Show that $\mathbb{E}\left|\hat{f_n}-f \right| \leq \frac{2}{n^{1/3}}$ where $\hat{f_n}$ is a density estimator for $f$

Question

Suppose we have a continuous probability density $f : \mathbb{R} \to [0,\infty)$ such that $\text{sup}_{x \in \mathbb{R}}(\left|f(x)\right| + \left|f'(x)\right|) \leq 1. \;$ Define the density estimator:

$$\hat{f_n} = \frac{1}{n^{2/3}}\sum^n_{i=1}\mathbf{1}\{-1/2 \leq n^{1/3}(x-X_i) \leq 1/2\}, \: x \in \mathbb{R}$$

Show that, for every $x \in \mathbb{R} \:$ and every $n \in \mathbb{N},$
$$\mathbb{E}\left|\hat{f_n}(x)-f(x)\right| \leq \frac{2}{n^{1/3}}$$


Attempt

It’s easy to show that $$\mathbb{E}\hat{f_n} = \frac{F(x+\frac{1}{2}n^{-1/3}) – F(x-\frac{1}{2}n^{-1/3})}{n^{-1/3}}$$
where $F$ is the CDF of $f$. From here, using the condition $\:\text{sup}_{x \in \mathbb{R}}(\left|f(x)\right| + \left|f'(x)\right|) \leq 1 \:$ and in particular, $\:\text{sup}_{x \in \mathbb{R}}\left|f'(x)\right|\leq 1\:$ we can show (geometrically) that,
$$\left|\mathbb{E}\hat{f_n}-f\right| \leq \frac{1}{4n^{1/3}}$$

However, I’m not sure how to get a bound on $\mathbb{E}\left|\hat{f_n}-f\right|$.

Answer

General

The density estimator is a special case of a kernel density estimator (KDE) with rectangular kernel $$K(x)=\mathbf{1}\{-1/2\leq x\leq 1/2\}$$ and bandwidth $h=n^{-1/3}$. It would therefore be possible to use the general theory of KDEs and deduce the result.

I will not follow that approach but instead derive properties of this specific estimator directly and link the results to the general KDE theory if possible.

Distribution of $\hat{f_n}$

As (apart from a multiplicative constant) the estimator is effectively counting the number of times an observation of $X_i$ is falling into the interval $\left[x-\frac{1}{2}n^{-1/3}, x+\frac{1}{2}n^{-1/3}\right]$ when calculating $\hat{f_n}(x)$ we can immediately see that $$n^{2/3} \hat{f_n}\sim\operatorname{Binomial}(n,p)$$ with $p$ equal to $$F\left(x+\frac{1}{2}n^{-1/3}\right)-F\left(x-\frac{1}{2}n^{-1/3}\right).$$ This secretly assumes that all the $X_i$ are independent and identically distributed with CDF $F$. From this we can infer directly that $\mathbb{E}\hat{f_n}(x)=n^{1/3}p$ and $\operatorname{Var}(\hat{f_n}(x))=n^{-1/3}p(1-p)$.

Asymptotics

To get the asymptotic behaviour of the expectation and variance we have to analyse $p$ from above. Using a Taylor approximation at $x$ we see that $$F\left(x\pm\frac{1}{2}n^{-1/3}\right)=F(x)\pm f(x)\frac{1}{2}n^{-1/3}+f'(\xi_\pm)\frac{1}{8}n^{-2/3}$$ for some $\xi_\pm$ between $x$ and $x\pm\frac{1}{2}n^{-1/3}$. Knowing that $|f’|\leq 1$ we get $$\left|\mathbb{E}\hat{f_n}(x)-f(x)\right|\leq\frac{1}{4}n^{-1/3}\tag1.$$

Remark
Assuming that also $f”$ exists and is continuous we could get a stronger result. In this case the bias of $\hat{f_n}$ would tend to zero with the order $n^{-2/3}$. This is in line with the more general result that under mild assumptions on the kernel $h$ we have that $$\left|\mathbb{E}\hat{f_n}(x)-f(x)\right|=\mathcal{O}(h^2)$$ for $n$ towards infinity.

For the variance we also use a shortened version of the Taylor approximation $$F\left(x\pm\frac{1}{2}n^{-1/3}\right)=F(x)\pm f(\zeta_\pm)\frac{1}{2}n^{-1/3}$$ for some $\zeta_\pm$ between $x$ and $x\pm\frac{1}{2}n^{-1/3}$. The variance can now be written as $n^{-1/3}p(1-p)=n^{-1/3}(p-p^2)$ with $$p-p^2=f(x)n^{-1/3}+(f'(\xi_+)-f'(\xi_-))\frac{1}{8}n^{-2/3} – (f(\zeta_+)+f(\zeta_-))^2\frac{1}{4}n^{-2/3}.$$ Again, using the given upper bounds for $f$ and $f’$ we can bound the variance from above by $$\operatorname{Var}(\hat{f_n}(x))\leq n^{-1/3}\left(n^{-1/3}+\frac{1}{4}n^{-2/3}\right)\leq\frac{5}{4}n^{-2/3}.\tag2$$

Remark
This result is similar to a result from the general theory of KDEs, which gives $$\operatorname{Var}(\hat{f_n}(x)) =\mathcal{O}((nh)^{-1})$$ for $n$ towards infinity.

Combining the results

To get the desired result we use a decomposition of the mean absolute error (MAE) similar to that of the mean squared error. We start with \begin{align}\mathbb{E}\left|\hat{f_n}(x)-f(x)\right|&\leq\mathbb{E}\left|\hat{f_n}(x)-\mathbb{E}\hat{f_n}(x)\right|+\mathbb{E}\left|\mathbb{E}\hat{f_n}(x)-f(x)\right|\\&\leq\left(\mathbb{E}\left|\hat{f_n}(x)-\mathbb{E}\hat{f_n}(x)\right|^2\right)^{1/2}+\left|\mathbb{E}\hat{f_n}(x)-f(x)\right|\\&=\operatorname{Var}\left(\hat{f_n}(x)\right)^{1/2}+\left|\mathbb{E}\hat{f_n}(x)-f(x)\right|\end{align} using the triangle and Hölder inequalities. Using bounds $(1)$ and $(2)$ we finally get $$\mathbb{E}\left|\hat{f_n}(x)-f(x)\right|\leq\left(\frac{\sqrt{5}}{2}+\frac{1}{4}\right)n^{-1/3}\leq 2n^{-1/3}$$ and have the result.

Attribution
Source : Link , Question Author : qwerty , Answer Author : TheSimpliFire

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