equal to $0$. This verifies the Lindeberg hypothesis for every $\varepsilon>0$.[/guided]

[step:Apply the triangular-array central limit theorem] The variables $Y_{n,i}-\mathbb E[Y_{n,i}]$ are independent within each row, centred, square-integrable, have total variance $s_n^2>0$ for all sufficiently large $n$, and satisfy Lindeberg's condition by the preceding step. Therefore the Lindeberg-Feller Central Limit Theorem for triangular arrays gives \begin{align*} \frac{\sum_{i=1}^n(Y_{n,i}-\mathbb E[Y_{n,i}])}{s_n}\xrightarrow{d}\mathcal N(0,1). \end{align*} Since \begin{align*} \sigma_{\mathrm{bc},n}(x)^2=\operatorname{Var}\left(\frac{1}{n}\sum_{i=1}^nY_{n,i}\right)=\frac{s_n^2}{n^2}, \end{align*} this is equivalent to \begin{align*} \frac{\hat f_{n,\mathrm{bc}}(x)-\mathbb E[\hat f_{n,\mathrm{bc}}(x)]}{\sigma_{\mathrm{bc},n}(x)}\xrightarrow{d}\mathcal N(0,1). \end{align*} [/step]

[step:Replace the centring and standard deviation] We decompose \begin{align*} \frac{\hat f_{n,\mathrm{bc}}(x)-f(x)}{\sigma_{\mathrm{bc},n}(x)}=\frac{\hat f_{n,\mathrm{bc}}(x)-\mathbb E[\hat f_{n,\mathrm{bc}}(x)]}{\sigma_{\mathrm{bc},n}(x)}+\frac{\mathbb E[\hat f_{n,\mathrm{bc}}(x)]-f(x)}{\sigma_{\mathrm{bc},n}(x)}. \end{align*} The first term converges in distribution to $\mathcal N(0,1)$, and the second term converges to $0$. By Slutsky's Theorem, \begin{align*} \frac{\hat f_{n,\mathrm{bc}}(x)-f(x)}{\sigma_{\mathrm{bc},n}(x)}\xrightarrow{d}\mathcal N(0,1). \end{align*} The assumption $\hat s_{\mathrm{bc},n}(x)/\sigma_{\mathrm{bc},n}(x)\xrightarrow{\mathbb P}1$ implies $\sigma_{\mathrm{bc},n}(x)/\hat s_{\mathrm{bc},n}(x)\xrightarrow{\mathbb P}1$ by the [continuous mapping theorem](/theorems/1847), applied to the reciprocal map on a neighbourhood of $1$. Applying Slutsky's Theorem again gives \begin{align*} \frac{\hat f_{n,\mathrm{bc}}(x)-f(x)}{\hat s_{\mathrm{bc},n}(x)}=\frac{\hat f_{n,\mathrm{bc}}(x)-f(x)}{\sigma_{\mathrm{bc},n}(x)}\frac{\sigma_{\mathrm{bc},n}(x)}{\hat s_{\mathrm{bc},n}(x)}\xrightarrow{d}\mathcal N(0,1). \end{align*} [/step]

[step:Convert the limiting pivot into confidence interval coverage] Fix $\tau\in(0,1)$, and let $Z$ be a real-valued [random variable](/page/Random%20Variable) with distribution $\mathcal N(0,1)$. Since $\hat s_{\mathrm{bc},n}(x)>0$, the event that the interval contains $f(x)$ is equivalent to \begin{align*} \left\{-z_{1-\tau/2}\le\frac{\hat f_{n,\mathrm{bc}}(x)-f(x)}{\hat s_{\mathrm{bc},n}(x)}\le z_{1-\tau/2}\right\}. \end{align*} The standard normal distribution function is continuous at $\pm z_{1-\tau/2}$. Hence the interval version of the [Portmanteau theorem](/theorems/1171) for convergence in distribution gives \begin{align*} \mathbb P\left(-z_{1-\tau/2}\le\frac{\hat f_{n,\mathrm{bc}}(x)-f(x)}{\hat s_{\mathrm{bc},n}(x)}\le z_{1-\tau/2}\right)\to\mathbb P(-z_{1-\tau/2}\le Z\le z_{1-\tau/2}). \end{align*} By the definition of the standard normal quantile and symmetry of $\mathcal N(0,1)$, \begin{align*} \mathbb P(-z_{1-\tau/2}\le Z\le z_{1-\tau/2})=1-\tau. \end{align*} Therefore \begin{align*} \mathbb P\left(f(x)\in\left[\hat f_{n,\mathrm{bc}}(x)-z_{1-\tau/2}\hat s_{\mathrm{bc},n}(x),\hat f_{n,\mathrm{bc}}(x)+z_{1-\tau/2}\hat s_{\mathrm{bc},n}(x)\right]\right)\to1-\tau. \end{align*} This proves both the asserted asymptotic normality and the asymptotic pointwise coverage. [/step]