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Let $X_1, ..., X_n$ be i.i.d. samples drawn from a pdf $f(x)$ on the real line. The kernel density estimator is defined as follows,

$$\hat{f_n}(x) = \frac{1}{nh}\sum_1^n K(\frac{x-X_k}{h})$$

where $K:\mathbb{R}\to [0, \infty)$ satisfying $\int_{-\infty}^{\infty}K(x) = 1$ and $h>0$.

How to prove that $\Vert \hat{f_n}(x) - f \Vert_1$ is sub-Gaussian w.r.t. $\frac{1}{\sqrt{n}}$ i.e.

$$\mathbb{P}(|\Vert \hat{f_n}(x) - f \Vert_1 - \mathbb{E}(\Vert \hat{f_n}(x) - f \Vert_1)|\geq t) \leq 2e^{-\frac{nt^2}{2}}$$

where

$$\Vert \hat{f_n}(x) - f \Vert_1:=\int_{-\infty}^{\infty}|\hat{f_n}(x)-f(x)|dx$$

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  • $\begingroup$ Why do think this is true? I don't think this is true in general. $\endgroup$ Commented Mar 12, 2023 at 1:25
  • $\begingroup$ I found it on one of the supplementary material from my advisor's High Dimension Probability Course. $\endgroup$
    – dc3506
    Commented Mar 12, 2023 at 3:03
  • $\begingroup$ Then you may want to ask your adviser about this. However, again, I don't think this is true in general. $\endgroup$ Commented Mar 12, 2023 at 3:38

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