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## Saturday, December 22, 2012

### Axiomatic definition of the center of mass

We define an effective divisor   on $\bR^N$ to be a   function with finite support $\mu:\bR^N\to\bZ_{\geq 0}$. Its mass, denoted by $\bm(\mu)$, is the  nonnegative integer

$$\bm(\mu)=\sum_{\bp\in\bR^N} \mu(\bp).$$

We denote by $\Div_+(\bR^N)$ the set of effective divisors.  Note that $\Div_+(\bR^N)$ has a natural structure of Abelian semigroup.

For any $\bp\in\bR^N$ we denote by $\delta_\bp$ the Dirac divisor of mass $1$ and supported at  $\bp$.   The  Dirac divisors generate  the  semigroup $\Div_+(\bR^N)$.     We have a natural  topology on  $\Div_+(\bR^N)$ where $\mu_n\to \mu$ if and only if

$$\bm(\mu_n)\to \bm(\mu),\;\; {\rm dist}\,\bigr(\;\supp(\mu_n),\; \supp(\mu)\;\bigr)\to 0,$$

where ${\rm dist}$ denotes the Haudorff distance.

A center of mass   is a map

$$\eC:\Div_+(\bR^N)\to\Div_+(\bR^N)$$

satisfying the following conditions.

1. (Localization) For any divisor $\mu$ the support of $\eC(\mu)$ consists of  a single point $\bc(\mu)$.

2.  (Conservation of mass)

$$\bm(\mu)=\bm\bigl(\;\eC(\mu)\;\bigr),\;\;\forall\mu \in\Div_+(\bR^N),$$

so that

$$\eC(\mu)=\bm(\mu)\delta_{\bc(\mu)},\;\;\forall\mu \in\Div_+(\bR^N).$$

3. (Normalization)

$$\bc(m\delta_\bp)=\bp,\;\;\bc(\delta_\bp+\delta_\bq)=\frac{1}{2}(\bp+\bq),\;\;\forall \bp,\bq\in\bR^N,\;\;m\in\bZ_{>0}.$$

$$\eC(\mu_1+\mu_2)= \eC\bigl(\,\eC(\mu_1)+\eC(\mu_2)\,\bigr),\;\;\forall \mu_1,\mu_2\in \Div_+(\bR^N).$$

For example, the   correspondence

$$\Div_+ \ni \mu\mapsto \eC_0(\mu)=\bm(\mu)\delta_{\bc_0(\mu)}\in\Div_+,\;\;\bc_0(\mu):=\frac{1}{\bm(\mu)}\sum_\bp \mu(\bp)\bp$$

is a center-of-mass  map.  I want to show that this is the only center of mass map.

Proposition  If $\eC:\Div_+(\bR^N)\to \Div_+(\bR^N)$ is a  center-of-mass map, then $\eC=\eC_0$.

Proof.      We carry the proof in several steps.

Step 1 (Rescaling).   We can write the additivity property as

$$\bc(\mu_1+\mu_2) =\bc\bigl(\; \bm(\mu_1)\delta_{\bc(\mu_1)} +\bm(\mu_2)\delta_{\bc(\mu_2)}\;\bigr).$$

In particular, this implies that the rescaling property

$$\bc( k\mu)=\bc(\mu),\;\;\forall\mu \in\Div_+,\;\; k\in\bZ_{>0}. \tag{R}\label{R}$$

This follows by induction $k$. For $k=1$ it is obviously true.  In general

$$\bc( k\mu)=\bc\bigl(\;(k-1)\bm(\mu)\delta_{\bc(\;(k-1)\mu)}+\bm(\mu)\delta_{\bc(\mu)}\;\bigr) =\bc\bigl(\; k\bm(\mu)\delta_{\bc(\mu)}\;\bigr)={\bc(\mu)}$$

Step 2. (Equidistribution)   For any $n>0$ and any collinear points $\bp_1,\dotsc,\bp_n$ such that

$$|\bp_1-\bp_2|=\cdots=|\bp_{n-1}-\bp_n|$$

we have

$$\eC\Bigl(\sum_{k=1}^n\delta_{\bp_k}\;\Bigr)=\eC_0\Bigl(\sum_{k=1}^n\delta_{\bp_k}\;\Bigr) \tag{E}\label{E}.$$

Equivalently, this means that

$$\bc\Bigl(\sum_{k=1}^n\delta_{\bp_k}\;\Bigr)=\bc_0\bigl(\sum_{k=1}^n\delta_{\bp_k}\;\bigr)={\frac{1}{n}(\bp_1+\cdots+\bp_n)}.$$

We  will prove  (\ref{E}) arguing by induction on $n$. For $n=1,2$ this follows from the normalization property. Assume that (\ref{E}) is true for any $n< m$. We want to prove it is true for $n=m$.

We distinguish two cases.

(a)  $m$ is even, $m= 2m_0$. We set

$$\mu_1=\sum_{j=1}^{m_0} \delta_{\bp_j},\;\;\mu_2=\sum_{j=m_0+1}^{2m_0}\delta_{\bp_j}.$$

Then

$$\bc(\mu_1+\mu_2)= \bc\bigl(\; m_0\delta_{\bc(\mu1)}+m_0\delta{\bc(\mu_2)}\;\bigr) =\bc\bigl( \delta_{\bc(\mu_1)}+\delta_{\bc(\mu_2)}\;\bigr). \tag{1}\label{2}$$

By induction

$$\bc(\mu_1)=\bc_0(\mu_1),\;\;\bc(\mu_2)=\bc_0(\mu_2).$$

The normalization  condition now implies that

$$\bc\bigl( \delta_{\bc(\mu_1)}+\delta_{\bc(\mu_2)}\;\bigr)=\bc_0\bigl( \delta_{\bc_0(\mu_1)}+\delta_{\bc_0(\mu_2)}\;\bigr).$$

Now run the  arguments in (\ref{2}) in reverse, with $\bc$ replaced by $\bc_0$.

(b) $m$ is odd, $m=2m_0+1$.  Define

$$\mu_1=\delta_{\bp_{m_0+1}},\;\;\mu_2'=\sum_{j<m_0+1}\delta_{\bp_j},\;\;\mu_2''=\sum_{j>m_0+1}\delta_{\bp_j},\;\;\mu_2=\mu_2'+\mu_2''.$$

(Observe that $\bp_{m_0+1}$ is the mid-point in the string of equidistant collinear points $\bp_1,\dotsc,\bp_{2m_0+1}$. ) We have

$$\eC(\mu_2'+\mu_2'')=\eC\bigl( \; \eC(\mu_2')+\eC(\mu_2'')\;\bigr).$$

By induction

$$\eC(\mu_2')+\eC(\mu_2'')= \eC_0(\mu_2')+\eC_0(\mu_2'') =m_0\delta_{\bc_0(\mu_2')}+m_0\delta_{\bc_0(\mu_2'')}=m_0\bigl(\;\delta_{\bc_0(\mu_2')}+\delta_{\bc_0(\mu_2'')}\;\bigr).$$

Observing that

$$\frac{1}{2}\bigl(\bc_0(\mu_2')+\bc_0(\mu_2'')\;\bigr)=\bp_{m_0+1}$$

we deduce

$$\eC(\mu_2)= \eC(\mu_2'+\mu_2'')=m_0\eC\bigl( \delta_{\bc_0(\mu_2')}+\delta_{\bc_0(\mu_2'')}\;\bigr)=m_0\eC_0\bigl( \delta_{\bc_0(\mu_2')}+\delta_{\bc_0(\mu_2'')}\;\bigr)=2m_0\delta_{\bp_{m_0+1}}=2m_0\mu_1.$$

Finally  we deduce

$$\eC(\mu)=\eC\bigl(\;\eC(\mu_1)+\eC(\mu_2)\;\bigr)=\eC\bigl(\;\eC(\mu_1)+2m_0\eC(\mu_1)\;\bigr)= (2m_0+1)\delta_{\bp_{m_0+1}}=\eC_0(\mu).$$

Step 3. (Replacement) We will show that for any distinct points $\bq_1,\bq_2$ and any positive integers $m_1,m_2$ we  can find  $(m_1+m_2)$ equidistant points  $\bp_1,\dotsc,\bp_{m_1+m_2}$ on the line determined by $\bq_1$ and $\bq_2$  such that

$$m_1\delta_{\bq_1}=\eC_0\Bigl(\sum_{j=1}^{m_1} \delta_{\bp_j}\Bigr)=\eC\Bigl(\sum_{j=1}^{m_1} \delta_{\bp_j}\Bigr),\;\;\;m_2\delta_{\bq_2}=\eC_0\Bigl(\sum_{j=m_1+1}^{m_1+m_2} \delta_{\bp_j}\Bigr)=\eC\Bigl(\sum_{j=m_1+1}^{m_1+m_2} \delta_{\bp_j}\Bigr).$$

This is elementary. Without  restricting the generality we can assume  that $\bq_1$ and $\bq_2$ lie on an axis (or geodesic) $\bR$ of $\bR^N$, $\bq_0=0$ and $\bq_2=q>0$.       Clearly we can find  real numbers $x_0, r$, $r>0$, such that
$$\frac{1}{m_1}\sum_{j=1}^{m_1}(x_0+j r)=0,\;\;\frac{1}{m_2}\sum_{j=m_1+1}^{m_1+m_2}(x_0+jr)=q.$$

Indeed, the above two equalities can be rewritten as

$$x_0+\frac{m_1+1}{2} r=0,$$

$$q=x_0 +m_1 r+\frac{m_2+1}{2}=x_0+\frac{m_1+1}{2} r+\frac{m_1+m_2}{2} r.$$

Now place the points  $\bp_j$ at the locations $x_0+jr$.

Step 4. (Conclusion)   We argue on  by induction on mass that

$$\eC(\mu)=\eC_0(\mu),\;\;\forall \mu\in \Div_+\tag{2}\label{3}$$

Clearly, the normalization condition  shows that (\ref{3})  is true if $\supp\mu$ consists of a single point, or if  $\bm(\mu)\leq 2$.

In general if $\bm(\mu)>2$ we write $\mu=\mu_1+\mu_2$ where $m_1=\bm(\mu_1),m_2=\bm(\mu_2)<\bm(\mu)$.

By induction we have

$$\eC(\mu)= \eC\bigl( \eC(\mu_1)+\eC(\mu_2)\bigr)=\eC(\;\eC_0(\mu_1)+\eC_0(\mu_2)\;\bigr).$$

If $\bc_0(\mu_1)=\bc_0(\mu_2)$  the divisors $\eC_0(\mu_1)$ $\eC_0(\mu_2)$ are supported at the same point and we are done. Suppose that  $\bq_1=\bc_0(\mu_1)\neq\bc_0(\mu_2)=\bq_2$. By Step 3, we can  find     equidistant points $\bp_1,\dotsc,\bp_{m_1+m_2}$ such that

$$m_1\delta_{\bq_1}=\eC\Bigl(\sum_{j=1}^{m_1} \delta_{\bp_j}\Bigr)= \eC_0\Bigl(\sum_{j=1}^{m_1} \delta_{\bp_j}\Bigr)$$
$$m_2\delta_{\bq_2}=\eC\Bigl(\sum_{j=m_1+1}^{m_1+m_2} \delta_{\bp_j}\Bigr)=\eC_0\Bigl(\sum_{j=m_1+1}^{m_1+m_2} \delta_{\bp_j}\Bigr).$$

We deduce that

$$\eC(\mu)=\eC\Bigl(\sum_{k=1}^{m+1+m_2}\delta_{\bp_k}\Bigr),\;\; \eC_0(\mu)=\eC_0\Bigl(\sum_{k=1}^{m+1+m_2}\delta_{\bp_k}\Bigr).$$

The conclusion now follows from  (\ref{E}).  q.e.d

Remark.    The above proof    does not really use the  linear structure. If we uses only the fact that any two points in $\bR^N$ determine a unique geodesic.  The  Normalization condition can be replaced by the equivalent one

$$\bc(\delta_\bp+\delta_\bq)= \mbox{the midpoint of the geodesic segment [\bp,\bq]}.$$

If we replace $\bR^N$ with a hyperbolic space the same arguments show that  there exists at most  one center of mass map.

## Tuesday, December 18, 2012

### The 11/8-conjecture was resuscitated back to life. Yep, there was a flaw

Apparently Tom Mrowka found a flaw in  Bauer's "proof" of the  11/8-th conjecture.

## Monday, December 17, 2012

### A really nifty linear algebra trick

I've been stuck on this statistics problem for quite a while, I could taste where it was going but I could never put into words my   intuition. Today I discovered how to neatly  bypass the obstacle.  Discovered is not the appropriate word, because someone else  figured it out  long before me.  It's a really, really elementary linear algebra trick  that I have never encountered in my travels. Very likely, more experienced statisticians than myself    would  smile at my ignorance.

Suppose that  $S$ is an invertible $n\times n$ matrix and $\bx,\by\in\bR^n$ are vectors which I regard as column vectors, i.e., column matrices.  Denote by $(-, -)$ the  natural inner product in  $\bR^n$

$$(\bx,\by)= \bx^\dagger\cdot \by,$$

where ${}^\dagger$ denotes the transpose of a matrix.

Let $r\in\bR$. The name of the game is to compute the scalar

$$r- (\bx, S^{-1} \by)= r-\bx^\dagger\cdot S\cdot \by.$$

Such computations are often required in the neck of the woods where I've been spend the best  part of the last three years namely,  geometric probability. So here is the trick. I'll name it after Belyaev because I am sure he was not the first to observe it. (He even refers to an old book by H. Cramer on statistics.) $\newcommand{\one}{\boldsymbol{1}}$

Belyaev's Trick.

$$r-\bx^\dagger\cdot S\cdot \by =\frac{\det\left[\begin{array}{cc} S &\by\\ \bx^\dagger & r \end{array}\right]}{\det S}=\frac{\det\left[\begin{array}{cc} r &\bx^\dagger\\ \by & S \end{array}\right]}{\det S}.$$

Here is the disappointingly simple proof. Note that

$$\left[ \begin{array}{cc}\one_n & S^{-1}\by\\ 0 & r-\bx^\dagger S^{-1} \by \end{array} \right]= \left[ \begin{array}{cc} \one_n & 0\\-\bx^\dagger & 1 \end{array} \right]\cdot \left[ \begin{array}{cc} S^{-1} & 0\\ 0 & 1 \end{array}\right] \cdot \left[ \begin{array}{cc} S &\by\\ \bx^\dagger & r \end{array}\right].$$

Now take the determinants of  both sides to obtain the first equality.  The second equality  follows  from the first by permuting the rows and columns of the matrix at numerator. $\DeclareMathOperator{\Cov}{\boldsymbol{Cov}}$ $\newcommand{\bsE}{\boldsymbol{E}}$

Here is how it works in practice.   Suppose that   $(X_0, X_1, \dotsc, X_n)\in \bR^{n+1}$ is a centered random  Gaussian, with covariance matrix

$$\Cov(X_0, X_1, \dotsc, X_n)= \Bigl( \;\bsE\bigl( X_i\cdot X_j\,\bigr)\;\Bigr)_{0\leq i,j\leq n}.$$

Assume  that the Gaussian vector $(X_1,\dotsc, X_n)$ is nondegenerate, i.e., the  symmetric matrix  $S=\Cov(X_1,\dotsc, X_n)$ is invertible.

We can then define in an unambiguous way the conditional random variable $\DeclareMathOperator{\var}{\boldsymbol{var}}$

$$(X_0|\; X_1=\cdots =X_n=0).$$

This is a  centered Gaussian random variable with variance given by the   the regression formula

$$\var(X_0|\; X_1=\cdots =X_n=0)= \var(X_0) - \bx^\dagger\cdot S\cdot \bx,$$

where $\bx^\dagger$ is the row vector

$$\bx^\dagger =\left(\; \bsE(X_0X_1),\cdots ,\bsE(X_0 X_n)\;\right).$$

If we now use  Belyaev's trick we deduce

$$\var(X_0|\; X_1=\cdots =X_n=0)=\frac{\det\Cov(X_0, X_1, \dotsc, X_n)}{\det\Cov(X_1,\dotsc, X_n)}.$$

In this form it is used in the related paper of Jack Cuzick (Annals of Probability, 3(1975), 849-858.)

### On a "Car-Talk" problem

While driving back home I heard an interesting math question from all places, the Car Talk show on NPR.  This   made me think of a generalization of the trick they used and in particular, formulate the following problem.  $\newcommand{\bR}{\mathbb{R}}$

Problem  Determine all, reasonably well  behaved compact domains  $D\subset \bR^2$  with the following  property: any line through the origin divides  $D$ into two regions of equal areas.  We will refer to this as property $\boldsymbol{C}$ (for cut).

I know that "reasonably well behaved" is a rather fuzzy   requirement.  At this moment I don't want to think of Cantor like weirdos.  So let's assume that $D$ is semialgebraic.

We say that a domain $D$ satisfies property $\boldsymbol{S}$ (for symmetry) if it is invariant with respect to the  involution

$$\bR^2\ni (x,y)\mapsto (-x,-y) \in \bR^2.$$

It is not hard to see that

$$\boldsymbol{S}\Rightarrow \boldsymbol{C}.$$

Is the converse true?

I describe below one situation when this happens.

A special case.    I'll assume that $D$ is semialgebraic,  star-shaped with respect to the origin and satisfies $\boldsymbol{C}$.  We can  then describe $D$ in polar coordinates by an inequality of the form

$$(r,\theta)\in D \Longleftrightarrow 0\leq f(\theta),\;\;\theta\in[0,2\pi],$$

where $f:[0,2\pi]\to (0,\infty)$ is a semialgebraic function  such that $R(0)=R(2\pi)$. We can extend $f$ by $2\pi$-periodicity to a function  $f:\bR\to [0,\infty)$ whose restriction to any finite interval is semialgebraic.

For any $\phi\in[0,2\pi]$ denote by $\ell_\phi:\bR^2\to \bR$ the linear functiondefined by

$$\ell_\phi (x,y)= x\cos\phi +y\sin\phi.$$

Denote by  $A(\phi)$ the area of the region $D\cap \bigl\{ \ell\phi\geq 0\bigr\}$.    Since $D$ satisfies $\boldsymbol{C}$ we deduce

$$A(\phi)=\frac{1}{2} {\rm area}\;(D),$$

so that

$$A'(\phi)=0,\;\;\forall \phi.$$

Observe that

$$A(\phi+\Delta \phi)-A(\phi) =\int_{\phi+\frac{\pi}{2}}^{\phi+\frac{\pi}{2}+\Delta\phi} \left(\int_0^{f(t)} r dr\right) dt -\int_{\phi+\frac{3\pi}{2}}^{\phi+\frac{3\pi}{2}+\Delta\phi} \left(\int_0^{f(t)} r dr\right) dt.$$

For simplicity we set $\theta=\theta(\phi)=\theta+\frac{\pi}{2}$. We can then rewrite the  above equality  as

$$A(\phi+\Delta \phi)-A(\phi)=\int_\theta^{\theta+\Delta\theta}\left(\int_0^{f(t)} r dr\right) dt -\int_{\theta+\pi}^{\theta+\pi+\Delta\theta} \left(\int_0^{f(t)} r dr\right) dt.$$

Hence

$$0=A'(\phi)= \frac{}{2}\Bigl( f(\theta)^2-f(\theta+\pi)^2\Bigr).$$

Hence $f(\theta)= f(\theta+\pi)$, $\forall \theta$. This shows that   $D$ satisfies the symmetry condition $\boldsymbol{S}$.

$$\ast\ast\ast$$

Here is a simple instance when $\boldsymbol{C}$   does not imply $\boldsymbol{S}$.

Suppose that $D$ is semialgebraic and has the annular description

$$f(\theta)\leq r\leq F(\theta). \tag{1}\label{1}$$.

Using the same notations as above we deduce that

$$0=A'(\phi)= \frac{1}{2} \Bigl(F^2(\theta)-f^2(\theta)\Bigr)- \frac{1}{2} \Bigl(F^2(\theta+\pi)-f^2(\theta+\pi)\Bigr).$$

Thus, the domain (\ref{1}) satisfies $\boldsymbol{C}$ iff the function $G(\theta)=F^2(\theta)-f^2(\theta)$ is $\pi$-periodic.  Note that

$$F(\theta)= \sqrt{f^2(\theta)+ G(\theta)}.$$

If we choose

$$f(\theta)= e^{\sin \theta},\;\; G(\theta)=e^{\cos 2\theta},$$

then we obtain the domain bounded by the two closed curves in the   figure below. This domain obviously violates the  symmetry condition $\boldsymbol{S}$.

## Wednesday, December 12, 2012

### 12.12.12-Once in a century

I had to do this. It the last time this century one can do this, and I could not pass this opportunity to immortalize it.

### Geometry conference in the memory of Jianguo Cao

It's been a bit over a year and a half now since my dear friend Jianguo Cao unexpectedly passed away. I miss him for many reasons. He   was  my gentle, wise and always wellcoming    Riemann geometry guru.   Our department is organizing a conference in his memory  (March 13-17, 2013).  The least I could do  is to spread the word. Unfortunately, I cannot attend. In any case here is a picture of Jianguo from 2004.  He is the leftmost person in the row, I am the  only bearded  guy.

### On conditional expectations.

Assume that $\bsU$ and $\bsV$ are two finite dimensional real vector spaces equipped with the $\si$-algebras of Borel subsets, $\eB_{\bsU}$ and respectively $\eB_{\bsV}$. Consider two random  variables  $X:\Omega\to \bsU$ and $Y:\Omega\to \bsV$ with probability   measures

$$p_X=X_*\bsP,\;\;p_Y=Y_*\bsP.$$

Denote by $p_{X,Y}$ the joint probability measure $\newcommand{\bsE}{\boldsymbol{E}}$

$$p_{X,Y}=(X\oplus Y)_*\bsP.$$

The expectation $\bsE(X|Y)$ is a new $\bsV$-valued  random variable $\omega\mapsto \bsE(X|Y)_\omega$,  but on a different probability space $(\Omega, \eA_Y, \bsP_Y)$ where $\eA_Y=Y^{-1}(\eB_\bsV)$, and $\bsP_Y$ is the restriction of $\bsP$ to $\eA_Y$. The events in $\eA_Y$ all have the form $\{Y\in B\}$, $B\in\eB_{\bsU}$.

This  $\eA_Y$-measurable random variable is defined uniquely by the  equality

$$\int_{Y\in B} E(X|Y)_\omega \bsP_Y(d\omega) = \int_{Y\in B}X(\omega) \bsP(d\omega),\;\;\forall B\in\eB_\bsV.$$

Warning: The truly subtle thing in the above   equality is  the integral in the left-hand-side which is performed with respect to the restricted measure $\bsP_Y$.

If we denote by $I_B$ the indicator function of  $B\in\eB_\bsV$, then we can rewrite the above  equality as

$$\int_\Omega \bsE(X|Y)_\omega I_B(Y(\omega)) \bsP_Y(d\omega)=\int_\Omega X(\omega) I_B(Y(\omega) )\bsP(d\omega).$$

In particular  we deduce that for any  step function $f: \bsV \to \bR$ we have

$$\int_\Omega \bsE(X|Y)_\omega f(Y(\omega)) \bsP_Y(d\omega) =\int_\Omega X(\omega) f(Y(\omega) )\bsP(d\omega).$$

The random variable  $\bsE(X|Y)$  defines  a   $\bsU$-valued  random variable $\bsV\ni y\mapsto \bsE(X|y)\in\bsU$ on the probability space $(\bsV,\eB_\bsV, p_Y)$   where

$$\int_B \bsE(X| y) p_Y(dy)=\int_{(x,y)\in B\times\bsV} x p_{X,Y}(dxdy).$$

Example 1.   Suppose that $A, B\subset \Omega$,  $X=I_A$, $Y=I_B$.   Then  $\eA_Y$ is the $\si$-algebra generated by $B$. The random variable $\bsE(I_A|I_B)$  has a constant value $x_B$ on $B$ and a constant value $x_{\neg B}$ on $\neg B :=\Omega\setminus B$. They are determined by the equality

$$x_B \bsP(B)= \int_B I_A(\omega)\bsP(d\omega) =\bsP(A\cap B)$$

so that

$$x_B=\frac{\bsP(A\cap B)}{\bsP(B)}=\bsP(A|B).$$

Similarly

$$x_{\neg B}= \bsP(A|\neg B).$$

$$\ast\ast\ast$$

Example 2.   Suppose $\bsU=\bsV=\bR$ and that $X$ and  $Y$ are discrete random variables with ranges $R_X$ and $R_Y$.   The random variable  $\bsE(X|Y)$ has a constant value $\bsE(X|y)$ on the set $\{Y=y\}$, $y\in R_Y$. It is determined from the equality

$$\bsE(X|Y=y)p_Y(y)=\int_{Y=y} \bsE(X|Y)_\omega \bsP_Y(\omega) =\int_{Y=y} X(\omega) d\bsP(\omega).$$

Then $\bsE(X|Y)$ can be viewed as a random variable $(R_Y, p_Y)\to \bR$,  $y\mapsto \bsE(X|Y)_y=\bsE(X|Y=y)$, where

$$\bsE(X|Y=y) =\frac{1}{p_Y(y)}\int_{Y=y} X(\omega) d\bsP(\omega).$$

For this reason  one should think of $\bsE(X|Y)$ as a function of $Y$.    From this point of view, a more appropriate notation would be $\bsE_X(Y)$.

The joint probability  distribution  $p_{X,Y}$ can be viewed as a function

$$p_{X,Y}: R_X\times R_Y\to \bR_{\geq 0},\;\;\sum_{(x,y)\in R_X\times R_Y} p_{X,Y}(x,y)= 1.$$

Then

$$\bsE(X|Y=y)= \sum_{x\in R_X} x\frac{P_{X,Y}(x,y)}{p_Y(y)}.$$

We introduce   new $R_X$-valued random variable  $(X|Y=y)$ with probability distribution $p_{X|Y=y}(x)=\frac{p_{X,Y}(x,y)}{p_Y(y)}$.

Then $\bsE(X|Y=y)$ is  the expectation of the  random variable $(X|Y=y)$.

$$\ast\ast\ast$$

Example 3. Suppose that  $\bsU, \bsV$ are equipped with  Euclidean metrics,  $X,Y$ are centered Gaussian random vectors with  covariance forms $A$ and respectively $B$.   Assume  that the covariance pairing between $X$ and $Y$ is $C$ so that the covariance  form of $(X, Y)$ is

$$S=\left[ \begin{array}{cc} A & C\\ C^\dagger & B \end{array} \right].$$

We have $\newcommand{\bsW}{\boldsymbol{W}}$

$$P_{X,Y}(dw) =\underbrace{\frac{1}{\sqrt{\det 2\pi S}} e^{-\frac{1}{2}(S^{-1}w,w)} }_{=:\gamma_S(w)}dw,\;\;w=x+y\in \bsW:=\bsU\oplus \bsV$$

$$P_Y(dy) = \gamma_B(y) dy),\;\;\gamma_B(Y)=\frac{1}{\sqrt{\det 2\pi B}} e^{-\frac{1}{2}(B^{-1}w,w)} .$$

For any  bounded measurable function $f:\bsU\to \bR$ we have

$$\int_{\Omega} f(X(\omega)) \bsP(d\omega)=\int_\bsV \bsE(f(X)|Y) \bsP_Y(d\omega)=\int_\bsV \bsE(f(X)| Y=y) dp_Y(y).$$

We deduce

$$\int_{\bsU\oplus \bsV} f(x) \gamma_S(x,y) dx dy= \int_\bsV \bsE(f(X)| Y=y) \gamma_B(y) dy.$$

Now observe that

$$\int_{\bsV}\left(\int_\bsU f(x) \gamma_S(x,y) dx\right) dy = \int_\bsV \bsE(f(X)| Y=y) \gamma_B(y) dy.$$

This implies that

$$\bsE(f(X)| Y=y) =\frac{1}{\gamma_B(y)} \left(\int_\bsU f(x) \gamma_S(x,y) dx\right).$$

We obtain a probability measure  $p_{X|Y=y}$ on the affine plane $\bsU\times \{y\}$ given by

$$p_{X|Y=y}(dx)= \frac{\gamma_S(x,y)}{\gamma_B(y)} dx.$$

This is a Gaussian    measure on $\bsU$. Its   statistics are described by  the regression formula.  More precisely, its mean is

$$m_{X|Y=y}= Cy,$$

and its covariance form is

$$S_{X|Y=y}= A- CB^{-1}C^\dagger.$$

$$\ast\ast\ast$$

In general, if we think of $p_{X,Y}$ as a density on $\bsU\oplus \bsV$,  of $p_Y$ as a density on $\bsV$ and we denote by $\pi_\bsV$ the natural projection $\bsU\oplus\bsV\to\bsV$, then the conditional probability distribution  $p_{X|Y=y}$ is a a probability density on $\pi^{-1}_\bsV(y)$. More precisely  it is the density $p_{X,Y}/\pi^*_Vp_Y$ defined as in  Section 9.1.1 of my lectures,  especially Proposition 9.1.8 page 350 of the lectures.

## Monday, December 3, 2012

### Degeneration of Gaussian measures

$$\bsV =\bsU\oplus \bsW. \tag{1}\label{1}$$

Suppose that $S_n: \bsU\to\bsU$ and $C_n:\bsW\to \bsW$ are symmetric positive definite  operators such that

$$S_n\to 0,\;\;C_n\to C,\;\;\mbox{as}\;\;n\to \infty$$

where $C$ is a symmetric positive definite operator on $\bsW$.     We set

$$A_n=S_n\oplus C_n :\bsV\to \bsV$$

and we think of $A_n$ as the covariance   matrix  of a  Gaussian measure on $\bsV$ $\newcommand{\bv}{\boldsymbol{v}}$

$$\gamma_{A_n}(|d\bv|)=\frac{1}{\sqrt{\det 2\pi A_n}} e^{-\frac{1}{2}(A_n^{-1} \bv,\bv)} |d\bv|.$$

Suppose that $f:\bsV\to \bR$ is a  locally Lipschitz function,   positively homogeneous of degree $k\geq 1$.

I am interested in the  behavior as $n\to \infty$ of the expectation

$$E_n(f):=\int_{\bsV} f(\bv)\gamma_{A_n}(|d\bv|).$$

Define

$$\bar{f}_n:\bsW\to [0,\infty),\;\; \bar{f}_n(\bw)= \int_{\bsU} f(\bu+\bw) \gamma_{S_n}(|d\bu|),$$

where $d\gamma_{S_n}$ denotes the Gaussian measure  on $\bsU$ with  covariance form $S_n$.  Then

$$E_n(f)=\int_{\bsW} \bar{f}_n(\bw) \gamma_{C_n}(|d\bw|). \tag{2}\label{2}$$

For $\bw\in \bsW$ and $r\in (0,1]$ we set

$$m(\bw, r) := \sup_{|\bu|\leq r}|f(\bw+u)- f(u)|.$$

Note that

$$\exists L>0: m(\bw,r)\leq Lr,\;\;\forall |\bw|= 1\tag{3}\label{3}$$

In general,   we set $\bar{\bw}:=\frac{1}{|\bw|} \bw$. If $|\bu|\leq r$  and we have
$$\bigl|\;f(\bw+\bu)-g(\bw) \;\bigr|= |\bw|^k \left| f\Bigl(\bar{\bw}+\frac{1}{|\bw|} \bu\Bigr) -f(\bar{\bw})\right| \leq L |\bw|^{k-1} r,$$
so that
$$m(w,r) \leq L|\bw|^{k-1} r,\;\;\forall \bw\in\bsW,\;\;r\in (0,1]. \tag{4}\label{4}$$

To proceed further, we need a vector counterpart for the Chebysev inequality.

Lemma 1.  Suppose $S:\bsU\to \bsU$ is a  symmetric, positive definite operator. We set $R:=S^{-\frac{1}{2}}$ and  denote by $\gamma_{S}$ the associated   Gaussian measure. Then for any $c,\ell>0$  we have $\newcommand{\bsi}{\boldsymbol{\sigma}}$

$$\int_{ |R \bu|\geq c} |\bu|^\ell d\gamma_S(|\bu|) \leq \sqrt{2^{\ell+m-\frac{3}{2}} \Gamma\Bigl(\; \ell+m-\frac{1}{2}\;\Bigr)} \frac{\bsi_{m-1}}{(2\pi)^{\frac{m}{2}}} \Vert S\Vert^{\frac{\ell}{2}}c^{-\frac{1}{2}}e^{-\frac{c^2}{4}} , \tag{5}\label{5}$$

where $m=\dim\bsU$ and  and $\bsi_N$ denote the area of the $N$-dimensional unit sphere.

Proof.   We make the change in variables $\newcommand{\bx}{\boldsymbol{x}}$  $\bx:=R\bu$ and we  deduce
$$\int_{ |R \bu|\geq c} |\bu|^\ell d\gamma_S(|\bu|)\leq \frac{1}{(2\pi)^{\frac{m}{2}}} \int_{|\bx|\geq c} |S^{\frac{1}{2}} \bx|^\ell e^{-\frac{1}{2}|\bx|^2} |d\bx|$$

$$\leq \frac{\Vert|S\Vert^{\frac{\ell}{2}}}{(2\pi)^{\frac{m}{2}}} \int_{|\bx|\geq c} |\bx|^\ell e^{-\frac{1}{2}|\bx|^2} |d\bx|=\frac{\bsi_{m-1}\Vert S\Vert^{\frac{\ell}{2}}}{(2\pi)^{\frac{m}{2}}}\int_{t>c} t^{\ell+m-1} e^{-\frac{1}{2} t^2} dt$$
$$\leq \frac{\bsi_{m-1}\Vert S\Vert^{\frac{\ell}{2}}}{(2\pi)^{\frac{m}{2}}} \left(\int_{t>c} e^{-\frac{1}{2} t^2} dt\right)^{\frac{1}{2}}\left(\int_{t>0} t^{2\ell+2m-2} e^{-\frac{1}{2} t^2} dt\right)^{\frac{1}{2}}$$
Now observe that we have
$$\int_{t>c} e^{-\frac{1}{2} t^2} dt \leq \frac{1}{c} e^{-\frac{c^2}{2}},$$
and  using the change of variables $s=\frac{t^2}{2}$ we  deduce

$$\int_{t>0} t^{2\ell+2m-2} e^{-\frac{1}{2} t^2} dt =2^{\ell+m-\frac{3}{2}}\int_0^\infty s^{\ell+m-\frac{1}{2}-1} e^{-s} ds= 2^{\ell+m-\frac{3}{2}} \Gamma( \ell+m-\frac{1}{2}).$$

This proves the lemma. q.e.d

We now want to compare $\bar{f}_n(\bw)$ and $f(\bw)$ for $\bw\in\bsW$.  We plan to use Lemma  1.   Set $R_n:=S_n^{-\frac{1}{2}}$ and $m:=\dim\bsU$.   Observe that

$$|\bu\|= |S_n^{\frac{1}{2}}R_n\bu|\leq \Vert S_n^{\frac{1}{2}}\Vert\cdot |R_n\bu|.$$

For simplicity   set $s_n:= \Vert S_n^{\frac{1}{2}}\Vert$.   Choose a sequence of positive numbers  $c_n$ such that $c_n\to\infty$ and  $s_n c_n\to 0$.  Later  we will add several requirements to this sequence.

$$\bigl|\;\bar{f}_n(\bw)-f(\bw)\;\bigr|=\left| \int_{\bsU} (\; f(\bw+\bu)- f(\bw)\; ) \gamma_{S_n}(|d\bu|)\right|$$
$$\leq \left| \int_{|R_n\bu|\leq c_n} (\; f(\bw+\bu)- f(\bw)\; ) \gamma_{S_n}(|d\bu|)\right|+\left|\int_{|R_n\bu|\geq c_n} (\; f(\bw+\bu)- f(\bw)\; ) \gamma_{S_n}(|d\bu|)\right|$$
$$\stackrel{(\ref{4})}{\leq} L|\bw|^{k-1}s_n c_n +C \int_{|R_n\bu|\geq c_n}(|\bw|^k+|\bu|^k) \gamma_{S_n}(|d\bu|)$$

$$\stackrel{(\ref{5})}{\leq} L|\bw|^{k-1}s_n c_n + Z(k, m)c_n^{-\frac{1}{2}} e^{-\frac{c_n^2}{4}}(1+s_n^k),$$
where $Z(k,m)$ is a constant that depends only on   $k$ and $m$.

We deduce that there exists a constant $C>0$ independent of   $n,w$  such that  for any sequence $c_n\to \infty$ such that $s_nc_n\to 0$, $s_n:=\Vert S_n\Vert^{\frac{1}{2}}$ we have

$$\bigl|\;\bar{f}_n(\bw)-f(\bw)\;\bigr| \leq C\bigl(\; |\bw|^{k-1}s_nc_n + e^{-\frac{c_n^2}{4}}\;\bigr). \tag{6}\label{6}$$
We deduce that

$$\Bigl|\; E_n(f) -\int_{\bsW} f(\bw) \gamma_{C_n}(|d\bw|)\;\Bigr| \leq C\left(s_nc_n\int_{\bsW} |\bw|^{k-1} \gamma_{C_n}(|d\bw|) + e^{-\frac{c_n^2}{4}}\;\right).\tag{7}\label{7}$$

Finally let us estimate

$$D_n:=\int_{\bsW} f(\bw) \gamma_{C_n}(|d\bw|)-\int_{\bsW} f(\bw) d\gamma_{C}(|d\bw|).$$

We have  $\newcommand{\one}{\boldsymbol{1}}$
$$D_n= \int_{\bsW} \left( f\bigl( C_n^{\frac{1}{2}}\bw\;\bigr)-f\bigl( C^{\frac{1}{2}}\bw\;\bigr) \;\right)\gamma_{\one}(|\bw|)$$
and we conclude that
$$\left|\; \int_{\bsW} f(\bw) d\gamma_{C_n}(|d\bw|)-\int_{\bsW} f(\bw) d\gamma_{C}(|d\bw|)\;\right| \leq L \Bigl\Vert \;C_n^{\frac{1}{2}}-C^\frac{1}{2}\;\Bigr\Vert \int_{\bsW}|\bw|^k \gamma_{\one}(|d\bw|). \tag{8}\label{8}$$

In (\ref{7}) we let $c_n:=s_n^{-\ve}$. If we denote by $A_\infty$ the limit of the covariance matrices $A_n$, $A=\lim_{n\to\infty} A_n =0\oplus C$, then we deduce from the above computations that for any $\ve>0$ there exists a  constant $C_\ve>0$ such that
$$\left|\; \int_{\bsV} f(\bv) \gamma_{A_n} (|d\bw|) -\int_{\bsV} f(\bv) \gamma_{A_\infty} (|d\bw|) \;\right|\leq C_\ve \left(s_n^{1-\ve}+ \Bigl\Vert \;C_n^{\frac{1}{2}}-C^\frac{1}{2}\;\Bigr\Vert\right)\leq C_\ve \Bigl\Vert A_n^{\frac{1}{2}}-A_\infty^{\frac{1}{2}}\Bigr\Vert^{1-\ve}.\tag{9}\label{9}$$
This can be generalized a bit. Suppose that $T_n:\bsU\to \bsU$ is a sequence of orthogonal operators such that  $T_n\to \one_{\bsU}$

Using (\ref{7})  we deduce

$$\left|\;\int_{\bsV} T^*_nf(\bv) \gamma_{A_n}(|d\bv|)-\int_{\bsV} f(\bv) \gamma_{A_n}(|d\bv|)\right|= \left| \int_{\bsV} f(T_n A_n^{\frac{1}{2}}\bx)- f( A_n^{\frac{1}{2}}\bx)\gamma_{\one}(|d\bx|) \right| \leq L \Bigl\Vert A_n^{\frac{1}{2}}\Bigr\Vert \Vert T_n-\one\Vert.$$
Observe that
$$\int_{\bsV} T^*_nf(\bv) \gamma_{A_n}(|d\bv|)=\int_{\bsV} f(\bv) \gamma_{B_n}(|d\bv|),$$
where
$$B_n= T_nA_nT_n^*.$$

Suppose that we are in the fortunate case when $f|_{\bsW}=0$.    Then

$$\int_{\bsW} f(\bw) d\gamma_{C_n}(|d\bw|)=\int_{\bsW} f(\bw) d\gamma_{C}(|d\bw|)=0$$

and (\ref{9})  can be improved to

$$\left|\; \int_{\bsV} f(\bv) \gamma_{A_n} (|d\bw|)\right|\leq C_\ve s_n^{1-\ve}.$$

### On the 11/8-conjecture

Stefan Bauer has just posted a proof for the 11/8- conjecture for simply connected $4$-manifolds

## Tuesday, November 27, 2012

### FRANK FEST: Workshop on High Dimensional Topology @ ND 2012

This is a conference in honor of Frank Connolly who is retiring this semester.

Workshop on High Dimensional Topology @ ND 2012

## Wednesday, November 21, 2012

### Sharp nondegeneracy estimates for a family of random Fourier series

Suppose that $w\in \eS(\bR)$ is an even, nonnegative  Schwartz  function.  Assume that $w\not\equiv 0$. $\newcommand{\hw}{\widehat{w}}$ We denote by $\hw(t)$   its Fourier transform

$$\hw(t)=\int_{\bR}e^{-\ii t x} w(x) dx.$$

For $n\in \bZ$ we  set $\newcommand{\be}{\boldsymbol{e}}$

$$\be_n(\theta) :=\frac{1}{\sqrt{\pi}}\begin{cases} \frac{1}{\sqrt{2}}, & n=0,\\ \sin n \theta , & n<0,\\ \cos n\theta , & n>0. \end{cases}$$

Observe that  the collection $\lbrace \be_n(\theta)\rbrace_{n\in\bZ}$ is an orthonormal  basis of $L^2(\bR/2\pi\bZ)$.   $\newcommand{\bT}{\mathbb{T}}$      For any positive integer $N$ we  denote by $\bT^N$ the $N$-dimensional torus

$$\bT^N:= (\bR/2\pi\bZ)^N$$.

Consider the random Fourier series

$$f_\ve(\theta)=\sum_{n\in \bZ} \sqrt{w(\ve n) } c_n \be_n(\theta),$$

$$\eE^{\ve}:\bT^1\times \bT^1\to \bR,\;\;\eE^\ve (\theta,\vfi)=\bsE\bigl( f_\ve(\theta)\cdot f_\ve(\vfi)) =\sum_{n\in \bZ} w(\ve n) \be_n(\theta)\be_n(\vfi)$$

$$=\frac{1}{2\pi} w(0)+\frac{1}{\pi}\sum_{n>0}w(\ve n)\cos n(\theta-\vfi)=\frac{1}{2\pi}\sum_{n\in\bZ}w(\ve n) e^{\ii n(\theta-\vfi)}=W_\ve(\theta-\vfi). \tag{1}\label{1}$$

Poisson formula.   For any $\phi\in \eS(\bR)$  and any $c\in\bR\setminus 0$ we have

$$\frac{2\pi}{c}\sum_{n\in\bZ} \phi\Bigl(\frac{2\pi n}{c}\Bigr)= \sum_{\nu\in\bZ} \widehat{\phi}(n c).$$

Suppose $\phi\in\eS(\bR)$ and $c$ are such that

$$\phi\Bigl(\;\frac{2\pi n}{c}\;\Bigr)= w(\ve n) e^{\ii n(\theta-\vfi)} .$$

If we formally replace $n =\frac{c x}{2\pi}$ we deduce from the above equality that

$$\phi(x)= w\Bigl(\frac{\ve c x}{2\pi}\Bigr) e^{\ii\frac{c(\theta-\vfi)x}{2\pi}}=w(ax)e^{\ii b x}, \;\; a:=\frac{\ve c}{2\pi},\;\;b :=\frac{c(\theta-\vfi)}{2\pi}.$$

Then

$$\widehat{\phi}(t) =\int_{\bR} e^{-\ii tx} w(ax) e^{\ii bx} dx = \frac{1}{a}\int_{\bR} e^{-\ii \frac{t-b}{a}x} w(y) dy = \frac{1}{a}\hw\Bigl( \frac{t-b}{a}\Bigr).$$

We now set $c:=2\pi$ so that $a=\ve$, $b=(\theta-\vfi)$. Using The Poisson formula in (\ref{1}) we deduce

$$W_\ve(\theta-\vfi)=\eE^\ve(\theta,\vfi) =\frac{1}{2\pi\ve} \sum_{n\in\bZ} \hw\Bigl(\frac{2\pi n-(\theta-\vfi)}{\ve}\Bigr) . \tag{2}\label{2}$$

Now consider the  random function

$$F_\ve:\bT^N\to \bR,\;\; F_\ve(\vec{\theta}) = \sum_{j=1}^n f_\ve(\theta_j).$$

The  correlation kernel of this  random function is

$$\eE_N^\ve(\vec{\theta},\vec{\vfi}) =\sum_{1\leq j,k\leq N} \eE^\ve(\theta_j-\vfi_k).$$

The differential of $F_\ve$ at a point $\vec{t}\in\bT^N$ is a Gaussian  random vector with covariance matrix $\newcommand{\pa}{\partial}$

$$S^\ve(\vec{t})= \Bigl( S^\ve_{jk}(\vec{t})\;\Bigr)_{1\leq j,k\leq N},\;\; S^\ve_{jk}(\vec{t})= \frac{\pa^2}{\pa\theta_j\pa \vfi_k} \eE_N^\ve\bigl(\;\vec{\theta},\vec{\vfi}\;\bigr)|_{\vec{\theta}=\vec{\vfi}=\vec{t}}=-W_\ve''(t_j-t_k)=\frac{1}{2\pi}\sum_{n\in\bZ} n^2w(\ve n) e^{\ii n(t_j-t_k)}\tag{3}\label{3}.$$

Definition. We say that $\vec{t}\in\bT^n$ is  nondegenerate if $t_j-t_k\in\bR\setminus 2\pi\bZ$, $\forall j\neq k$. We denote by $\bT^N_*$ the collection of nondgenerate  points in $\bT^N$.

$$\ast\ast\ast$$

We have the following result similar to the one in our   previous post.

Proposition 1.  There exists $\ve_0=\ve_0(w,N)>0$ such that if $\ve \in (0,\ve_0)$ and  $\vec{t}\in \bT^N$ is nondegenerate, then the  matrix $S^\ve(\vec{t})$ is positive  definite.

Proof.     Set

$$Z_\ve:=\bigl\{ n\in\bZ;\;\;w(\ve n)\neq 0\;\bigr\}.$$

Consider the space $H_\ve$ consisting of functions $\newcommand{\bC}{\mathbb{C}}$

$$u: Z_\ve \to \bC,\;\;\sum_{n\in Z_\ve} |u(n)|^2 n^2 w(\ve n) <\infty.$$

This is a separable  Hilbert space with inner product

$$(u,v)_\ve= \frac{1}{2\pi} \sum_{n\in\bZ} u(n)\cdot \overline{v(n)}\; n^2w(\ve n).$$

We denote by $\Vert-\Vert_\ve$ the associated norm.

For $t\in\bT^1$ consider the truncated character $\chi^\ve_t:Z_\ve\to \bT^1$, $\chi^\ve_t(n)=e^{\ii tn}$.  For $\vec{z}\in \bC^N$ $\newcommand{\vez}{{\vec{z}}}$  and $\vec{t}\in \bT^N$ consider $T_{\vez,\vec{t}}\in H_\ve$

$$T_{\vez,\vec{t}}(n)=\sum_{j=1}^n z_j \chi^\ve _{t_j}(n)=\sum_{j=1}^N z_j e^{\ii t_j n},\;\;n\in Z_\ve.$$

From the equality (\ref{3}) we deduce that

$$\sum_{j,k=1}^n S^\ve_{jk}(\vec{t}) z_j\bar{z}_k = \Vert T_{\vez,\vec{t}}\Vert_\ve^2.$$

Thus, the matrix $S^\ve(\vec{t})$ has a kernel if and only if the truncated characters $\chi^\ve_{t_1},\dotsc, \chi^\ve_{t_N}$ are linearly dependent.  We show that this is not possible if  $\vec{t}$ is  nondegenerate and  $\ve$ is sufficiently small.

Fix  $\ve_0=\ve_0(N,w)$ such that  if $\ve<\ve_0$ the support   of $x\mapsto w(\ve x)$ contains a long  interval of the form $[\nu_\ve, \nu_\ve+N-1]$, for some integer $\nu_\ve>0$. (Recall that $w$ is even.) In other words $\nu_\ve,\nu_\ve+1,\cdots,\nu_\ve+N-1\in Z_\ve$.

Let $\ve\in (0,\ve_0)$  and suppose that  $\vez\in\bC^N\setminus 0$ and $\vec{t}\in\bT^N$ are such that  such that  $T_{\vez,\vec{t}}=0$.   Thus

$$(T_\vez, u)_\ve =0,\;\;\forall u\in H_\ve$$

For any $m\in Z_\ve$ consider the Dirac function $\delta_m: Z_\ve\to \bC$,  $\delta_m(n)=\delta_{mn}=$ the Kronecker  delta.

We deduce that for any $m=\nu_\ve,\nu_\ve+1,\dotsc, \nu_\ve+N-1$ we have

$$0 = (T_\vez, \delta_m)_\ve=m^2w(\ve m) \sum_{j=1}^N z_j e^{\theta_j m} .$$

This can happen if and only if

$$0= \det\left[ \begin{array}{cccc} e^{\ii \nu_\ve t_1} & e^{\ii \nu_\ve t_2} & \cdots & e^{\nu_\ve t_N}\\ e^{\ii(\nu_\ve+1)t_1} & e^{\ii(\nu_\ve+1)t_2} & \cdots & e^{\ii (\nu_\ve+1) t_N}\\ \vdots & \vdots &\vdots &\vdots\\ e^{\ii(\nu_\ve+N-1)t_1} & e^{\ii(\nu_\ve+N-1)t_2} &\cdots & e^{\ii(\nu_\ve+N-1)t_N} \end{array} \right] = e^{\ii\nu_\ve(t_1+\cdots +t_N)} \prod_{j<k} \Bigl( e^{\ii t_k}-e^{\ii t_j}\Bigr) .$$

This shows that  $S^\ve(\vec{t})$ has a kernel if and only of $\vec{t}$ is degenerate.    Q.E.D.

Remark.    Here is an alternate proof of  Proposition 1 that yields a bit more. The above proof shows that

$$S^\ve_{jk}(\vec{t})= (\chi_{t_j},\chi_{t_k})_\ve.$$

Suppose for simplicity that $0\in Z_\ve$, i.e.,    $w(0)>0$. Then for $\ve>0$ sufficiently small we have $1,\dotsc, N\in Z_\ve$.  Observe that

$$(\delta_j,\delta_k)_\ve= \frac{k^2w(\ve k)}{2\pi}\delta_{jk},\;\;j,k=1,\dotsc, N.$$

We have a Cauchy-Schwartz inequality

$$\Bigl|\; \bigl( \chi_{t_1}\wedge\cdots \chi_{t_n}, \delta_1\wedge \cdots \wedge \delta_N\;\bigr)_\ve\;\Bigr| \leq \bigl|\; \chi_{t_1}\wedge\cdots \wedge\chi_{t_n}\;\bigr|_\ve\cdot \bigl|\;\delta_1\wedge \cdots \wedge \delta_N\;\bigr|_\ve.$$

This translates to

$$\Bigl| \det\Bigl(\; (\chi_{t_j},\delta_k)_\ve\;\Bigr)_{1\leq i,j\leq N}\;\Bigr| \leq \sqrt{\det\Bigr( \; (\chi_{t_j},\chi_{t_k})_\ve\;\Bigr)_{1\leq i,j\leq N} } \cdot \sqrt{\det\Bigr( \; (\delta_j,\delta_k)_\ve\;\Bigr)_{1\leq i,j\leq N} },$$
or, equivalently

$$\prod_{j<k} \Bigl| e^{\ii t_j}-e^{\ii t_k}\Bigr|^2 \leq \frac{1}{(2\pi)^N}\Bigl(\prod_{j=1}^N j^2w(\ve j)\Bigr) \det S^\ve(\vec{t}). \tag{4} \label{4}$$

$$\ast\ast\ast$$

The basic question that interests me is the following: what  happens to $S^\ve(\vec{t})$ as $\ve\to 0$, and $\vec{t}$ is nondegenerate.

Observe that (\ref{2}) implies that

$$W_\ve''(t)=\frac{1}{2\pi\ve^3}\sum_{n\in\bZ}\hw''\Bigl(\frac{2\pi n-t}{\ve}\Bigr).$$

We make the change in variables $t=\ve \tau$, we set

$$C^\ve(\vec{\tau}) := S^\ve(\ve\vec{\tau})$$

and we deduce

$$C^\ve_{jk}(\vec{\tau})=\frac{1}{2\pi}\sum_{n\in\bZ}n^2w(\ve n) e^{\ii\ve n(\tau_j-\tau_k)} =-\frac{1}{2\pi\ve^3}\sum_{n\in\bZ}\hw''\Bigl(\tau_j-\tau_k-\frac{2\pi n}{\ve}\Bigr),\;\;0\leq \tau_j <\frac{2\pi}{\ve},\;\;j=1,\dotsc, N. \tag{5}\label{5}$$

$$\bde_k=\bde_k^\ve :=\frac{\sqrt{2\pi}}{k\sqrt{w(\ve k)}}\delta_k.$$

By construction,  the collection $\bde_1,\dotsc,\bde_N$ is an $(-,-)_\ve$-orthonormal basis of $\bsD_N$.

The  $(-,-)_\ve$-orthogonal projection $P_\ve=P_\ve(\vec{t}): X(\vec{t})\to \bsD_n$ is given by

$$P_\ve \chi_{t_j} =\sum_{k=1}^n (\chi_{t_j},\bde_k)_\ve \bde_k = \sum_{k=1}^N e^{\ii kt_j} \delta_k.$$

With respect to the natural bases $\chi_{t_1},\dotsc,\chi_{t_N}$ of $X(\vec{t})$ and $\delta_1,\dotsc, \delta_N$ of $\bsD_N$ the    projection is therefore given by the  Vandermonde matrix

$$V= V(\vec{t}),\;\; V_{kj}= e^{\ii k t_j},\;\; V=\left[\begin{array}{cccc} e^{\ii t_1} & e^{\ii t_2} &\cdots & e^{\ii t_N}\\ e^{2\ii t_1} & e^{2\ii t_2} & \cdots & e^{2\ii t_N}\\ \vdots &\vdots &\vdots &\vdots\\ e^{N\ii t_1} & e^{N\ii t_2} &\cdots & e^{N\ii t_N} \end{array} \right].$$

## Wednesday, November 14, 2012

### On a family of symmetric matrices

$$\widehat{w}(\theta)=\int_{\bR} e^{-\ii x\theta}w(\theta) dx.$$

For any $\vec{\theta}\in\bR^n$ we form the  complex Hermitian  $n\times n$ matrix

$$A_w(\vec{\theta})= \bigl(\; a_{ij}(\vec{\theta})\;)_{1\leq i,j\leq n},\;\; a_{ij}(\vec{\theta})=\widehat{w}(\theta_i-\theta_j) .$$

Observe that  for any $\vec{z}\in\mathbb{C}^n$  we have $\newcommand{\bC}{\mathbb{C}}$

$$\bigl(\; A_w(\vec{\theta})\vec{z},\vec{z}\;\bigr)=\sum_{i,j} \widehat{w}(\theta_i-\theta_j) z_i\bar{z}_j =\int_{\bR} | T_{\vec{z}}(x,\vec{\theta})|^2 w(x) dx,$$

where  $T_{\vec{z}}( x)$ is  is the trigonometric polynomial $\newcommand{\vez}{\vec{z}}$

$$T_{\vez}(x,\vec{\theta})= \sum_j z_j e^{\ii \theta_j x}.$$

We denote by $(-,-)_w$ the inner product

$$(f,g)_w=\int_{\bR} f(x) \bar{g(x)} w(x) dx,\;\;f,g:\bR\to \bC.$$

We see that $A_w(\vec{\theta})$  is the Gramm-Schmidt matrix

$$a_{ij}(\vec{\theta})= (E_{\theta_i}, E_{\theta_j})_w,\;\; E_\theta(x)=e^{\ii\theta x}.$$

We see that $\sqrt{\;\det A_w(\vec{\theta})\;}$ is equal to the $n$-dimensional volume   of the parallelepiped   $P(\vec{\theta})=L^2(\bR, wdx)$ spanned  by the  functions $E_{\theta_1},\dotsc, E_{\theta_n}$. We observe  that if these exponentials are  linearly  dependent,  then this volume is  zero.  Here is a first elementary result.

Lemma  1.   The exponentials  $E_{\theta_1},\dotsc, E_{\theta_n}$ are linearly dependent  (over $\bC$) if and only if $\theta_j=\theta_k$ for  some $j\neq k$.

Proof.    Suppose that

$$\sum_{j=1}^n z_j E_{\theta_j}(x)=0,\;\;\forall x\in \bR.$$

Then for any $f\in \eS(\bR)$ we have

$$\sum_{j=1}^n z_j E_{\theta_j}(x)f(x)=0,\;\;\forall x\in \bR.$$

By taking the Fourier Transform of the last equality we deduce

$$\sum_{j=1}^n z_j \widehat{f}(\theta-\theta_j) =0. \label{1}\tag{1}$$

If we now choose $\newcommand{\ve}{{\varepsilon}}$ a  family $f_\ve(x)\in\eS(\bR)$ such that, as $\ve\searrow 0$,  $\widehat{f}_\ve(\theta)\to\delta(\theta)=$ the Dirac  delta function concentrated at $0$, we  deduce  from (\ref{1})  that

$$\sum_{j=1}^n z_j\delta(\theta-\theta_j)=0. \tag{2}\label{2}$$

Clearly this can happen if and only if  $\theta_j=\theta_k$ for  some $j\neq k$.  q.e.d.

If we set

$$\Delta(\vec{\theta}) :=\prod_{1\leq j<k\leq n} (\theta_k-\theta_j),$$

then we deduce from the above lemma that

$$\det A_w(\vec{\theta})= 0 \Leftrightarrow \Delta(\vec{\theta})=0.$$

A more precise statement is true.

Theorem 2.  For any integrable weight $w:\bR\to [0,\infty)$  such that $\int_{\bR} w(x) dx >0$ there  exists a  constant $C=C(w)>0$ such that for any $\theta_1,\dotsc, \theta_n\in [-1,1]$ we have

$$\frac{1}{C}|\Delta(\vec{\theta})|^2 \leq \det A_w(\vec{\theta}). \tag{E}\label{E}$$

Proof.          We regard $A_w(\vec{\theta})$ as a hermitian operator

$$A_w(\vec{\theta}):\bC^n\to \bC^n.$$

We denote by $\lambda_1(\vec{\theta})\leq \cdots \leq \lambda_n(\vec{\theta})$ its eigenvalues so that

$$\det A_w(\vec{\theta})=\prod_{j=1}^n \lambda_j(\vec{\theta}) \tag{Det}\label{D}.$$

$$\vec{z}\in \ker A(\vec{\theta}) \Lra \sum_{j=1}^n z_j E_{\theta_j}(x) =0,\;\;\forall x\in{\rm supp}\; w \Lra \sum_{j=1}^n z_j E_{\theta_j}(x) =0,\;\;\forall x\in\bR. \tag{Ker}\label{K}$$

We want to give a  more precise description    of $\ker A_w(\vec{\theta})$.    Set

$$I_n:=\{1,\dotsc, n\},\;\; \Phi_{\vec{\theta}}=\{ \theta_1,\dotsc,\theta_n\}\subset \bR.$$

We want to emphasize that $\Phi{\vec{\theta}}$ is not a multi-set so that $\#\Phi(\vec{\theta})\leq n.$  $\newcommand{\vet}{{\vec{\theta}}}$.

Example 3.  For example  with $n=6$ and $\vet=(1,2,3,2,2,4)$ we have

$$\Phi_\vet=\Phi_{(1,2,3,2,2,4)}=\{1,2,3,4\}.$$

For $\newcommand{\vfi}{{\varphi}}$ $\vfi\in\Phi_\vet$ we set

$$J_\vfi=\bigl\{ j\in I_n;\;\; \theta_j=\vfi\;\bigr\}.$$

In the example above for $\vet=(1,2,3,2,2,4)$ and $\vfi=2$ we have $J_\vfi=\{2,4,5\}$.   $\newcommand{\vez}{\vec{z}}$  For $J\subset I_n$ we set

$$S_J:\bC^n\to \bC,\;\;S_J(\vez)=\sum_{j\in J} z_.$$

In particular, for  any $\vfi\in\Phi_\vet$ we define

$$S_\vfi:\bC^n\to \bC,\;\; S_{\vfi}(\vec{z})=S_{J_\vfi}(\vez)=\sum_{j\in J_\vfi} z_j.$$

We deduce

$$\sum_{j\in I_n} z_jE_{\theta_j}=\sum_{\vfi\in \Phi_\vet} S_\vfi(\vec{z}) E_\vfi.$$

Using (\ref{K}) we deduce

$$\vez\in\ker A(\vet)\Lra \sum_{\vfi\in \Phi_\vet} S_\vfi(\vec{z}) E_\vfi\Lra S_\vfi(\vez)=0,\;\;\forall \vfi\in \Phi_\vet . \tag{3}\label{3}$$

In particular we deduce

$$\dim \ker A(\vet)=n-\#\Phi_\vet.$$

Step 1.   Assume  that $w$ has compact support so that $\widehat{w}(\theta)$  is real analytic over $\bR$.   We will show  that we have the two-sided estimate

$$\frac{1}{C}|\Delta(\vec{\theta})|^2 \leq \det A_w(\vec{\theta}) \leq C |\Delta(\vec{\theta})|^2. \tag{E_*}\label{Es}$$

In this case  $\det A_w(\vet)$ is real analytic and symmetric in the variables $\theta_1,\dotsc, \theta_n$  and vanishes   if and only if $\theta_j=\theta_k$ for some $j=k$.  Thus $\det A_w(\vet)$ has a  Taylor series expansion (near $\vet=0$)

$$\det A_w(\vet)= \sum_{\ell\geq 0} P_\ell(\vet),$$

where $P_\ell(\vet)$ is a  symmetric polynomial  in $\vet$ that vanishes   when $\theta_j=\theta_k$ for some $j\neq k$. Symmetric  polynomials of this type  have the form,

$$\Delta(\vet)^{2N} \cdot Q(\vet)$$

where $N$ is some positive integer  and $Q$ is a symmetric polynomial.  We deduce from  the \Lojasewicz inequality  for  subanalytic functions that there exists  $C=C(w)>0$, a positive integer $N$  and a rational number and $r>0$ such that

$$\frac{1}{C} |\Delta(\vet)|^{r}\leq \det A_w(\vet) \leq C \Delta(\vet)^{2N},\;\;\forall |\vet|\leq 2\pi. \tag{4} \label{4}$$

We want to show  that in (\ref{4})  we have $2N=r=2$.   We argue by contradiction, namely we assume that $r\neq 2$ or $N\neq 1$.      Let

$$\vet(t)= (0, t, \theta_3, \dotsc, \theta_n), \;\; 0\leq |t| < \theta_3<\cdots < \theta_n.$$

Set $A_w(t)=A_w\bigl(\,\vet(t)\;\bigr)$.   Denote its eigenvalues by

$$0\leq \lambda_1(t)\leq \lambda_2(t)\cdots \leq \lambda_n(t).$$

The eigenvalues are so arranged so that the functions $\lambda_k(t)$ are real analytic for $t$ in a neighborhood of $0$. We deduce from (\ref{3}) that $\ker A_w(0)$ is one dimensional so that $\lambda_1(0) =0$,  $\lambda_k(0)>0$, $\forall k>1$. Hence

$$\det A_w(t) \sim \lambda_1(t) \prod_{k=2}^n \lambda_k(0)\;\;\mbox{as t\searrow 0}. \tag{5}\label{5}$$

On the other hand

$$\Delta(\vet(t))^2 \sim Zt^2 \;\;\mbox{as}\;\; t\searrow 0$$

for some positive constant $Z$.  Using this estimate in (\ref{4}) we deduce $r=2N$.  On the other hand, using the above estimate in (\ref{5}) we deduce

$$\lambda_1(t) \sim Z_1 t^{2N} \;\;\mbox{as}\;\;t\searrow 0. \tag{6}\label{6},$$

for another positive constant $Z_1$.

The   kernel of $A_w(0)$ is spanned by the  unit vector

$$\vez(0)= (\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0,\dotsc 0).$$

We can find a real analytic family of vectors $t\mapsto \vec{z}(t)$ satisfying

$$|\vez(t)|=1,\;\; A_w(t) \vez(t)=\lambda_1(t)\vez(t),\;\;\lim_{t\to 0}\vez(t)=\vez(0).$$

In particular, we deduce

$$\dot{A}_w(0)\vez(0)+A_w(0)\dot{\vez}(0)=\dot{\lambda}_1(0)\vez(0)+\lambda_1(0)\dot{\vez}(0)=0.$$

A simple computation  shows that $\dot{A}_w(0) \vez(0)=0$ so we deduce  $A_w(0)\dot{\vez}(0)=0$.  This shows that

$$\dot{z}_1(0)+\dot{z}_2(0)=0,\;\;\dot{z}_k(0)=0,\;\;\forall k>2.$$

$$\lambda_1(t)= (A_w(t) \vez(t),\vez(t))= \int_{\bR} \Bigl| \;\underbrace{\sum_{j=1}^n z_j(t) e^{\theta_j(t) x}}_{=:f_t(x)}\;\Bigr|^2 w(x) dx.$$

Observe that

$$f_t(x):= \sum_{j=1}^n z_j(t) e^{\theta_j(t) x}= \frac{1}{\sqrt{2}}(1-e^{\ii t x}) +\sum_{j=1}^k \ve_j(t) e^{\ii\theta_j(t) x},\;\;\ve_j(t)=z_j(t)-z_j(0).$$

We   deduce that

$$\lim_{t\to 0} \frac{1}{t}f_t(x) = -\frac{\ii x}{\sqrt{2}} + \sum_{k=1}^n \dot{z}_k(0)= -\frac{\ii x}{\sqrt{2}}\tag{7}\label{7}$$

uniformly  for  $x$   on compacts. Since  $w$ has compact support  we deduce that (\ref{7}) holds for uniformly for $x$ in the support of $w$.  We deduce that

$$\lambda_1(t)\sim \frac{1}{2}\;\underbrace{\left(\int_{\bR} x^2 w(x)dx \right)}_{=\widehat{w}''(0)}\;t^2\;\;\mbox{as t\to 0}.$$

Using the last equality in (\ref{6}) we obtain $2N=2$ which proves  (\ref{Es}) .

Step 2.    We will show that if (\ref{E}) holds for $w_0$ and $w_1(x) \geq w_0(x)$,  $\forall x$,  then (\ref{E}) holds for  $w_1$ as well.       For any weight $w$  and any $\vet$ such that the $\Delta(\vet)\neq 0$ consider the ellipsoid

$$\Sigma_w:=\bigl\{\vez\in\bC^n;\;\; (A_w\vez,\vez)\leq 1\bigr\}.$$

Then

$${\rm vol}\, \bigl(\;\Sigma_w(\vet)\;\bigr)=\frac{\pi^n}{n!\det A_w(\vet)}.$$

Observe that if $w_0\leq w_1$ then $\Sigma_{w_0}(\vet)\subset \Sigma_{w_1}(\vet)$ and we  deduce

$$\det A_{w_0}(\vet) \leq \det A_{w_1}(\vet).$$

This proves our claim.

Step 3. We show that (\ref{E}) holds for any integrable weight.   At  least one of the level sets $\{w\geq \ve\}$, $\ve>0$ is nonempty.  We can find a compact set of nonzero measure  $K \subset \{w\geq \ve \}$. Now define $w_0=I_{K}$. Clearly $I_K\leq w$.   From  Step 1 we know that (\ref{E}) holds for $w_0$. Invoking Step 2 we deduce that (\ref{E}) holds for  $w$.  Q.E.D.

### The half-life of a theorem, or Arnold's principle at work - MathOverflow

This is a very interesting thread.

The half-life of a theorem, or Arnold's principle at work - MathOverflow

## Tuesday, November 13, 2012

### Supersymmetry in doubt

About a dozen of years ago, at a Great Lakes Conference dinner at Northwestern  I asked  Witten  what parts of high energy physics he thinks will be  confirmed  experimentally in our lifetime.  Super-symmetry was one of the first things he mentioned. Now comes  this news from the Large Hadron Collider casting some doubt on the supersymmetry premise.

BBC News - Popular physics theory running out of hiding places

A word of caution though.   About a year ago people thought that the Large Hadron Collider detected a particle traveling faster than the speed of light.

News - Popular physics theory running out of hiding places

### Influence By Degree, Episode 1

An interesting BBC documentary on how private donors  influence academic life.

BBC World Service - The Documentary, Influence By Degree, Episode 1

## Friday, November 9, 2012

### Tim Gowers recovering after a heart intervention

I'll let the Great Man tell the story.

## Thursday, November 1, 2012

### Analele Stiintifice Iasi

I thought I ought to do some advertising for the math journal published by my Alma Mater.  I am talking of course of the  Analele Stiintifice ale Universitatii  "Al.I. Cuza" Iasi. Sectia Matematica.

## Wednesday, October 31, 2012

### Separable random functions

In applications $X$ is a metric space and $\bsT$ is a  locally closed subset of some Euclidean space $\bR^N$. (Example to keep in mind:  $\bsT$ an open subset of $\bR^N$ or $\bsT$ a properly embedded submanifold of $\bR^N$. Often $X$ is a vector space.)

A random function on $\bsT$ is then a function

$$f:\bsT\times\Omega\to X,\;\; \bsT\times\Omega\ni (t,\omega)\mapsto f(t,\omega) \in X,$$

such that, for any $t\in\bsT$,   the   correspondence

$$\Omega\ni \omega\mapsto f_t(\omega) :=f(t,\omega)\in X$$

is measurable  with respected to the $\si$-algebra of Borel subsets  of $X$. In other words,  a random function on $\bsT$ is a family of random variables (on the same probability space) parameterized by $\bsT$.

Observe that we have a natural  map $\Phi: \Omega\to X^{\bsT}$,

$$\Omega\ni \omega\mapsto f_\omega\in X^\bsT,\;\;f_\omega(t)=f(t,\omega).$$

The pushforward via $\Phi$ of $(\eS,P)$ induces  structure of probability space on $X^{\bsT}$.  The functions $f_\omega$, $\omega\in \Omega$ are called the   sample functions  of the given random function.

Let us observe that  there are certain  properties of functions which a priori may not  measurable subsets of $\Omega$.  For example the set of $\omega$'s such that $f_\omega$ is continuous on $\bsT$   may not be measurable if $\bsT$   is uncountable.  To deal  with such issues  we will restrict our attention to certain  classes of  random functions, namely the  separable ones.

Definition 1.   Suppose that  $\bsT$ is a locally closed subset  of $\bR^N$ and $X$ is a Polish space, i.e., a  complete, separable  metric space. Fix a countable, dense subset $S\subset \bsT$. A random function $f:\bsT\times\Omega\to X$ is called $S$-separable    if  there exists a negligible subset $N\subset \Omega$, with the following property: for any closed subset $F\subset X$, any open subset $U\subset \bsT$ the  symmetric  difference of the sets

$$\Omega(U,F):=\Bigl\{ \omega\in \Omega;\;\; f_\omega(t)\in F,\forall t\in U\;\Bigr\},\;\;\Omega_S(U, F):=\Bigl\{ \omega\in \Omega;\;\; f_\omega(t)\in F,\;\;\forall t\in U\cap S\;\Bigr\} \tag{1}\label{1}$$

is a subset of $N$, i.e.,

$$\Omega(U,S)\setminus \Omega_S(U,S),\;\;\Omega_S(U,F)\setminus \Omega(U,F)\subset N.$$

Definition 2.  Let $\bsT$ and $X$ be as in  Definition 1.  A random function  $g: \bsT\times \Omega\to X$ is called a  version of the  random function $f:\bsT\times \Omega\to X$  if

$$P(g_t=f_t)=1,\;\;\forall t\in\bsT.$$

Let me   give an application of separability.      We say that a random function $g:\bsT\times \Omega\to X$ is  a.s.  continuous if

$$P\bigl(\;\lbrace \omega;\;\; f_\omega: \bsT\to X\;\;\mbox{is continuous} \rbrace\;\bigr)=1.$$

Proposition 3.   Suppose that $f$ is an $S$-separable  random function $\bsT\times \Omega\to \bR$ and $g$ is a version of $f$.  If $g$ is a.s. continuous, then
$P\bigl(\lbrace \omega; \;\;g_\omega=f_\omega\rbrace\bigr)=1.$
In particular,  $f$ is a.s. continuous.

Proof.    Consider the set $N\subset \Omega$ in the definition of $S$-separability of $f$.   Define

$$\Omega_*:=\bigl\lbrace \;\omega\in \Omega\setminus N;\;\;g_\omega\;\mbox{is continuous},\;\;g(s,\omega)=f(s,\omega),\;\;\forall s\in S\;\bigr\rbrace.$$

Observe that $P(\Omega_*)=1$.   We will prove that
$g_\omega(t)=f_\omega(t),\;\;\forall \omega\in \Omega_*,\;\; t\in\bsT.\tag{\ast}\label{ast}$

Fix an open set $U\subset \bsT$. For any $\omega\in\Omega_*$ set

$M_\omega(U, S):= \sup_{t\in S\cap U} f_\omega(t).$
Invoking the definition of separability with $F=(-\infty, M_\omega(U, S)]$ we deduce  that
$f_\omega(t)\leq M_\omega(U,S),\;\;\forall t\in U,$
so that
$\sup_{t\in U} f_\omega(t)\leq \sup_{t\in S\cap U} f_\omega(t)\leq \sup_{t\in U} f_\omega(t).$
In other words, for any open set $U\subset\bsT$ we have
$\sup_{t\in U}f_\omega(t)=\sup_{t\in S\cap U} f_\omega(t),\;\;\forall \omega\in\Omega_*\tag{2}\label{2}.$
A variation of the above argument shows
$\inf_{t\in U}f_\omega(t)=\inf_{t\in S\cap U} f_\omega(t),\;\;\forall \omega\in\Omega_*\tag{3}\label{3}.$
Now let $\omega\in\Omega_*$,  $t_0\in T$. Given   $\newcommand{\ve}{\varepsilon}$ $\ve>0$, choose  an neighborhood $U=U(\ve \omega)$ of $t_0$ such that
$g_\omega(t_0)-\ve\leq g_\omega(t)\leq g_\omega(t_0)+\ve,\;\;\forall t\in U(\ve,\omega).$
Since $g_\omega(t)=f_\omega(t)$ for $t\in S\cap U$ we deduce from (\ref{2}) and (\ref{3}) that
$g_\omega(t_0)-\ve \leq \inf_{t\in U(\ve,\omega)} f(t) \leq \sup_{t\in U(\ve,\omega)} f_\omega(t)\leq g_\omega(t_0)+\ve.$
In particular, we deduce
$g_\omega(t_0)-\ve \leq f_\omega(t_0)\leq g_\omega(t_0)+\ve,\;\;\forall \ve>0.$
This proves (\ref{ast}).     Q.E.D.

We have the following result.

Theorem 4.  Suppose that $f:\bsT\times \Omega\to X$ is a random function, where $\bsT$ are Polish spaces. If $X$ is compact, then  $f$ admits a separable version.

Proof.     We follow the approach in Gikhman-Skhorohod. $\DeclareMathOperator{\cl}{\mathbf{cl}}$  Fix a countable dense subset $S\subset \bsT$. $\newcommand{\eV}{\mathscr{V}}$ Denote by $\eV$ the collection of open balls in $\bsT$ centered at points in $S$ and with rational radii. For any  $\omega\in\Omega$  and any  open set $U\subset \bsT$ we set

$$R(U,\omega):=\cl\bigl\lbrace f_\omega(t);\;\;t\in S\cap U\,\bigr\rbrace,$$

$$R(t,\omega)=\bigcap_{t\in V\in\eV} R(V,\omega),$$

where $\cl$ stands for the closure of a set.   Observe that $R(t,\omega)\neq\emptyset$ because it is the intersection of a family of compact sets such that any finitely many sets in the family have nonempty intersection.

Lemma 5.  The following statements are equivalent.

(a) The random function  $f$ is $S$-separable.

(b) There exists $N\subset \Omega$ such that $P(N)=0$ and for any $\omega\in\Omega\setminus N$ and any $t\in\bsT$ we have $f_\omega(t)\in R(t,\omega)$.

Proof of the lemma.  (a) $\Rightarrow$ (b)   We know that $f$ is $S$-separable. Choose $N$ as in the definition of $S$-separability.   Fix $\omega_0\in \Omega\setminus N$ and $t_0\in \bsT$.   For any ball $V\in \eV$ that contains $t_0$  we have

$\bigl\lbrace \omega\in \Omega\setminus N;\;\;f_\omega(t)\in R(V,\omega_0)\;\;\forall t\in S\cap B\;\bigr\}=\bigl\lbrace\; \omega\in \Omega\setminus N;\;\;f_\omega(t)\in R(V,\omega_0)\;\;\forall t\in B\;\bigr\rbrace.$
Observe that $\omega_0$ belongs to the set in the left-hand-side of the above equality and so it must belong to the set in the right-hand-side.  Hence

$f_{\omega_0}(t)\in R(B,\omega_0),\;\;\forall t\in B$
and therefore $f_{\omega_0}(t_0)\in R(B,\omega_0)$ for any $B\in\eV$ that contains $t_0$. Thus $f_{\omega_0}(t_0)\in R(t_0,\omega_0)$ which finishes the proof of the implication (a) $\Rightarrow$ (b).

(b) $\Rightarrow$ (a)   Set $\Omega_*=\Omega\setminus N$.  Suppose that $F\subset X$ is closed.  For any $B\in\eV$  and $\omega\in \Omega_*$ we have

$$f_\omega(t)\in F\;\;\forall t\in S\cap B \Leftrightarrow F\supset R(B,\omega).$$

Since $R(t,\omega)\subset R(B,\omega)$ for any $t\in B$ we deduce  that

$$\Omega(B,F):=\bigl\{ \omega\in \Omega_*;\;\; f_\omega(t)\in F\;\;\forall t\in B\;\bigr\} =\bigl\{ \omega\in \Omega_*;\;\; f+\omega(t)\in F\;\;\forall t\in S\cap B\;\bigr\}=\Omega_S(B,F).$$

If $U$  an open set then we can write $U$ as a countable union of balls in $\eV$

$$U=\bigcup_n B_n.$$

Then

$$\Omega(U,F)=\bigcap_n \Omega(B_n, F)=\bigcap_n\Omega_S(B_n, F)= \Omega_S(U,F).$$

This finishes the proof of the lemma.  q.e.d.

Lemma 6.  For any  Borel set $B\subset X$ there exists a countable subset $C_B\subset \bsT$ such that  for any $t\in\bsT$ the set

$$N(t, B):=\bigl\{ \omega\in \Omega; \;\;f_\omega(\tau)\in B,\;\;\forall \tau\in C_B,\;\;f_\omega(t)\in \bsT\setminus B\;\bigr\}$$

has probability $0$.

Proof of the lemma.   We construct $C_B$ recursively.  Choose $\tau_1\in\bsT$ arbitrarily and set $C_B^1:=\{\tau_1\}$.   Suppose that we have  constructed  $C_B^k=\{\tau_1,\dotsc,\tau_k\}$.    Set

$$N_k(t):=\bigl\{\omega;\;\; g_\omega(\tau)\in B\;\;\forall \tau\in C_B^k,\;\;f_\omega(t)\in\Omega\setminus B\;\bigl\},\;\;p_k=\sup_{t\in\bsT} P\bigl(\;N_k(t)\;\bigr).$$

Observe that  $p_1\geq p_2\geq \cdots \geq p_k$. If $p_k=0$   we stop and we set $C_B:=C_B^k$.

If this is not the case, there exists $\tau_{k+1}\in \bsT$ such that

$$P\bigl(N_k(\tau_{k+1}\bigr)\geq \frac{1}{2}p_k.$$

Set $C_B^{k+1}:=C_B^k\cup\{\tau_{k+1}\}$.  Observe that the events $N_1(\tau_2),\dotsc , N_k(\tau_{k+1})$ are mutually  exclusive  and thus

$$1\geq \sum_{j=1}^k P(N_j(\tau_{j+1})) \geq \frac{1}{2}\sum_{j=1}^k p_{j+1}.$$

Hence $\lim_{n\to\infty} p_n=0$. Now  set

$$N(t, B):=\bigcap_{k\geq 1} N_k(t).$$
q.e.d.

Lemma 7.   $\newcommand{\eB}{\mathscr{B}}$ Suppose that $\eB_0$ is a countable family  of Borel subsets of $X$ and $\eB$  is the family obtained by taking the intersections of  all the subfamilies of $\eB_0$.  Then there exists a countable subset $C\subset \bsT$, and for each $t$ a subset $N(t)$ of probability zero  such that for any $B\in \eB$ we have

$$\bigl\{ \omega;\;\; f_\omega(\tau)\in B\;\;\forall \tau\in C\;\; f_\omega(t)\not\in B\;\bigr\}\subset N(t).$$

Proof.   For any $t\in \bsT$ we define

$$C:=\bigcup_{B\in\eB_0}C_B, \;\; N(t) :=\bigcup_{B\in\eB_0} N(t,B),$$

where $C_B$  and $N(t,B)$ are  constructed as in Lemma 6. Clearly $C$ is countable.

If $B'\in\eB$ and $B\in\eB_0$ are such that $B\supset B'$, then

$$\bigl\{ \omega;\;\; f_\omega(\tau)\in B'\;\;\forall \tau\in C\;\; f_\omega(t)\not\in B\;\bigr\}\subset \bigl\{ \omega;\;\; f_\omega(\tau)\in B\;\;\forall \tau\in C\;\; f_\omega(t)\not\in B\;\bigr\}\subset N(t,B)\subset N(t).$$

If

$$B'=\bigcap_{k\geq 1} B_k,\;\;B_k\in\eB_0\;\;\forall k,$$

then

$$\bigl\{ \omega;\;\; f_\omega(\tau)\in B'\;\;\forall \tau\in C\;\; f_\omega(t)\not\in B'\;\bigr\}\subset \bigcup_{k\geq 1} \bigl\{ \omega;\;\; f_\omega(\tau)\in B'\;\;\forall \tau\in C\;\; f_\omega(t)\not\in B_k\;\bigr\}\subset \bigcup_{k\geq 1} N(t, B_k)\subset N(t).$$
q.e.d.

The proof of Theorem 4 is now within reach.  Suppose that $S$ is a countable and dense  set of points in $\bsT$ and $D$ is a countable dense subset of $X$.  Denote by $\eV$ the collection of open balls in $\bsT$ centered at points in $S$. Denote by $\eB_0=\eB_0(D)$ the collection of open balls with  in $X$ of rational radii centered at points in $D$.  As in Lemma 7, denote by $\eB$ the collection of sets obtained by taking intersections of arbitrary families in   $\eB_0$. Clearly $\eB$ contains all the closed subsets of $X$.

Fix a ball $V\in \eV$.  Lemma 7  applied to the restriction of $f$ to $V$ implies the existence of a countable set

$$C(V)\subset V$$

and  of a family of  negligible sets

$$N_V(t)\subset \Omega,\;\;t\in V$$

such that  for any $B\in eB$

$$\{ \omega;\;\; f_\omega(\tau)\in B,\;\;\forall \tau\in C,\;\;f_\omega(t)\in V\setminus B\;\bigr\}\subset N_V(t).$$

Set

$$C=\bigcup_{V\in\eV}C(V),$$

while for $t\in \bsT$ we set

This follows from  (\ref{3}) and (\ref{4}}).


text:
## Chemistry: The Central Science (13th Edition)

$C_8H_8(l)+10O_2(g)\rightarrow 8CO_2+4H_2O$
Combustion means $O_2$ is the other reactant. The unbalanced equation is: $C_8H_8(l)+O_2(g)\rightarrow CO_2+H_2O$ Add a coefficient of 8 to $CO_2$ to balance $C$. Add a coefficient of 4 to $H_2O$ to balance $H$. $C_8H_8(l)+O_2(g)\rightarrow 8CO_2+4H_2O$ The right side now has 20 $O$ atoms. Add a coefficient of 10 to $O_2$ to balance. $C_8H_8(l)+10O_2(g)\rightarrow 8CO_2+4H_2O$

format: plain


text:
Home What is the purpose of Node.js module.exports and how do you use it?
 What is the purpose of Node.js module.exports and how do you use it? I can't seem to find any information on this, but it appears to be a rather important part of Node.js as I often see it in source code. According to the Node.js documentation: module A reference to the current module. In particular module.exports is the same as the exports object. See src/node.js for more information. But this doesn't really help. What exactly does module.exports do, and what would a simple example be?


text:
I had a job interview with SNOLAB last week, and the interviewer asked me if I had ‘good lab hands’. I told him ‘not yet, but hopefully I will develop them next week’. That is because this week I have been assembling the experimental apparatus for the experiment I am running on my current NSERC USRA term.

The goal of my project is to characterize the transfer function of electrodes in my laboratory’s low temperature vaccuum system, by using the spin-sample we are experimenting on as a detection tool. In our lab, we attach indium-phosphide spin samples to the end of silicon nanowires, which are protruding from a silicon chip off which they are grown. These nanowires are sensitive to electric fields, and oscillate. The oscillations are measured with an interferometer.

For our proof of concept test of our measurement scheme, we are using a microscale cantilever as a stand in for the silicon nanowire. These cantilevers are 160 microns long, and protrude from a 3.14mm x 4mm chip. We plate the cantilever with a thing layer of gold for conduction, and then position it 50um above a 200um wide copper strip (the ‘gate’).

The idea is that the force felt by the cantilever is proportional to the product of the gate voltage and the external voltage of the electrodes we want to characterize. This is because the cantilever and gate act as a capacitor, which puts a charge on the cantilever with magnitude proportional to the gate voltage; $q = CV_g$ and then the force felt by the cantilever is $F = qE = CV_g \nabla \phi \sim CV_g V_e$ We can then give our gate voltage a frequency such that one of the product termshas a resonant frequency with the cantilever, such that we can recontruct the electrode voltage from the cantilever oscillations.

It all sounds good on paper, but a 160um cantilever really is very small in everyday terms. When holding the chip in your palm, you cannot see the cantilever with your naked eye. Not only that but touching the cantilever at all will very likely break it right off of the chip. So you can imagine that it is pretty difficult to put it 50um above a copper gate without breaking it. This is why I needed to develop my laboratory hands.

After having an aluminum holder machined and printing a circuit board with the copper gate on it, as well as a pocket for the chip to sit at the correct depth, I was ready to put the cantilever chip into the board. I had a chip which already lost it’s cantilevers which I had been using for measurements, so I decided to try and mount this chip in first as a practice run.

I got a pair of tweezers, a small screwdriver (00-90 screws! they are tiny!) and sat down at our microscope station to try and mount this chip. With the microscope on 2x magnification, I tried to lift the chip out of it’s tray and place it on my aluminum mount beside the copper gate. I grabbed it at the wrong angle with the tweezers and it flew off and landed on the table. This would be the first of many such failures.

After an hour of putting the practice chip in and taking it back out, trying out different techniques and occasionally sending the chip flying, the poor chip was totally scratched and dented all over. However, I had mastered the technique of handling this chip and gently screwing it down into the circuit board. So I got a chip with cantilevers remaining, and loaded it in - successfully as you can see from the image. Now I am just waiting for the previous experiment in the vacuum chamber to finish so I can run my experiment!

Next time somebody asks if I have laboratory hands I can confidently say yes.


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# zbMATH — the first resource for mathematics

Deformations of Diophantine systems for quadratic forms of the root lattices $$A_n$$. (Russian, English) Zbl 1076.11026
Zap. Nauchn. Semin. POMI 314, 5-14, 285 (2004); translation in J. Math. Sci., New York 133, No. 6, 1605-1610 (2006).
The matrix equation $$Q[X] = {^tT}QT = A$$ is studied, where $$Q$$ and $$A$$ are positive definite symmetric integral matrices of order $$n$$ and $$m$$, respectively, and $$T$$ ranges over the set of $$n\times m$$ integer matrices. When $$A = A' \oplus A''$$, where the summands have squarefree and coprime determinants, a formula is found for the weighted average of representations of $$A''$$ by a suitable quadratic form $$Q'$$. This computation depends on the approach due to V. G. Zhuravlev [Algebra Anal. 8 , No. 1, 21–112 (1996; Zbl 0860.11018)]. Two concrete numerical cases are considered in detail.
##### MSC:
 11E12 Quadratic forms over global rings and fields 11D85 Representation problems 11E20 General ternary and quaternary quadratic forms; forms of more than two variables
##### Keywords:
embeddings of lattices; weight of representations
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# Tag Info

6

This is indeed a subtle point. What is generally meant with this statement is that correlation is going up in bear markets, so it is not so much the "turmoil" part (i.e. volatility per se) but the "trend" (i.e. negative in this case) part. Putting it another way is that when you control for volatility not the correlation but the covariance (which is the part ...

6

As a practitioner, I have worked on the following Maximize Yield/OAS for a Fixed Income Portfolio keeping the Rates Duration (Key Rate Durations) and Spread duration in a constrained range . There are other constraints such as No short selling Max amount you can buy is X% of Max outstanding amount in market Maximum exposure to a perticular country , ...

5

This optimization is trivial $$w^{T,J}_i = \begin{cases} 1 \quad \text{if } i=\arg \max_i R^{T,J}(S_i) \\0 \quad \text{otherwise} \end{cases}$$ That is to say, when you optimize only one weight will be nonzero. That's because these ratios incorporate no notion of distributional width, and therefore do not reward diversification. With no concentration ...

5

The weak EMH states that it is impossible to earn an excess return given publicly known information such as past prices. Clearly, different securities have different expected returns. For example: the bond and the stock of one company or a security that generates twice the return of another one. This difference in expected return is explained by a ...

5

To clarify notation, you have an universe of $n=2000 \space$ stocks and two portfolio vectors $\mathbf{a},\mathbf{b}\in\mathbb{R}^{n}$ with $\left\|\mathbf{a}\right\|_{1}=\left\|\mathbf{b}\right\|_{1}=1$. Further, you have Estimators for the true Variance $\operatorname{Var}\left[\mathbf{a}\right]$ resp. $\operatorname{Var}\left[\mathbf{b}\right]$ and the ...

5

Accurately stated: Diversification helps during turmoil, but helps less as what would be expected by using $w^T \Omega w$ as the portfolio variance where the off-diagonal covariances are estimated during tranquil periods. This is because correlations and covariances change during turmoil, typically increasing. This reduces the benefit of diversification ...

4

With respect to issue one, it can be simpler to consider the case where the constraint on the expected return is an equality. In that case, the first problem can be transformed to Minimize with respect to $\left\{ x,\lambda_{1},\lambda_{2}\right\}$: $x'\Sigma x + \lambda_{1} (\mu'x - r) + \lambda_{2} (1'x - 1)$ by the technique of Lagrangian multipliers, ...

4

The initial investment is the capital in the account used to support the portfolio, not the cost of the assets in the portfolio. For example, when you sell a stock or bond short, your account doesn't actually accrue any cash. Instead you start receiving a regular cash flow. There isn't necessarily a difference between these quantities in a long-only ...

4

You can obtain the covariance between 2 portfolios by multiplying the row vector, containing the weights of portfolio A with the variance-covariance matrix of the assets and then multiplying with the column vector, containing the weights of assets in portfolio B. Equally you can set up a new portfolio A+B by creating a new column vector that contains the ...

4

There are plenty of books on portfolio issues built according to formula "some theory + some R code (or Matlab, or S - which is very similar to R)". See for example Pfaff B. Financial Risk Modelling and Portfolio Optimization with R.// 2013. Best M.J. Portfolio Optimization. Chapman & Hall, 2010. Würtz D. et al. Portfolio Optimization with ...

4

The Efficient Market Hypothesis (EMH) states that you cannot beat the market on a risk-adjusted basis by looking at past prices. You can certainly earn higher returns than the market if you take on more risk (by leveraging, for example). Modern Portfolio Theory allows you to construct portfolios that are efficient. According to this theory, you still cannot ...

4

Lots of wealth management firms still use MPT; in my experience regulators like it because they understand it. If asset returns are normally distributed, the standard deviation of the portfolio is a coherent risk measure (this can be seen by noting that the normal distribution's CVaR, which is a coherent risk measure, can be written as $$\mu+c \sigma$$ ...

3

Initial capital is not a real constraint in theoretical analysis, but might be a practical constraint in reality. The objective function you gave defines the efficient frontier corresponding to a given risk tolerance $q \in [0, \infty]$: $$\min\{w^T\Sigma w-qR^Tw\}$$ This criterion is among the other popular optimization criteria, such as minimum variance, ...

3

There are many papers on this subject (try googling portfolio optimization skewness kurtosis) that can describe the assumptions of including skewness and kurtosis in a utility function (if that's what you're interested in). I would highlight two main points. Mean-variance optimization does not make an assumption of normality. Assume returns are ...

2

Check out following link. In page 23 you'll find the derivation. http://faculty.washington.edu/ezivot/econ424/portfolioTheoryMatrix.pdf

2

The Lyxor white paper Regularization of Portfolio Allocation contains a lot on this topic. The head of quant research there, Thierry Roncalli, also held a talk about this recently.

2

The concrete (general) answer to part (ii) of my question seems to be contained in Equation 8 of the following link: http://www.columbia.edu/~ks20/FE-Notes/4700-07-Notes-portfolio-I.pdf In particular, interpreting $\sigma$ as volatility, take for example $E_A=0.10,\sigma_A=0.15,E_B=0.25,\sigma_B=0.40$ and $\rho =−0.2$. I get that about 83 percent of the ...

2

Pretty good explanation is in Schweser CFA Study Notes for CFA level III. Books 3 and 5, at least from 2009, if I remember right. See also Tsay R.S. Analysis of Financial Time Series (Wiley Series in Probability and Statistics). // 2010. - good example with implementation in R.

2

Let there be n stocks, 2 portfolio a and b. c is a combined portfolio of portfolio a and portfolio b. $\Sigma$ is variance-covariance matrix of the n assets. Weight vectors for portfolios a and b are $$w_{pa},w_{pb}\in\mathbb{R}^{n} ,$$ $$\left\|w_{pa}\right\|_{1}=\left\|w_{pb}\right\|_{1}=1$$ then $$Var(a)= w_{pa}' \Sigma w_{pa}$$ $$Var(b)= w_{pb}' ... 2 EMH says that one can not earn excess return using some information. This is known as joint-hypothesis problem: to test for market efficiency one have to determine first what is "normal" market return, i.e. what type of information is normally priced by the market. Usually to test for EMH they use CAPM or 3-factor Fama-French model (which is a kind of ... 2 I assume you're talking about this formula:$$U(w) = w'\mu - \frac{1}{2} \lambda w' \Sigma w = w'\mu - \frac{1}{2} \lambda \sigma_\omega^2$$where \sigma_\omega^2 denotes the portfolio variance for a portfolio with weights \omega. Dividing by two is purely done for convenience, optimizing this formula requires taking the derivative with respect to ... 2 I am engineer studying Finance, therefore Im not an expert in Math/Stat, but not noob. I disagree with the previous answer. In fact, I know portfolio managers and hedge fund assesors that usses MPT. It must be said that you need to know what that represents, and also not only focus your investment in MPT, but consider other methods. Like in every other ... 2 Sure a lot of traditional (mutual) buy side funds use MPT. They also mostly subscribe to the efficient market hypotheses. And they also do not hide the fact that they have no interest to lobby many retirement investment and savings schemes to allow for long/short investments but hold on to long-only. And finally, most of them underperform simple benchmark ... 2 Well, you are asking something very subjective. In addition it should be mentioned that S&P500 are the companies with higher capitalization of S&P1500. Therefore a huge weight of S&P1500 is set by S&P500. In fact, as it can be seen in 2008 both went down a 37%, in the other hand S&P500 has 80% of the total of the US equity Market. After ... 2 You are asking two different questions: what would be the model result, and what would be the actual performance of an actual portfolio. The optimal model results with the S&P 1500 will be at least as good as the model results with the S&P 500. The S&P is a proper subset of the S&P 1500, so you can get the results of the S&P 500 model by ... 1 MPT should be called Medieval Portfolio Theory, it is a theory from 50 years ago with huge theoretical flaws (mean-variance utility, use of Pearson's correlation that is not coherent, based on historical data). Come on, it is an error maximizer. The least one could do is Michoud resampling, but it is patented. Or a bayesian Black-Litterman would be more ... 1 a) The formula for Beta is:$$\beta_i=\frac{\sigma_{i,M}^2}{\sigma_M^2}=\frac{0.165^2}{0.11^2}=2.25$$b) So by the CAPM equation, the expected return for the asset is:$$E(R_i)=r_f+\beta(R_M-r_f)=0.04+2.25(0.12-0.04)=0.22=22\% c) If the variance of the stock is $0.22^2$, since this variance was multiplied by $\beta=2.25$, we get: ...

1

MPT uses expected values for its parameters. How these expected (future) parameters are estimated, is another question. Usually one takes historic averages when its the only information available, but one could for example also use analysts forecasts or other advanced estimation methods.

1

Only certain aspects of the risks that you bear in power markets given exposure to variable quantity swaps can be hedged. To your point, you have to have some expectation of what the load will look like. Even if you immediately go out and buy power against this expected qty you are subject to the risk that the load will deviate from said qty. There is no ...

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