[proofplan]
We minimise the two-term AMISE expression as a one-variable calculus problem on $(0,\infty)$. The variance term decreases like $h^{-1}$, while the squared-bias term increases like $h^4$, so the minimiser occurs where these two derivative contributions balance. After solving the critical point equation, we prove that this critical point is the unique global minimiser by checking the sign of the derivative on both sides.
[/proofplan]
[step:Differentiate the AMISE function on the positive bandwidth domain]
The Kernel Density AMISE formula writes the asymptotic mean integrated squared error as
\begin{align*}
\operatorname{AMISE}(h)=\frac{R(K)}{nh}+\frac{h^4}{4}\mu_2(K)^2R(f''),
\end{align*}
where $R(g)$ denotes the roughness functional applied to a function $g$ and $\mu_2(K)$ denotes the second moment of the kernel $K$. Define the constants $A$ and $B$ by
\begin{align*}
A := \frac{R(K)}{n}, \qquad B := \frac{\mu_2(K)^2 R(f'')}{4}.
\end{align*}
Since $R(K) > 0$, $n \in \mathbb{N}$, $\mu_2(K) \neq 0$, and $R(f'') > 0$, we have $A > 0$ and $B > 0$. The AMISE function is therefore the map $\operatorname{AMISE}: (0,\infty) \to (0,\infty)$ given by $\operatorname{AMISE}(h) = A h^{-1} + B h^4$.
It is differentiable on $(0,\infty)$, and differentiation of powers gives
\begin{align*}
\operatorname{AMISE}'(h) = -A h^{-2} + 4B h^3.
\end{align*}
Substituting the definitions of $A$ and $B$ gives
\begin{align*}
\operatorname{AMISE}'(h) = -\frac{R(K)}{n h^2} + h^3 \mu_2(K)^2 R(f'').
\end{align*}
[guided]
We first recall the one-variable expression supplied by the Kernel Density AMISE formula:
\begin{align*}
\operatorname{AMISE}(h)=\frac{R(K)}{nh}+\frac{h^4}{4}\mu_2(K)^2R(f''),
\end{align*}
where $R(g)$ is the roughness functional applied to a function $g$ and $\mu_2(K)$ is the second moment of the kernel $K$. We isolate the two positive constants that do not depend on the bandwidth by defining $A$ and $B$ through
\begin{align*}
A := \frac{R(K)}{n}, \qquad B := \frac{\mu_2(K)^2 R(f'')}{4}.
\end{align*}
The assumptions give $R(K)>0$ and $n \in \mathbb{N}$, hence $A>0$. They also give $\mu_2(K)\neq 0$, so $\mu_2(K)^2>0$, and $R(f'')>0$, hence $B>0$.
With this notation, the AMISE function becomes the map $\operatorname{AMISE}: (0,\infty) \to (0,\infty)$ given by $\operatorname{AMISE}(h) = A h^{-1} + B h^4$.
This form makes the competing effects visible: $A h^{-1}$ decreases as $h$ increases, while $B h^4$ increases as $h$ increases. Since the function is a sum of differentiable power functions on $(0,\infty)$, differentiating term by term gives
\begin{align*}
\operatorname{AMISE}'(h) = \frac{d}{dh}\left(Ah^{-1}\right)+\frac{d}{dh}\left(Bh^4\right).
\end{align*}
Evaluating the two power derivatives gives
\begin{align*}
\operatorname{AMISE}'(h) = -A h^{-2} + 4B h^3.
\end{align*}
Substituting the definitions of $A$ and $B$ gives
\begin{align*}
\operatorname{AMISE}'(h) = -\frac{R(K)}{n h^2} + h^3 \mu_2(K)^2 R(f'').
\end{align*}
[/guided]
[/step]
[step:Solve the critical point equation]
A point $h \in (0,\infty)$ is critical exactly when
\begin{align*}
-\frac{R(K)}{n h^2} + h^3 \mu_2(K)^2 R(f'') = 0.
\end{align*}
Since $h>0$, multiplying by $h^2$ gives the equivalent equation
\begin{align*}
\frac{R(K)}{n} = h^5 \mu_2(K)^2 R(f'').
\end{align*}
Because $\mu_2(K)^2 R(f'') n > 0$, this equation has the unique positive solution
\begin{align*}
h_* = \left(\frac{R(K)}{\mu_2(K)^2R(f'')n}\right)^{1/5}.
\end{align*}
Thus $\operatorname{AMISE}$ has exactly one critical point in $(0,\infty)$, namely $h_*$.
[/step]
[step:Show the critical point is the unique global minimiser]
For $h \in (0,\infty)$, factor the derivative as
\begin{align*}
\operatorname{AMISE}'(h) = \frac{1}{h^2}\left(h^5\mu_2(K)^2R(f'')-\frac{R(K)}{n}\right).
\end{align*}
Since $h^{-2}>0$, the sign of $\operatorname{AMISE}'(h)$ is the sign of
\begin{align*}
h^5\mu_2(K)^2R(f'')-\frac{R(K)}{n}.
\end{align*}
The function $h \mapsto h^5$ is strictly increasing on $(0,\infty)$, and $h_*$ is defined by
\begin{align*}
h_*^5\mu_2(K)^2R(f'')=\frac{R(K)}{n}.
\end{align*}
Therefore $\operatorname{AMISE}'(h)<0$ for $0<h<h_*$ and $\operatorname{AMISE}'(h)>0$ for $h>h_*$. The map $\operatorname{AMISE}: (0,\infty) \to (0,\infty)$ is continuous on each compact subinterval of $(0,\infty)$ and differentiable on the interior of each such subinterval because it is a sum of power functions there. By the [Mean Value Theorem](/theorems/186) applied to $\operatorname{AMISE}$ on each compact subinterval of $(0,h_*)$ and $(h_*,\infty)$, a negative derivative implies strict decrease on $(0,h_*)$, and a positive derivative implies strict increase on $(h_*,\infty)$. Hence $h_*$ is the unique global minimiser on $(0,\infty)$. Substituting the value of $h_*$ gives
\begin{align*}
h_{\operatorname{AMISE}}=\left(\frac{R(K)}{\mu_2(K)^2R(f'')n}\right)^{1/5}.
\end{align*}
[/step]
