User zhoraster - MathOverflow

Expected inverse determinant with independent rows

2010-11-04T16:10:12Z

Let $a_1,a_2,\dots,a_n$ be independent identically distributed random vectors in $\mathbb R^n$. I need a bound for $E[|\det A|^{-1}]$, where $A$ is the matrix composed out of these vectors.

More specifically, these vectors take their values on a curve.

And more generally, I will be happy even if there is an estimate for a "non-square" determinant, precisely, for $E[G^{-1/2}]$, where $G$ is the Grammian determinant of $n$ iid vector in $\mathbb R^m$ and $m>n$.

Update: yes, I need an upper bound.

Precisely, I have something like $a_i = (f_1(\xi_i),f_2(\xi_i),\dots,f_n(\xi_i))$, where $f_j(x) = |x|^{-\alpha_i}\sin(\beta_i x)$ with $\alpha_i\in(1,2)$ and look for an estimate in terms of $|\alpha_i-\alpha_j|$ and $|\beta_i-\beta_j|$.

I'm flexible with the choice of distribution for $\xi$, but it should be the same for all sets of $\alpha$'s and $\beta$'s.

Answer by zhoraster for Pseudonyms of famous mathematicians

2010-11-07T20:15:42Z

I'm surprised that no one named Leonardo of Pisa, known as Fibonacci to us (though he didn't use this nickname, and its origin is not completely clear).

Al-Khoresmi is apparently a nickname as well (though this time used by the author), meaning his origin.

(Maybe not exactly an answer to the original question, because these are rather nicknames, not pseudonyms.)

An analogue of Hilbert-Schmidt theorem for multilinear forms

2011-03-03T10:49:31Z

Let $H$ be a (the) real separable Hilbert space. The Hilbert--Schmidt theorem says that a compact self-adjoint operator $A$ has an eigenfunction expansion. Instead of operator, we can think of a symmetric bilinear form and write $$ A = \sum_{k\ge 1} \lambda_k \varphi_k\otimes \varphi_k \tag{1} $$

My question is:

Are there any multilinear analogues of (1)? Which $n$-linear symmetric forms can be represented in a form $$ A = \sum_{k\ge 1} \lambda_k \varphi_k^{\otimes n}\ ?$$

(There is no compactness notion for multilinear forms, but we can assume that they are e.g. Hilbert--Schmidt.)

Answer by zhoraster for Brownian local time density

2011-02-26T14:59:31Z

Another useful way to study local time is related to the very useful occupation formula (its meaning is obvious if you think a little bit): $\int_0^t g(W_s) ds = \int_{\mathbb R} g(x) L(x,t) dx $.

Putting $g(x) = e^{izx}$: $$ \int_{\mathbb R} e^{izx} L(x,t) dx = \int_0^t e^{iz W_s}ds. $$ Now the lhs is the Fourier transform of $L$; inverting it, we get $$ L(x,t) = \frac{1}{2\pi} \int_{\mathbb R} e^{-izx} \int_0^t e^{iz W_s} ds\, dz = \frac{1}{2\pi} \int_{\mathbb R} \int_0^t e^{iz (W_s-x)} ds\, dz. $$ From here one can e.g. more or less easily find the moments of $L$.

Answer by zhoraster for What is the shortest Ph.D. thesis?

2011-02-08T17:15:05Z

Edmund Landau's thesis was 13 pages long.

Answer by zhoraster for What would be a fractional Poisson Process like

2011-01-26T18:36:30Z

Here is a thesis containing (in Section 2) an overview of different definitions of fPP.

My personal favorite is the "Standard Fractional generalization I" defined in 2.2. The reason is that there seems to be (I failed to find any relevant results) an isomorphism between this version and fBm similar to the (usual) Wiener-Poisson isomorphism.

Answer by zhoraster for Mathematical "urban legends"

2011-01-25T17:51:20Z

(A rather sad story)

For obvious reasons, I won't give the place and/or names.

On a thesis defence (we have here a procedure very different from Europe or US; for instance, the committee is more or less fixed) one member of the committee rose and asked to vote against the thesis because of plagiarism: the thesis contained (almost verbatim!) definitions from a book X.

Answer by zhoraster for Commuting supremum and expectation

2010-12-03T19:19:49Z

Clearly, $M(\omega) = \sup_{a\in U} g(a,S_t)$ is $\mathcal F_t$-measurable.

Define for $\delta>0$ $$ \mathfrak A_\delta = \{(a,\omega)\in U\times \Omega\mid g(a,\omega)>M(\omega)-\delta\} $$ This set is in $\mathcal B(\mathbb R)\otimes \mathcal F_t$, and it has a full projection onto $\Omega$. By a measurable selection theorem (which I think one can find in Bogachev Measure Theory) there is an $\mathcal F_t$-measurable $A_\delta$ such that $(A_\delta(\omega),\omega)\in \mathfrak A_\delta$ almost surely. Hence $E[g(A_\delta,S_t)]\ge E[M(\omega)]-\delta$. We get the desired statement by letting $\delta\to 0$.

(One can also use Kuratowski--Ryll-Nardzewski theorem to prove the existence of a measurable $A_\delta$.)

Answer by zhoraster for Two geometric probability questions (one answered, one more to go)

2010-11-11T10:47:52Z

2) This is, of course, the same as saying about spacings between uniform points on a segment (you can say that $Y_1=0$, for example). Let it be the segment $[0,1]$.

Now the joint distribution of $I_1,\dots, I_{n}$ is the same as of $E_1/E,\dots, E_n/E$, where $E_1,\dots, E_n$ are iid exponential distributed, $E=\sum_{k=1}^n E_k$ (see Devroye Non-Uniform Random Variate Generation, p.208). So the distribution of $I_{(1)},\dots, I_{(n)}$ is the same as of $E_{(1)}/E,\dots, E_{(n)}/E$. But the joint distribution of $\{E_{(k)}-E_{(k-1)},k=1,\dots,n\}$ ($E_{(0)}:=0$) is the same as of $\{(n-k+1)^{-1} E_k,k=1,\dots,n\}$ (ibid, p.211).

So the distribution of $J_1,\dots, J_n$ is the same as of $\{(n+k-1)^{-1} E_k/E,k=1,\dots,n\}$, where $E_1,\dots, E_k$ are iid exponential rv's, $E=\sum_{k=1}^n E_k$. And this is, by the previous paragraph, equivalent to saying that the distribution is the same as of $\{(n-k+1)^{-1} I_k,k=1,\dots,n\}$.

These are not independent, but very close to be, and from here you can find the distribution of maximum and minimum (but nothing very pleasant there, as the variables in question are not identically distributed; a formula for the expectation looks extremely ugly).

How to get distribution of $J$ omitting $E$. In fact, this is simple owing to the fact that the ordering map on the simplex $\{(t_1,\dots,t_n)|t_j\ge 0,\sum_j t_j=1\}$ (the support of $I$) is picewise linear, and moreover each image has the same number of preimages due to the apparent symmetry. So the distribution of ${I_{(1)},\dots,I_{(n)}}$ is uniform on its support. Now we have a one-to-one linear map to $J$. So $J$ is also uniformely distributed. So it's only about finding its support, which is simple, as John Jiang noted.

Answer by zhoraster for Vector spaces of random variables having zero expectation

2010-11-08T10:11:49Z

One can, of course, think of $L^2(\Omega)$ as of a Hilbert space with a scalar product $E[\xi\eta]$. But for random variables much more important is the covariance $E[\xi\eta]-E[\xi]E[\eta]$. Though it looks at first sight as a scalar product, unfortunately it's not, as $\mathrm{cov}(\xi,\xi)=0$ doed not imply $\xi=0$. However, on the space of centered r.v.'s it is a scalar product. And this Hilbertian structure fully determines the laws in some cases, like a Gaussian case, as Shvai Covo already mentioned. And also this Hilbertian structure plays a very important role for (weakly) stationary processes (also noted by Shvai Covo).

Vector spaces of non-centered random variables are not so popular. One of applications which I think about is financial mathematics, though there you more often work with some cones rather than full vector spaces. Still, a lot of machinery is based (especially in discrete time) on some vector space techniques.

Answer by zhoraster for Are there more connected or disconnected graphs on n vertices?

2010-11-04T23:01:26Z

Connectedness wins by a knockout: the proportion of disconnected graphs is about $n2^{-n+1}$. See Flajolet, Sedgewick "Analytic Combinatorics", p. 138.

Answer by zhoraster for Coordinatizing the disk via Brownian motion

2010-10-29T07:08:43Z

Let $B$ be the two-dimensional Brownian motion, $D$ be the unit disk, $\tau$ be the moment of hitting the circle (equivalently, of escaping the interior of the disk), $A$ be one of the arcs.

You're speaking about $u(z) = E^z[\mathbf{1}_{A}(B(\tau))]$, where $E^z$ is the expectation given $B(0)=z$. By Kakutani, this is exactly the solution to the following Dirichlet problem for Laplace's equation: \begin{equation}\tag{LE} \begin{cases} \Delta u = 0 & \quad\text{inside }D, \\ u=\mathbf{1}_A & \quad\text{on }\partial D. \end{cases} \end{equation}

For example, it can be expressed via the Poisson integral formula.

Update: a solution of (LE)

(This must be known and written somewhere, though I don't know any reference.)

Let $O$ be the disk center, $A$ be the arc from $e_1$ to $e_2$ (counterclockwise) and $|A|$ its length.

Start by the following simple observation. Let us be given a point $e$ on the circle. Define the function $$u_e(z) = \angle(zeO),$$ which to a point $z$ assigns an angle in $(-\pi/2,\pi/2)$ by which one should rotate the ray $ez$ counterclockwize around $e$ to get $eO$. Then $u_e$ is harmonic inside $D$.

Now on the circle, $u_{e_1} -u_{e_2} = \pi -|A|/2$ inside the arc $A$ and $u_{e_1}-u_{e_2} = -|A|/2$ outside. Thus we can immediately write the solution of (LE): $$ u(z) = \frac{1}{\pi}(u_{e_1}(z)-u_{e_2}(z)+|A|/2). $$

Consequently, the map you are interested in is invertible.

Hope I didn't mess with those angles. Made it community wiki so one can correct if needed.

When is it possible to construct a joint law from its two-dimensional marginals?

2010-10-27T15:18:32Z

My question is much more specific than the title:

Given a symmetric distribution $\Xi$ on $\mathbb R^2$, when is it possible to construct a sequence $\xi_1,\xi_2,\dots$ of random variables such that the joint distribution of any two of them is $\Xi$?

For example, if the pdf of $\Xi$ is decomposable: $p_\Xi(x,y) = p(x) p(y)$, then one can just take a sequence of independent r.v.'s.

(To construct certain counterexample related to fractional Brownian motion) I am particularly interested in the pdf $p_\Xi(x,y) = \frac{(a+1)(a+2)}2 |x-y|^{a}1_{[0,1]}(x)1_{[0,1]}(y)$, $a\in(-1,0)$.

Answer by zhoraster for An Upper Bound for the Average of Top Order Statistics

2010-10-22T10:23:43Z

Write $$ E[T_k]=E[E[T_k|X_{(n-k)}]] $$ The distribution of $X_{(n-k+1)},\dots,X_{(n)}$ given $X_{(n-k)}=x$ is the same as the conditional distribution of a monotone arrangement of $n-k$ independent r.v.'s with cdf $F$ given they all are not less than $x$. (This can be proved easily e.g. by using the quantile transform.)

Now the mean value of monotone arrangement of a sequence is equal to the mean value of the sequence itself. So by linearity of conditional expectation we can write $$ E[E[T_k|X_{(n-k)}]]=E[E[X|X\ge X_{(n-k)}]], $$ where $X$ is an independent of $X_{(n-k)}$ r.v. with cdf $F$.

Now it is again about the quantile transform. Let $U=F(X)$, $U_{(n-k)}=F(X_{(n-k)})$. We have (by the independence of $U$ and $U_{(n-k)}$) $$ E[X|X\ge X_{(n-k)}]=E[F^{(-1)}(U)|U>U_{(n-k)}] = \frac{1}{1-U_{(n-k)}}\phi(U_{(n-k)}), $$ where $\phi(u) = \int_u^1 F^{(-1)}(x)dx$. But we do know the density of $U_{(n-k)}$, so the expectation of this expression is equal to $$ \frac{n!}{(n-k-1)k!}\int_0^1 u^{n-k-1}(1-u)^{k-1}\phi(u)du = \frac{n}{k}E[\phi(B)], $$ where $B$ has $\mathrm{Beta}(n-k,k)$ distribution. But $\phi(x)$ is easily seen to be concave, so by Jensen's inequality $E[\phi(B)]\le \phi(E[B]) = \phi(1-k/n)$, as required.

Answer by zhoraster for Examples of common false beliefs in mathematics.

2010-10-20T18:20:40Z

A projection of a measurable set is measurable. Not only students believe this. I was asked once (the quote is not precise): "Why do you need this assumption of a measurability of projection? It follows from ..."

A polynomial which takes integer values in all integer points has integer coefficients.

Another one seems to be more specific, I just recalled it reading this example. A sub-$\sigma$-algebra of a countably generated $\sigma$-algebra is countably generated.

Answer by zhoraster for The question about Kolmogorov tightness criterion

2010-09-30T10:52:10Z

$n_0$ must be independent of $t_1$ and $t_2$, of course. If it's not, the processes might be even discontinuous. For instance, $X_n$ is a Poisson process with parameter $1/n$. Then $$E(|X_n(t_1)-X_n(t_2)|^2)\le |t_1-t_2|^2$$ for all $n>|t_1-t_2|^{-1}$ (for all $n\ge 1$ if $t_1=t_2$).

And the same answer works for the second question: when $\alpha=1$, the processes need not to be continuous. In some special cases, where you have higher moments controlled by a lower one polynomially, it may help (e.g. in the Gaussian case $\gamma=2$ and $\alpha=1$ is enough).

Answer by zhoraster for Looking for a version of Itô's Lemma

2010-08-04T10:38:15Z

Let $F(t,m-b,m)=G(t,m,b)$ (this way it's easier to write). As long as $M$ has bounded variation, we can happily write (I skip arguments, which are $t,M_t,B_t$) $$ dG(t,M_t,B_t) = G'_t dt + G'_m dM_t + G'_b dB_t + \frac12 G''_{bb} dt. $$

Answer by zhoraster for Approximation of the law of a stochastic process

2010-08-04T09:58:09Z

This is wrong. Let for simplicity $\omega = 1$. Write $$ E[V_T^2] = \frac 1T \int _0^T E[\exp(2W_t-t)]dt = 1/T\int_0^Te^{t} dt = (e^{T}-1)/T. $$ This is inconsistent with their claim.

(In fact, by considering expectations of greater powers of $V_T$, one can see that $V_T$ cannot be log-Gaussian with any parameters.)

Answer by zhoraster for Probability of random permutation having certain cycles

2010-08-04T09:35:23Z

The exponential generating function of the number of permutations of length $n$ such than all their cycle sizes are in a certain set $A\subset \mathbb N$ is $$ P(z) = \exp\left(\sum_{n\in A}\frac{z^n}{n}\right). $$

Comment by zhoraster

2012-07-06T22:18:43Z

And what is known about $g$ and $f$? Obviously, we can't deduce anything like that if they are arbitrary.

Comment by zhoraster

2012-07-06T21:56:58Z

Note quite what I needed. Still I voting it up so you will be able to add a comment rather than posting an answer next time.

Comment by zhoraster

2011-05-13T07:45:02Z

Looks like a homework.

Comment by zhoraster

2011-04-18T18:43:16Z

Does not seems like a research level question to me. Try asking at math.stackexchange.com.

Comment by zhoraster

2011-03-22T15:29:58Z

Some further scholar googling gave a similar result: archive.numdam.org/ARCHIVE/CM/CM_1986__60_1/… for a dimension $\ge 3$.

Comment by zhoraster

2011-03-22T15:28:21Z

Quick scholar googling gave this reference: jstor.org/stable/2244253, which cites a similar result for general two-dimensional Riemann manifold (though the number 2 is missing from the rhs there).

Comment by zhoraster

2011-03-04T06:33:12Z

Yes, I am missing something, and I see what. Thank you for answer, accepting it.

Comment by zhoraster

2011-03-04T04:44:55Z

However, $u_k$ here cannot be arbitrary, and, as far as I see, they must be symmetric forms as well. So your answer gives the desired statement for $n=2^r$, or I'm missing something? And for odd $n$ the question remains. You wrote "the $u_k$'s will not in general be in the form...", but can you give any example when they are not?

Comment by zhoraster

2011-03-04T04:03:30Z

Thank you, though it is not what my wishful thinking imagined, this may be helpful.

Comment by zhoraster

2011-02-26T17:29:49Z

@Paul: Aha, I see now. Your post unfortunately lacks some signs in it, so it was not very clear. You can check "community wiki" so that everyone is able to edit it.

Comment by zhoraster

2011-02-26T17:00:42Z

The argument is not completely correct. However, it is possible to make it correct. E.g. saying that $\inf \theta_a > 0$ a.s. implies $\inf \theta_a>c$ with a positive probability for some $c>0$ and further speaking of the variation on the interval $[0,c]$.

Comment by zhoraster

2011-02-26T16:44:17Z

@Didier thanks, indeed.

Comment by zhoraster

2011-02-26T16:41:50Z

It is not true even if all $\sigma^a$ are deterministic.

Comment by zhoraster

2011-02-12T21:15:16Z

I think math.stackexchange.com is more appropriate for such questions.

Comment by zhoraster

2011-02-12T21:12:55Z

$(A\Rightarrow B)\Leftrightarrow (\overline B \rightarrow \overline A)$.