Basic properties of the kernel density estimator This is a question from a mathematical statistics textbook, used at the first and most basic mathematical statistics course for undergraduate students. This exercise follows the chapter on nonparametric inference. Part 1 is quite straightforward, however, with part 2-6 I am stuck. I've looked into van der Vaart's book Asymptotic Statistics pages 344-346, but this seems to be the course book in a more advanced course. An attempt at a solution is given. Any help is appreciated.
Exercise

Suppose $x_1, ..., x_n$ are independent and identically distributed (i.i.d.) observations of a random variable $X$ with unknown distribution function $F$ and probability density function $f\in C^m$, for some $m>1$ fixed. Let
$$f_n(t)=\frac{1}{n}\sum_{i=1}^n \frac{1}{h}k\left(\frac{t-x_i}{h}\right)$$
be a kernel estimator of $f$, with $k\in C^{m+1}$ a given fixed function such that $k\geq 0$, $\int_{\mathbb{R}} k(u)\mathrm{d}u=1$, $\mathrm{supp} (k)=[-1,1]$ and bandwidth $h=h(u)$ (for the time being unspecified).

*

*Show that $\mathbb{E}[f_n(t)]=\int_{\mathbb{R}} k(u) f(t-hu)\mathrm{d}u$.

*Make a series expansion of $f$ around $t$ in terms of $hu$ in the expression for $\mathbb{E}[f_n(t)]$. Suppose that $k$ satisfies $\int_{\mathbb{R}} k(u)\mathrm{d}u=1$, $\int_{\mathbb{R}} k(u)u^l\mathrm{d}u=0$ for all $1<l<m$ and $\int_{\mathbb{R}} k(u)u^m\mathrm{d}u<\infty$. Determine the bias $\mathbb{E}[f_n(t)]-f(t)$ as a function of $h$.

*Suppose that $\mathrm{Var}[k(X_1)]<\infty$ and determine $\mathrm{Var}[f_n(t)]$ as a function of $h$.

*Determine the mean square error $\mathrm{mse}[f_n(t)]$ from 2 and 3 as a function of $h$.

*For what value of $h$, as a function of $n$, is $\mathrm{mse}[f_n(t)]$ smallest?

*For the value of $h=h(n)$ obtained from 5, how fast does $\mathrm{mse}[f_n(t)]$ converge to 0, when $n$ converges to $\infty$?


Note; $h=h(u)$ and $1<l<m$ are most likely typos for $h=h(n)$ and $1\leq l<m$ respectively.
Attempt

*

*By linearity of the expectation, identical distribution of $x_1,...,x_n$, the law of the unconscious statistician and the change of variables $u=(t-x)/h$,
\begin{align}
\mathbb{E}[f_n(t)]&=\frac{1}{n}\sum_{i=1}^n \mathbb{E}\left[\frac{1}{h}k\left(\frac{t-x_i}{h}\right)\right]\\
&=\mathbb{E}\left[\frac{1}{h}k\left(\frac{t-x}{h}\right)\right]\\
&=\int_{\mathbb{R}}\frac{1}{h}k\left(\frac{t-x}{h}\right)f(x)\mathrm{d}x\\
&=\int_{\mathbb{R}}\frac{1}{h}k(u)f(t-hu)h\mathrm{d}u\\
&=\int_{\mathbb{R}}k(u)f(t-hu)\mathrm{d}u.
\end{align}

*From $f\in C^m$, it follows that $$f(t-hu)=\sum_{l=0}^m \frac{f^{(l)}(t)}{l!} (-hu)^l+o((hu)^m).$$
Then from part 1 and linearity of integration, \begin{align}
\mathbb{E}[f_n(t)]&=\int_{\mathbb{R}}k(u)\left(\sum_{l=0}^m \frac{f^{(l)}(t)}{l!} (-hu)^l+o((hu)^m)\right)\mathrm{d}u \\
&=\sum_{l=0}^m\int_{\mathbb{R}}k(u)\frac{f^{(l)}(t)(-hu)^l}{l!}\mathrm{d}u+\int_{\mathbb{R}}k(u)o((hu)^m)\mathrm{d}u. \label{remain}
\end{align}
From the given conditions on $k$, the $l=0$ term reads
\begin{equation}
    \int_{\mathbb{R}} k(u)f(t)\mathrm{d}u=f(t)\int_{\mathbb{R}} k(u) \mathrm{d}u=f(t).
\end{equation}
The $1\leq l<m$ terms are
$$\int_{\mathbb{R}} k(u)\frac{f^{(l)}(t)}{l!} (-hu)^l\mathrm{d}u=\frac{f^{(l)}(t)(-h)^l}{l!}\int_{\mathbb{R}} k(u)u^l\mathrm{d}u=0.$$
Finally, the $l=m$ term is $$ \frac{f^{(m)}(t)(-h)^m}{m!}\int_{\mathbb{R}} k(u)u^m\mathrm{d}u<\infty.$$
The remainder term is given in Misius's answer (+1). Putting it all together:
$$\mathbb{E}[f_n(t)] = f(t) + \frac{f^{(m)}(t)(-h)^m}{m!} \int_{\mathbb{R}}k(u)u^m \mathrm{d}u + o(h^m),$$ and thus $$\mathbb{E}[f_n(t)]-f(t)=\frac{f^{(m)}(t)(-h)^m}{m!} \int_{\mathbb{R}}k(u)u^m \mathrm{d}u + o(h^m)=A(t)h^m+o(h^m),$$ where $A(t)=\frac{f^{(m)}(t)(-1)^m}{m!} \int_{\mathbb{R}}k(u)u^m \mathrm{d}u<\infty.$

*See Misius's answer.

*

\begin{align} 
    \mathrm{mse}[f_n(t)]&=\mathrm{Var}[f_n(t)]+\mathrm{Bias}^2[f_n(t)] \\
    &=\left(\frac{f(t)}{nh}\int_{\mathbb{R}}k^2(u)\mathrm{d}u+o\left(\frac{1}{nh}\right)\right)+
    \left(A(t)h^m+o(h^m)\right)^2 \\
    &=\left(\frac{f(t)}{nh}\int_{\mathbb{R}}k^2(u)\mathrm{d}u+o\left(\frac{1}{nh}\right)\right)+\left(A(t)^2h^{2m}+2A(t)h^mo(h^m)+o(h^{2m})\right) \\
    &=\left(\frac{f(t)}{nh}\int_{\mathbb{R}}k^2(u)\mathrm{d}u+o\left(\frac{1}{nh}\right)\right)+\left(A(t)^2h^{2m}+o(h^{2m})+o(h^{2m})\right)\\
    &=\left(\frac{f(t)}{nh}\int_{\mathbb{R}}k^2(u)\mathrm{d}u+o\left(\frac{1}{nh}\right)\right)+\left(A(t)^2h^{2m}+o(h^{2m})\right)\\
    &\approx \frac{f(t)}{nh}\int_{\mathbb{R}}k^2(u)\mathrm{d}u+A(t)^2h^{2m}.
\end{align}


*From the approximation obtained in part 4, it follows that $\mathrm{mse}[f_n(t)](h)$ has an absolute minimum for $h\in(0,\infty)$, since $\mathrm{mse}[f_n(t)](h)\to\infty$ for $h\to 0$ and $h\to \infty$. The absolute minimum is found by differentiating $\mathrm{mse}[f_n(t)](h)$ and solving for $h$ when the derivative equals $0$, that is
\begin{equation}
    h=\left(\frac{f(t)\int_{\mathbb{R}}k^2(u)\mathrm{d}u}{A^2(t)2mn}\right)^{1/(2m+1)}.
\end{equation}


*Plugging in the value of $h$ obtained in part 5 into the approximation obtained in part 4, one finds that
\begin{equation}
    \mathrm{mse}[f_n(t)]\propto n^{-2m/(2m+1)},
\end{equation}
since both $1/nh$ and $h^{2m}$ reduce to $n^{-2m/(2m+1)}$ for $h\propto n^{-1/(2m+1)}$.
 A: You did everything correctly, you are just missing the last step. First of all, you can write $o((uh)^m) = u^m o(h^m)$, which will lead to
$$ \int_\mathbb{R} k(u) o((uh)^m)du = o(h^m)\int_\mathbb{R} k(u) u^mdu = o(h^m)$$
by the properties of the kernel function given in the assignment. Hence,
$$\mathbb{E}[f_n(t)] = f(t) + \frac{f^{(m)}(t)(-h)^m}{m!} \int_{\mathbb{R}}k(u)u^m du + o(h^m).$$
In the simplest case, when the underlying density is twice continuously differentiable, we have
$$Bias(f_n(t)) = \mathbb{E}[f_n(t)] - f(t) = h^2 \frac{\nu_2(k)}{2}{}f^{\prime\prime}(t) + o(h^2),$$
where $\nu_2(k) = \int_{\mathbb{R}}k(u)u^2$. This is the most common expression for the bias and for many usual kernel functions.
Edit:
Note that the Bias depends on $t$. Hence, in your notation it should be $A(t)$, not just $A$. It exists and is finite, because $f^{(m)}$ is $m$-times continuously differentiable, and $f^{(m)}(t)$ is just a number.
3.
\begin{align*}
\mathrm{Var}[f_n(t)]&=\frac{1}{nh}\left(h\int_{\mathbb{R}}k^2(u)f(t-hu)\mathrm{d}u-\left(h\int_{\mathbb{R}}k(u)f(t-hu)\mathrm{d}u\right)^2\right)\\
&=\frac{1}{nh^2}\left(h\int_{\mathbb{R}}k^2(u)\left(f(t)-(hu)f^{(1)}(t)+o(hu)\right)\mathrm{d}u +O(h^2) \right)\\
&=\frac{1}{nh^2}\left(h\cdot f(t)\int_{\mathbb{R}}k^2(u)du - h^2 f^{(1)}(t)\int_{\mathbb{R}}k^2(u)udu + o(h^2) +O(h^2) \right)\\
&=\frac{1}{nh^2}\left(h\cdot f(t)\int_{\mathbb{R}}k^2(u)du +O(h^2) \right)\\
&=\frac{f(t)}{nh}\int_{\mathbb{R}}k^2(u)du +o\left(\frac{1}{nh}\right).
\end{align*}
Here, we used the following facts:
a. $o(h^2) + O(h^2) = O(h^2)$
b. $\int |k^2(u)u|du \le \int k^2(u)du< \infty$ because $k(u)$ has support on $[-1, 1]$. In the kernel density estimation, it is usually assumed that $\int k^2(u)du$ is finite and sometimes is denoted by $R(k)$ (roughness of the kernel, though the term is rarely used). It is possible that this condition can vbe derived from the assumptions that are given in the exercise but I have not tried it.
c. $O(h^2) \cdot \frac{1}{nh^2} = O(h) \cdot\frac{1}{nh} = o(1)\cdot \frac{1}{nh} = o\left(\frac{1}{nh}\right)$.


*Everything is correct, just do not forget that $A$ depends on $t$.


*You will have different order for the variance. If I understand the question correctly, you can "forget" about the remainder terms and find $h$ such that the main part of MSE is the smallest. The sum of these two terms are often called asymptotic mean squared error.


*And here you will need also to substitute the optimal bandwidth from 5. into the formula for MSE from 4.
