User attar reda - MathOverflow

Answer by Attar Reda for What is known about the Gaussian measure of the unit ball in a Hilbert Space?

2010-08-25T06:39:39Z

I think you can see the articles entitled "concentration of measure phenomenon ". The idea is as follows: Let $(X,d,\mu)$ be a metric measure space, such as $\mu(X)=1$. Let $$\alpha(\epsilon) = \sup {\mu(X \backslash A_\epsilon) \ | \ \mu(A) = 1/2 }$$ where $$A_\epsilon = { x \ | \ d(x, A) < \epsilon }$$ is the $\epsilon$-extension of a set $A$. The function $\alpha(.)$ is called the concentration rate of the space $E$. The following equivalent definition has many applications:$$\alpha(\epsilon) = \sup { \mu( { F >= M + \epsilon }) },$$ where the supremum is over all $1$-Lipschitz functions $F: X \to \mathbb{R}.$ For example the median (or Levy mean) $M = \mathop{Med}(F) $ is defined by the inequalities $$\mu ( F \geq M ) >= 1/2, \ \mu ( F <= M ) \geq 1/2.$$ More precisely, the space $X$ exhibits a concentration phenomenon if $\alpha(\epsilon)$ decays very fast as $\epsilon$ grows. More formally, a family of metric measure spaces $(X_n,d_n,\mu_n)$ is called a Levy family if the corresponding concentration rates $\alpha(\epsilon)$ satisfy $$\forall \epsilon > 0 \ \ \alpha_n(\epsilon) \to 0,$$ and a normal Levy family if $$ \forall \epsilon \to 0 \ \ \alpha_n(\epsilon) = O(\exp(-C n \epsilon^2))$$ for $C$ some positive constant. the last inequality is obtained bay applying the "Hoeffding inequality" and in the case of Hilbert space with concentration in small balls we do : $$\forall (x_1,x_2)\in X^2, \ \ d(x_1,x_2)=\|x_1,x_2\|< r.$$

Answer by Attar Reda for The density of x_1^n+x_2^n where x_i are Gaussian

2010-08-24T15:05:21Z

Firstly you forgot to multiply the density $f(x^n=y)$ by $1/\sqrt{2\pi}$. I think if you obtained the density of the random variable $X_{1,2}=X_1^n+X_2^n$ by the convolution method, the problem no more posed , because for $X_1^n+X_2^n+X_3^n=X^n_{1,2}+X_3^n=X^n_{1,2,3}$, and you have the density of $X^n_{1,2}$, the density of $X^n_{3}$, you can calculate there convolution, i.e the density of $X^n_{1,2,3}$. If the calculation is very difficult with the convolution (I think) you can use the characteristic function. You calculate the function characteristic of the variable $X_1^n$ that one noted $\psi_{X_1^n}(t)$. As the two variables $X_1^n$ and $X_2^n$ are i.i.d, then $\psi_{X_1^n+X_2^n}(t)=\psi_{X_1^n}(t)\cdot \psi_{X_2^n}(t)=(\psi_{X_1^n}(t))^2$ and so on for variable $X_1^n+X_2^n+...+X_k^n$ we will have $\psi_{X_1^n+X_2^n+\ldots +X_k^n}(t)=(\psi_{X_1^n}(t))^k$. Just well calculate $\psi_{X_1^n}(t)$.

Answer by Attar Reda for Kernel width in Kernel density estimation

2010-08-22T06:51:46Z

Robin is right, you can use this site. Let me ask a question, you estimate the density by the kernel estimator (Nadaraya-Watson), your writing of the estimator is not enough correct, when I copy your source and compile it with LaTeX writing is not correct, any ways the kernel estimator of the density of a functional regressor is $$ \widehat{f}(x)= \frac{\displaystyle\sum_{i=1}^{n}K\Big(h_K^{-1}d(x,X_i)\Big)Y_i}{\displaystyle\sum_{i=1}^{n}K\Big(h_K^{-1}d(x,X_i)\Big)} . $$ where $d (.,.)$ is the metric . You can refer to Ferraty et al of Sabatier University, I think you know people there (I saw your CV). So we must choose the metric $d$, to estimate the smoothing parameter $h$, I think they found a way to choose, by cross-validation, they are obtained the optimal $h$. So before starting, I would advise you to see the work of the team probability for Laboratory Toulouse. For the robustness you can see the book of Huber (1981): Robust Statistics, it is very good, or Hambel et al (1986). Good lunk.

Answer by Attar Reda for connection between the Gaussian and the Cauchy distribution

2010-08-21T14:27:12Z

$$ F_Z(z)=P(Z\leq z)=P(Y/X\leq z)=P(Y\leq zX)\ =P(Y\leq zX,\ X> 0)+ P(Y\geq zX,\ X< 0)$$ that implies $$ f_Z(z)= \frac{dF_Z(z)}{dz}=\int_{-\infty}^{+\infty}|x|f_Y(zx)f_X(x)\, dx\ =\frac{1}{2\pi}\int_{-\infty}^{+\infty}|x|e^{-(z^2+1)x^2/2}\ dx=\frac{1}{\pi(x^2+1)}. $$ Againt, The difficulty I encountered is how to prove that the characteristic function of the variable YX is the same as the Cauchy distribution ?

Answer by Attar Reda for connection between the Gaussian and the Cauchy distribution

2010-08-21T14:20:41Z

$$ F_Z(z)=\BBp(Z\leq z)=\BBp(Y/X\leq z)=\BBp(Y\leq zX)\ =\BBp(Y\leq zX,\,X> 0)+ \BBp(Y\geq zX,\,X< 0),\,\, \mbox{that implies}\ f_Z(z)= \frac{dF_Z(z)}{dz}=\int_{-\infty}^{+\infty}|x|f_Y(zx)f_X(x)\, dx\ =\frac{1}{2\pi}\int_{-\infty}^{+\infty}|x|e^{-(z^2+1)x^2/2}\, dx=\frac{1}{\pi(x^2+1)}. $$

I believe that the jsmath is the same as LaTeX, but I think really have a problem with jsmath, I have no time to learn it. So if my source does not work you can take this source .tex and compile it in LaTeX, you have my answer .. Thank you

Answer by Attar Reda for connection between the Gaussian and the Cauchy distribution

2010-08-21T13:57:33Z

Jon Peterson, you are correct. But why does it directly the calcul of cdf and pdf of $Y/X$, ie

\begin{eqnarray*} F_Z(z)&=&\BBp(Z\leq z)=\BBp(Y/X\leq z)=\BBp(Y\leq zX)\ &=&\BBp(Y\leq zX,\,X> 0)+ \BBp(Y\geq zX,\,X< 0),\,\, \mbox{that implies}\ f_Z(z)&=& \frac{dF_Z(z)}{dz}=\int_{-\infty}^{+\infty}|x|f_Y(zx)f_X(x)\, dx\ &=&\frac{1}{2\pi}\int_{-\infty}^{+\infty}|x|e^{-(z^2+1)x^2/2}\, dx=\frac{1}{\pi(x^2+1)}. \end{eqnarray*}

The difficulty I encountered is how to prove that the characteristic function of the variable $ Y / X $ is the same as the Cauchy distribution ?

Answer by Attar Reda for Estimating the mean of a truncated gaussian curve

2010-08-21T06:26:59Z

For the estimate of m, I think you can us the nonparametric approach (Nadaraya-Watson estimator), is beater to the Bayesian approach. See http://en.wikipedia.org/wiki/Kernel_regression

Comment by Attar Reda

2010-08-25T06:50:43Z

Or, the site : books.google.fr/…

Comment by Attar Reda

2010-08-25T06:47:37Z

For more detail, you can see the book of Michel Ledoux - "The Concentration of Measure Phenomenon.