User joe neeman - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T00:34:39Z http://mathoverflow.net/feeds/user/21652 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/111233/multivariate-bernstein-polynomials-for-approximation-of-derivatives Multivariate Bernstein polynomials for approximation of derivatives. Joe Neeman 2012-11-02T01:04:13Z 2013-03-04T05:10:59Z <p>If I have a $C^\infty$ function $f: [0,1]^n \to \mathbb{R}$ then its Bernstein polynomials $$B_m(x) = \sum_{k_1,\dots,k_n=0}^m f\left(\frac{k_1}{m}, \dots, \frac{k_m}{m}\right) \prod_{i=1}^n \binom{m}{k_i} x^{k_i} (1-x_i)^{m-k_i}$$ converge uniformly to $f$, and all of the partial derivatives of $B_m$ converge uniformly to the corresponding partial derivatives of $f$.</p> <p>My question is whether anyone knows of a reference for this. The univariate statement can be found practically everywhere, while I found the multivariate statement for $f$ (but not its derivatives) <a href="http://www.iue.tuwien.ac.at/phd/heitzinger/node130.html" rel="nofollow">here</a>.</p> http://mathoverflow.net/questions/93615/concentration-of-gaussian-vectors Concentration of Gaussian vectors Joe Neeman 2012-04-09T23:44:57Z 2012-04-16T05:18:08Z <p>If $f: \mathbb{R}^n \to \mathbb{R}$ is a Lipschitz function and $X$ is a standard $n$-dimensional Gaussian vector with $\mathbb{E} f(X) = 0$, then $f(X)$ is subgaussian (in a way that does not depend on $n$). If $f$ is $\mathcal{C}^1$, this is equivalent to saying that $|\nabla f|$ bounded implies $f(X)$ is subgaussian.</p> <p>There seem to be two natural generalizations of this. The first is to ask for weaker bounds on $|\nabla f|$. For example, if $|\nabla f|$ is subgaussian, then $f$ should be subexponential. The second generalization concerns functions $f: \mathbb{R}^n \to \mathbb{R}^k$. If I want to control $|f|$ independently of $k$, it is no longer enough to assume that $f$ is Lipschitz, since for the function $f(x) = (x_1, \dots, x_k)$, $|f|$ concentrates around $\sqrt k$. The natural condition seems to be a bound on the Frobenius norm of $D f$ (the matrix of partial derivatives).</p> <p>The following statement contains both generalizations simultaneously (and is not hard to prove): If $f: \mathbb{R}^n \to \mathbb{R}^k$ is continuously differentiable and $\mathbb{E} f(X) = 0$ then $$\big(\mathbb{E} |f(X)|^p\big)^{1/p} \le c \sqrt p \big(\mathbb{E} \|Df\|_F^p\big)^{1/p}.$$</p> <p>My question is whether a statement like this is known and (if so) where I can find a reference.</p> http://mathoverflow.net/questions/93615/concentration-of-gaussian-vectors/94161#94161 Answer by Joe Neeman for Concentration of Gaussian vectors Joe Neeman 2012-04-15T23:35:19Z 2012-04-16T05:18:08Z <p>To answer my own question, this follows from a more general result that is mentioned in <a href="http://webmail.impan.gov.pl/cgi-bin/ba/pdf?ba55-3-07" rel="nofollow">"On measure concentration of vector valued maps"</a> by Ledoux and Oleszkiewicz, Theorem 4: for any convex function $\Psi: \mathbb{R}^k \to \mathbb{R}$, $$\mathbb{E} \Psi(f(X)) \le \mathbb{E} \Psi(\frac{\pi}{2} Y \cdot Df(X))$$ where $X$ and $Y$ are independent standard Gaussians. If you condition the right hand side on $X$ and integrate $Y$, a standard result on the moments of order-2 Gaussian chaos gives $$\mathbb{E} (\frac{\pi}{2} Y \cdot Df(X))^p \le (cp)^{p/2} \mathbb{E} \|Df\|_F^p$$ which is what I claimed above. (By following the references a little more carefully, you can even get the sharp constant.)</p> http://mathoverflow.net/questions/93917/a-question-about-the-tail-of-an-absolutely-integrable-function/93920#93920 Answer by Joe Neeman for A question about the tail of an absolutely integrable function Joe Neeman 2012-04-13T00:09:47Z 2012-04-13T00:09:47Z <p>You can't say anything in general. For a simple example showing that you can't get any polynomial rate, consider the measure space $X = [1, \infty)$ with the Lebesgue measure and the function $f(x) = x^{-(1 + \epsilon)}$. Assuming I integrated it correctly, you get $E(\delta) = \frac{1}{\epsilon} \delta^{\epsilon/(1 + \epsilon)}$ which gets arbitrarily slow as $\epsilon \to 0$.</p> <p>You can also do a more complicated construction that gives a bit more: for any function $e(\delta)$ with $e(\delta) \to 0$, you can find an $f$ such that $E(\delta_k) \ge e(\delta_k)$ for some sequence $\delta_k \to 0$. In particular, you can get arbitrarily slow rates. For this construction, choose a decreasing sequence $\delta_k \to 0$ with $e(\delta_k) \le 2^{-k}$. Then define $f$ such that $\mu(f = \delta_k) = 2^{-k}/\delta_k$. (If $X$ is $[0, \infty)$ with Lebesgue measure, you can do this by setting $f(x) = \delta_k$ if $\sum_{j \le k-1} 2^{-j}/\delta_j \le x &lt; \sum_{j \le k} 2^{-j}/\delta_j$.) Then $f$ is integrable because $$\int f = \sum_{k} \delta_k \mu(f = \delta_k) = \sum_k 2^{-k} &lt; \infty.$$ But you have $$E(\delta_k) = \int_{f \le \delta_k} f = \sum_{j \ge k} \delta_j \mu(f = \delta_j) \ge \delta_k \mu(f = \delta_k) = 2^{-k} = e(\delta_k).$$</p> http://mathoverflow.net/questions/78044/what-is-the-maximum-diameter-of-n-steps-of-a-random-walk/93808#93808 Answer by Joe Neeman for What is the maximum diameter of $N$ steps of a random walk? Joe Neeman 2012-04-11T22:34:57Z 2012-04-11T22:47:27Z <p>Sorry, disregard what is below. The LIL gives $\max_{i \le N} |x_i| \approx \sqrt{2 N \log \log N}$ for infinitely many $N$, but for any particular $N$, $\max_{i \le N} |x_i|$ should be of the order $\sqrt N$.</p> <p>If you only care about bounds up to a constant factor, then I think you're after the law of the iterated logarithm (LIL). As Cardinal indicated, it's enough to consider the 1-dimensional problem (if you don't care about losing a factor of 2). Moreover, $$\max_i|x_i| \le \max_{i,j} |x_i - x_j| \le 2\max_i |x_i|$$ and so you may as well consider $\max |x_i|$ instead. By the LIL, $\max_{i \le N} |x_i| \sim \sqrt{2 N \log \log N}$ almost surely.</p> <p>The same argument works if the steps are distributed on the unit circle, since the LIL doesn't require Gaussian variables.</p> <p>If you want to try to get the sharp constant, there are also multi-dimensional versions of the LIL available. You can search for them on Google; I don't really know that area...</p> http://mathoverflow.net/questions/111233/multivariate-bernstein-polynomials-for-approximation-of-derivatives/111257#111257 Comment by Joe Neeman Joe Neeman 2012-11-15T19:56:30Z 2012-11-15T19:56:30Z I've marked your answer as accepted because it's a very nice explanation. I would still like to have a reference, though: to put it in context, I'm writing a paper which needs some polynomial approximation, but the topic of the paper is very much unrelated to polynomial approximation and the intended audience is probably not too familiar with it. Right now I just have a proof in an appendix, but it would be nice to replace it with a reference. http://mathoverflow.net/questions/111233/multivariate-bernstein-polynomials-for-approximation-of-derivatives Comment by Joe Neeman Joe Neeman 2012-11-03T01:26:49Z 2012-11-03T01:26:49Z Thanks, I believe it's fixed now. http://mathoverflow.net/questions/93615/concentration-of-gaussian-vectors/94161#94161 Comment by Joe Neeman Joe Neeman 2012-04-16T05:18:53Z 2012-04-16T05:18:53Z Thanks, link fixed (and I've also written the authors, etc.)