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I encountered the following excercise in a book:

Exercise: Given a Lipschitz continuous function $f$ on $[0,1]$, with Lipschitz constant $c$.

Show that $|B_{n,f}(p) - f(p)|\le \frac{c}{2\sqrt{n}}$ for all $p\in[0,1]$, where $B_{n,f}$ is the Bernstein polynomial of order $n$ constructed from $f$.

The exercise also gives the following hint:

$B_{n,f}(p) = \mathbf{E}[f(X/n)]$, where $X$ is a random variable with binomial distribution for $n$ trials with success probability $p$.

I got this far:

$$ \begin{align} \left\lvert B_{n,f}(p) - f(p)\right\rvert = \left\lvert \mathbf{E}\left[f\left(\frac{X}{n}\right)\right] - f(p)\right\rvert &= \left\lvert \mathbf{E}\left[f\left(\frac{X}{n}\right) - f(p)\right]\right\rvert \label{1} \\ &\le \mathbf{E}\left[\left\lvert f\left(\frac{X}{n}\right) - f(p)\right\rvert\right] & (|\cdot|\,\text{ is convex}) \\ &\le \mathbf{E}\left[c\left\lvert \frac{X}{n} - p\right\rvert\right] & (\text{Lipschitz continuity}) \\ &=c\cdot \mathbf{E}\left[\left\lvert \frac{X}{n} - p\right\rvert\right] \end{align} $$

So it remains to show that $\mathbf{E}\left[\left\lvert \frac{X}{n} - p\right\rvert\right] \le \frac{1}{2\sqrt{n}}$. I could not figure out how to proceed from this point. Any help will be appreciated.

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1 Answer 1

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The variance of $X$ is $np(1-p) \le n/4$. And since the variance is $E[|X-np|^2]$, it follows that $$E[|X/n -p|^2] =\frac1{n^2}E[|X-np|^2] \le \frac{1}{4n}$$ Then apply the Cauchy-Schwarz inequality.

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