[proofplan] For each $n\in\mathbb N$, the plug-in bandwidth is obtained from the stated AMISE bandwidth by replacing the unknown curvature functional $R(f'')$ with the positive estimator $\widehat{R(f'')}_n:(\Omega,\mathcal F)\to(0,\infty)$. Dividing the two closed-form bandwidths cancels all kernel and sample-size factors, leaving only the ratio \begin{align*} \left(\frac{R(f'')}{\widehat{R(f'')}_n}\right)^{1/5}. \end{align*} Since $\widehat{R(f'')}_n$ converges in probability to the positive constant $R(f'')$, we prove directly from the definition of convergence in probability that this fifth-root ratio converges in probability to $1$. [/proofplan]

[step:Cancel the common AMISE factors in the ratio] Let $(\Omega,\mathcal F,\mathbb P)$ be the probability space from the theorem statement, and define the positive constant \begin{align*} \theta:=R(f'')\in(0,\infty). \end{align*} For each $n\in\mathbb N$, the theorem statement defines \begin{align*} h_{\operatorname{AMISE},n}&=\left(\frac{R(K)}{\mu_2(K)^2\theta n}\right)^{1/5}, & \hat h_{\operatorname{PI},n}&=\left(\frac{R(K)}{\mu_2(K)^2\widehat{R(f'')}_n n}\right)^{1/5}. \end{align*} The assumptions $R(K)>0$, $\mu_2(K)\neq 0$, $\theta>0$, $\widehat{R(f'')}_n:(\Omega,\mathcal F)\to(0,\infty)$, and $n\in\mathbb N$ make both bandwidths positive. Hence division is valid. First, \begin{align*} \frac{\hat h_{\operatorname{PI},n}}{h_{\operatorname{AMISE},n}}=\frac{\left(\frac{R(K)}{\mu_2(K)^2\widehat{R(f'')}_n\,n}\right)^{1/5}}{\left(\frac{R(K)}{\mu_2(K)^2\theta\,n}\right)^{1/5}}. \end{align*} Since both quantities inside the fifth roots are positive, the quotient of fifth roots equals the fifth root of the quotient. Cancelling the positive factors $R(K)$, $\mu_2(K)^2$, and $n$ gives \begin{align*} \frac{\hat h_{\operatorname{PI},n}}{h_{\operatorname{AMISE},n}}=\left(\frac{R(K)}{\mu_2(K)^2\widehat{R(f'')}_n\,n}\cdot\frac{\mu_2(K)^2\theta\,n}{R(K)}\right)^{1/5}=\left(\frac{\theta}{\widehat{R(f'')}_n}\right)^{1/5}. \end{align*} [/step]

[step:Show the fifth-root ratio converges in probability to $1$]For each $n\in\mathbb N$, define the [random variable](/page/Random%20Variable) $Y_n:(\Omega,\mathcal F)\to(0,\infty)$ on the probability space fixed in the theorem statement by \begin{align*} Y_n(\omega)=\left(\frac{\theta}{\widehat{R(f'')}_n(\omega)}\right)^{1/5}. \end{align*} This map is measurable because $\widehat{R(f'')}_n$ is a positive real-valued random variable and because the function $\varphi:(0,\infty)\to(0,\infty)$ defined by \begin{align*} \varphi(t)=\left(\frac{\theta}{t}\right)^{1/5} \end{align*} is continuous. We prove $Y_n\xrightarrow{\mathbb P}1$. Fix $\varepsilon>0$. Since $\varphi$ is continuous at $t=\theta$ and satisfies $\varphi(\theta)=1$, there exists $\delta\in(0,\theta/2)$ such that \begin{align*} |t-\theta|<\delta \quad\Longrightarrow\quad \left|\left(\frac{\theta}{t}\right)^{1/5}-1\right|<\varepsilon \end{align*} for every $t\in(0,\infty)$. Therefore \begin{align*} \{|Y_n-1|\geq\varepsilon\} \subset \{|\widehat{R(f'')}_n-\theta|\geq\delta\}. \end{align*} Taking probabilities and using $\widehat{R(f'')}_n\xrightarrow{\mathbb P}\theta$, we obtain \begin{align*} 0 \leq \mathbb P(|Y_n-1|\geq\varepsilon) \leq \mathbb P(|\widehat{R(f'')}_n-\theta|\geq\delta) \longrightarrow 0. \end{align*} Since $\varepsilon>0$ was arbitrary, $Y_n\xrightarrow{\mathbb P}1$.[/step]

[guided]The only probabilistic input is the consistency assumption \begin{align*} \widehat{R(f'')}_n\xrightarrow{\mathbb P}\theta, \qquad \theta:=R(f'')>0. \end{align*} We want to transfer this convergence through the function $t\mapsto(\theta/t)^{1/5}$. Because the estimator is positive, this function is evaluated only on $(0,\infty)$, where it is well-defined. Define the function $\varphi:(0,\infty)\to(0,\infty)$ by \begin{align*} \varphi(t)=\left(\frac{\theta}{t}\right)^{1/5}. \end{align*} The assumption $\theta>0$ is essential here: it places the limiting value inside the domain $(0,\infty)$, away from the singular point $t=0$. The function $\varphi$ is continuous at $\theta$, and \begin{align*} \varphi(\theta)=\left(\frac{\theta}{\theta}\right)^{1/5}=1. \end{align*} Therefore, for the fixed tolerance $\varepsilon>0$, there exists $\delta\in(0,\theta/2)$ such that every $t\in(0,\infty)$ with $|t-\theta|<\delta$ satisfies \begin{align*} \left|\left(\frac{\theta}{t}\right)^{1/5}-1\right|<\varepsilon. \end{align*} Now apply this deterministic implication pointwise to the random quantity $\widehat{R(f'')}_n(\omega)$. If \begin{align*} |\widehat{R(f'')}_n(\omega)-\theta|<\delta, \end{align*} then \begin{align*} \left|\left(\frac{\theta}{\widehat{R(f'')}_n(\omega)}\right)^{1/5}-1\right|<\varepsilon. \end{align*} Equivalently, the event where the transformed estimator is at least $\varepsilon$ away from $1$ can occur only when the original estimator is at least $\delta$ away from $\theta$: \begin{align*} \left\{ \left|\left(\frac{\theta}{\widehat{R(f'')}_n}\right)^{1/5}-1\right|\geq\varepsilon \right\} \subset \{|\widehat{R(f'')}_n-\theta|\geq\delta\}. \end{align*} Taking probabilities gives \begin{align*} 0 \leq \mathbb P\left( \left|\left(\frac{\theta}{\widehat{R(f'')}_n}\right)^{1/5}-1\right|\geq\varepsilon \right) \leq \mathbb P(|\widehat{R(f'')}_n-\theta|\geq\delta). \end{align*} The right-hand side tends to $0$ by the assumed convergence in probability of $\widehat{R(f'')}_n$ to $\theta$. Hence the transformed sequence converges in probability to $1$.[/guided]

[step:Identify the transformed estimator with the bandwidth ratio] From the algebraic identity in the first step, \begin{align*} \frac{\hat h_{\operatorname{PI},n}}{h_{\operatorname{AMISE},n}} = \left(\frac{\theta}{\widehat{R(f'')}_n}\right)^{1/5} = Y_n. \end{align*} The second step proved $Y_n\xrightarrow{\mathbb P}1$. Therefore \begin{align*} \frac{\hat h_{\operatorname{PI},n}}{h_{\operatorname{AMISE},n}} \xrightarrow{\mathbb P}1, \end{align*} which is the desired consistency of the plug-in AMISE bandwidth. [/step]