User john jiang - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T15:19:50Z http://mathoverflow.net/feeds/user/4923 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/127428/inner-product-with-normalized-gaussian/127443#127443 Answer by John Jiang for Inner product with normalized Gaussian John Jiang 2013-04-13T05:12:48Z 2013-04-13T14:17:25Z <p>$X/|X|$ is almost surely a uniformly random element on the unit sphere of dimension $n-1$; this is not the same thing as a rotation, which is a matrix. Since a multivariate standard Gaussian vector is invariant in law under a fixed rotation, its law is certainly invariant under a random rotation as well: thus if $A$ is a uniformly random orthogonal matrix, then $A Y$ is again standard Gaussian. Now let $Z = (X/|X|)^T Y$. Then $Z$ has the same law as the first component of $AY$, which is clearly univariate standard Guassian. To verify its independence with $X$: if you rotate $X$ by an orthogonal matrix $A$, you can absorb $A$ into $Y$, whose law is invariant under rotation. So conditional law of $Z$ under two different $X$ values differing by a rotation stays the same. More obviously, if you scale $X$, the conditional law of $Z$ remains the same.</p> <p>Another thing I noticed is that $Z$ and $Y$ are not jointly normal! Notice that conditioning on $Y$ clearly has an effect on $Z$. Now assuming they are jointly normal, we can show $E Z v^T Y = 0$ for all vector $v$. In fact we can show $E (X/|X|)^T u v^T u = 0$ for fixed $u,v$. In fact, this follows simply from $E (X/|X|)^T u = E (-X/|-X|)^T u = 0$. This is a contradiction. This is similar to the situation $B X$ and $X$, where $X$ is standard 1d gaussian and $B$ is an independent Bernoulli $\pm 1$ variable. They are not jointly normal either.</p> http://mathoverflow.net/questions/20688/whats-a-natural-candidate-for-an-analytic-function-that-interpolates-the-tower-f What's a natural candidate for an analytic function that interpolates the tower function? John Jiang 2010-04-08T04:55:45Z 2013-04-10T19:18:04Z <p>I know that there are analytic functions whose composition with itself is the exponential function, so called functional square root of the exponential functions, with the additional property that it is real on the real line. Is similar property possible for a holomorphic function that interpolates the tower function? Tower function on the positive integers is defined recursively by f(n+1) = exp(f(n), f(1) = 1. </p> http://mathoverflow.net/questions/88475/first-passage-time-brownian-motion/126816#126816 Answer by John Jiang for first passage time, brownian motion John Jiang 2013-04-08T02:27:05Z 2013-04-08T02:27:05Z <p>While the expected value of first hitting time of circle of radius $r$ is certainly not directly related to the expected value of the maximum radius achieved up to time $t$, the following formula I found in Pitman and Yor's article does show the distributions of the two processes are related in a nice way: $$\displaystyle \mathcal{L}(M_{\delta*}^{-2}) = \mathcal{L}(\tau_\delta)$$ where $M_{\delta*} = M_{\delta*}(1)= \sup_{0 &lt; t &lt; 1} B_\delta(t)$ is the running maximum of the Bessel process of dimension $\delta$ up to time $1$, started at $0$, and $\tau_\delta = \inf{t: B_\delta(t) =1}$ is the first hitting time of $1$ of the same Bessel process. See this following linked article:</p> <p><a href="http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.38.596" rel="nofollow">http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.38.596</a></p> <p>The formula can be seen by realizing that $P(\tau_\delta > t) = P(M_{\delta*(1)} > t) = P(M_{\delta*}(t) > t^{-1/2})$ by Brownian scaling: $c^{1/2} B_t$ has the same law as $B_{ct}$. </p> <p>Now the distribution function of $\tau_\delta$ is given in the following article as an infinite series, coming from Laplace transforms.</p> <p><a href="http://arxiv.org/pdf/1106.6132v3.pdf" rel="nofollow">http://arxiv.org/pdf/1106.6132v3.pdf</a></p> <p>The expected value of $\tau_\delta$ has a closed formula in terms of Bessel functions (hence the name; see formula 2.1 in the linked article 2). I don't expect $\mathbb{E} M_{\delta*}$ to have a closed form formula since it amounts to $\mathbb{E} \tau_{\delta}^{-1/2}$, i.e., a negative fractional moment. </p> http://mathoverflow.net/questions/126739/sum-k0-infty-frac1km-k-equiv-1-for-m2/126744#126744 Answer by John Jiang for $\sum _{k=0}^{\infty } \frac{1}{(k+m) k!} \equiv 1$ for $m=2$ John Jiang 2013-04-07T01:00:01Z 2013-04-07T01:00:01Z <p>Another way is to write $1/(k+2)$ as $1/(k+1) - 1/((k+1)(k+2))$.</p> http://mathoverflow.net/questions/126677/linear-recurrence-relations-in-2-variables-with-variable-coefficients/126685#126685 Answer by John Jiang for Linear Recurrence Relations in 2 Variables with Variable Coefficients John Jiang 2013-04-06T01:36:05Z 2013-04-06T01:36:05Z <p>I don't think constant is the only tempered growth solution to your 2d recurrence. Essentially the recurrence needs a 1-dimensional subspace in $\mathbb{Z}^2$ of boundary conditions. Then your recurrence is saying the value at any point $(x,y)$ is the average of values at $(x-1,y)$ and $(x,y-1)$, which has the effect of smoothing things out as we move to the upper right corner of the 2d lattice.</p> <p>For instance, take the subspace $S = {(i,-i): i \in \mathbb{Z}}$. Then you can prescribe any values on points in $S$ and get a consistent solution first on the part of the integer lattice to the up-left of $S$. If you choose the values on $S$ to be say bounded, then the solution on the upper left side of $S$ will also be bounded, hence certainly "tempered". Now to extend this solution below and to the left of $S$, just assign $0$ to all points in $T ={(i,i): i &lt; 0}$ and all the other points in the bottom left corner of the lattice are uniquely determined. Again it's easy to see they are all bounded. </p> <p>Your second recurrence can be treated in the same way. One trivial solution is to assign all points constant at $0$. On the other hand I imagine it can be quite interesting to consider 2d recurrences where the coefficients do not add up to 0 (or cannot be arranged on the right and left of the equality so that the moduli of two sides add up to the same number) hence can exhibit exponential growth. But maybe that case can be reduced to 1d recurrence. </p> http://mathoverflow.net/questions/88075/universality-of-macdonald-polynomials universality of Macdonald polynomials John Jiang 2012-02-10T05:25:44Z 2013-03-30T17:40:29Z <p>I have been recently learning a lot about Macdonald polynomials, which have been shown to have probabilistic interpretations, more precisely the eigenfunctions of certain Markov chains on the symmetric group. </p> <p>To make this post more educational, I will define these polynomials a bit. Consider the 2-parameter family of Macdonald operators (indexed by powers of the indeterminate $X$) for root system $A_n$, on a symmetric polynomial $f$ with $x = (x_1, \ldots, x_n)$:</p> <p>$$D(X;t,q) = a_\delta(x)^{-1} \sum_{\sigma \in S_n} \epsilon(\sigma) x^{\sigma \delta}\prod_{i=1}^n (1 + X t^{(\sigma \delta)_i} T_i),$$</p> <p>(mathoverflow doesn't seem to parse $T_{q,x_i}$ in the formula above, so I had to use the shorter symbol $T_i$, which depends on q).</p> <p>where $\delta$ is the partition $(n-1,n-2,\ldots, 1,0)$, $a_\delta(x) = \prod_{1 \le i &lt; j \le n} (x_i - x_j)$ is the Vandermonde determinant (in general $a_\lambda(x)$ is the determinant of the matrix $(a_i^{\lambda_j})_{i,j \in [n]}$). </p> <p>$x^{\sigma \delta}$ means $x_1^{(\sigma \delta)_1} x_2^{(\sigma \delta)_2} \ldots x_n^{(\sigma \delta)_n}$.</p> <p>Also $(\sigma \delta)_i$ denotes the $\sigma(i)$-th component of $\delta$, namely $n-i$. </p> <p>Finally the translation operator $T_i = T_{q,x_i}$ is defined as $$T_{q,x_i}f(x_1, \ldots, x_n) = f(x_1, \ldots, x_{i-1}, q x_i , x_{i+1} ,\ldots, x_n).$$</p> <p>I like to think of the translation operator as the quantized version of the differential operator $I + \partial_i$, where $q-1$ is analogous to the Planck constant(?). </p> <p>If we write $D(X;q,t) = \sum_{r=0}^n D_{n-r}(q,t) X^r$, then Macdonald polynomials $p_\lambda(q,t)$ are simply simultaneous eigenfunctions of these operators. When $q=t$ they become Schur polynomials, defined by $s_\lambda = a_{\delta +\lambda} / a_\delta$. When $q= t^\alpha$ and $t \to 1$, we get Jack symmetric polymomials, which are eigenfunctions of a Metropolis random walk on the set of all partitions that converge to the so-called Ewens sampling measure, which assigns probability proportional $\alpha^{\ell(\lambda)} z_\lambda^{-1}$. When $q = 0$, they become the Hall-Littlewood polynomials and when $t=1$ they become the monomial symmetric polynomials etc. </p> <p>I was told repeatedly by experts that Macdonald polynomials exhaust all previous symmetric polynomial bases in some sense. Does anyone know a theorem that says that every family of symmetric polynomial under some conditions can be obtained from Macdonald polynomials by specializing the $q$ and $t$? </p> http://mathoverflow.net/questions/126006/is-it-true-that-the-geodesics-on-son-and-sun-are-closed/126009#126009 Answer by John Jiang for Is it true that the geodesics on SO(n) and SU(n) are closed? John Jiang 2013-03-30T13:14:59Z 2013-03-30T13:21:23Z <p>Clearly there are geodesics which are not periodic. Take the maximal torus of say $SO(4)$, and let $$X = \begin{pmatrix} J &amp; 0 \ 0 &amp; \alpha J \end{pmatrix}$$</p> <p>be in block diagonal form, where $J = \begin{pmatrix} 0 &amp; 1 \ -1 &amp; 0 \end{pmatrix}$, with $\alpha$ irrational. Then the geodesic generated by $X$ will be dense in the set of 2x2 block diagonal elements of $SO(4)$, but is not the whole set, hence can't be closed. </p> http://mathoverflow.net/questions/48724/convergence-of-finite-difference-method-for-boundary-value-ode convergence of finite difference method for boundary value ODE John Jiang 2010-12-09T07:37:03Z 2012-06-22T22:06:54Z <p>Suppose we have to solve $d^2y/dx^2= f(y,x)$ where $f$ is Lipschitz and $y(0) =a, y(1) =b$, using finite difference method, i.e., by discretizing the problem into $y_{i+1} - 2y_i + y_{i-1} = h^2f(y_i,x_i)$, with equal spacing. How do we show it converges, i.e., $\lim_n \max_{i \le n} |y(x_i) - y_i| = 0$? You can assume there is a unique twice differentiable solution, but I would like to know what happens when the solution is not unique as well. This seems quite a bit harder than the initial value problem situation.</p> <p>I understand this is probably a very basic question in numerical analysis, but I just couldn't find a good reference that covers this general case. There has been a paper on this subject, <a href="http://www.jstor.org/stable/2004339?seq=3" rel="nofollow">http://www.jstor.org/stable/2004339?seq=3</a>, but I found their proof for the boundary value problem difficult to follow (note that the proof does not mention a or b at all). </p> http://mathoverflow.net/questions/36816/representation-theoretic-interpretation-of-jack-polynomials representation theoretic interpretation of Jack polynomials John Jiang 2010-08-26T22:15:48Z 2012-04-04T12:34:37Z <p>Monomial symmetric polynomials on $n$ variables $x_1, \ldots x_n$ form a natural basis of the space $\mathcal{S}_n$ of symmetric polynomials on $n$ variables and are defined by additive symmetrization of the function $x^{\lambda} = x_1^{\lambda_1} x_2^{\lambda_2} \ldots x_n^{\lambda_n}$. Here $\lambda$ is a sequence of $n$ nonnegative numbers, arranged in non-increasing order, hence can also be viewed as partition of some integer with number of parts $l(\lambda) \le n$.</p> <p>Power sum polynomials $p_\lambda$ on $n$ variables also form a basis for $\mathcal{S}_n$ </p> <p>and are defined as $p_\lambda = \prod_{i=1}^n p_{\lambda_i}$, where $p_r = \sum_{i=1}^n x_i^r$. </p> <p>Schur functions $s_\lambda$ (polynomials) form a basis of the space of symmetric polynomials, indexed by partitions $\lambda$ of at most $n$ parts, and are characterized uniquely by two properties:</p> <ol> <li><p>$\langle s_\lambda, s_\mu \rangle = 0$ when $\lambda \neq \mu$, where the inner product is defined on the power sum basis by $\langle p_\lambda, p_\mu \rangle = \delta_{\lambda,\mu} z_\lambda$, and $z_\lambda = \prod_{i=1}^n i^{\alpha_i} \alpha_i !$, where $\alpha_i$ is the number of parts in $\lambda$ whose lengths all equal $i$. Notice $n!/z_\lambda$ is the size of the conjugacy class in the symmetric group $S_{\sum \lambda_i}$ whose cycle structure is given precisely by $\lambda$. </p></li> <li><p>If one writes $s_\lambda$ as linear combination of $m_\mu$'s, then the $m_\lambda$ coefficient is $1$ and $m_\mu$ coefficients are all $0$ if $\mu > \lambda$, meaning the partial sums inequality $\sum_{i=1}^k \mu_i \ge \sum_{i=1}^k \lambda_i$ hold for all $k$ and is strict for at least one $k$. Thus one can say the transition matrix from Schur to monomial polynomial basis is upper triangular with $1$'s on the diagonal.</p></li> </ol> <p>Jack polynomials generalize Schur polynomials in the theory of symmetric functions by replacing the inner product in the first characterizing condition above with $\langle p_\lambda, p_\mu \rangle = \delta_{\lambda, \mu} \alpha^{l(\lambda)} z_{\lambda}$. The second condition remains the same. It can be thought of as an exponential tilting of the Schur polynomials, and in fact it is intimately connected with the Ewens sampling distribution with parameter $\alpha^{-1}$, a 1-parameter probability measure on $S_n$ or on the set of partitions of $n$ that generalize the uniform measure and the induced measure on partitions respectively.</p> <p>It turns out that the theory of Schur polynomials has connections with classical representation theory of the symmetric group $S_n$. For instance the irreducible characters of $S_n$ are related to the change of basis coefficient from Schur polynomials to power sum polynomials in the following way:</p> <p>if we write $s_\lambda = \sum_{\mu} c_{\lambda,\mu} p_\mu$, then $$\chi_\lambda(\mu) = c_{\lambda,\mu} z_\lambda^{-1.}$$. </p> <p>These are eigenfunctions of the so-called random transposition walk on $S_n$, when viewed as a walk on the space of partitions. The eigenfunctions of the actual random transposition walk on $S_n$ are proportional to the diagonal elements of $\rho$, $\rho$ ranges over all irreducible representations of $S_n$. </p> <p>The characters $\chi_\lambda$ admit natural generalization in the Jack polynomial setting: simply take the transition coefficients from the Jack polynomials to the poewr sum polynomials. And these when properly normalized indeed gives the eigenfunctions for the so-called metropolized random transposition walk that converges to the Ewens sampling distribution, which is an exponentially tilted 1-parameter family of uniform measure on $S_n$. </p> <p>My question is, what is the analogue of the diagonal enties of the representations of $\rho$ in the Jack case? Certainly they will be functions on $S_n$. </p> http://mathoverflow.net/questions/92195/orbit-of-the-identity-matrix-under-lie-group-algebra-actions Orbit of the identity matrix under Lie group algebra actions John Jiang 2012-03-25T19:33:36Z 2012-03-25T20:06:42Z <p>I would like an explicit description of $\mathbb{R} SO(n) I_n$, i.e., the image of the identity under the action of the group algebra of $SO(n)$ by left multiplication. Equivalently, what is an explicit description of the vector space $\{\sum_i c_i A_i: A_i \in SO(n)\}$? Presumably this is well-known and in the context of all compact simple Lie groups? I need this to understand some continuous state space Markov chain of particle systems. </p> <p>Edit: to those who vote to close, please state reason. If you have a one-liner answer, why not give it a shot? I will close it myself when I see a satisfactory response. </p> http://mathoverflow.net/questions/92118/identify-a-curious-subgroup-in-un identify a curious subgroup in $U(n)$ John Jiang 2012-03-24T23:47:52Z 2012-03-25T19:26:02Z <p>Consider the following element $A$ in $U(n)$: $$\begin{pmatrix} 1/2(1+z) &amp; 1/2(1-z) &amp; \\ 1/2(1-z) &amp; 1/2(1+z) &amp; \\ &amp; &amp;I_{n-2} \end{pmatrix},$$ where $|z| = 1$.</p> <p>Now conjugate $A$ by permutation matrices $S$, i.e., $S(i,j) =1$ if $\sigma(i) = j$ for a fixed $\sigma \in S_n$. What group does $S A S^{-1}$ generate? What is its dimension?</p> <p>Finally let $A' = \begin{pmatrix} i &amp; -i &amp; \\ -i &amp; i &amp; \\ &amp; &amp; 0_{n-2} \end{pmatrix}$ be the associated Lie algebra element to $A$ (i.e., derivative with respect to $z$ at $z = 1$; notice the condition $|z|=1$). Can one give an explicit basis of the Lie algebra closure generated by $S A S^{-1}$, with each element $B$ in the basis of the form $Ad(Ad(\ldots Ad(Ad(Ad(Ad(S_1,A),Ad(S_2,A))\ldots, Ad(S_k,A)),Ad(S_0,A'))$, where $S_i$, $i=0,1,\ldots, k$ are permutation matrices. </p> <p>Edit: I now have the following conjecture regarding the Lie algebra: It is simply given by $\{A \in \mathfrak{u}(n): \sum_j A_{ij} = 0, \text{ for all }i \in [n]\}$, so it looks like the dimension should be $n^2 - 2n+1$. If so the group will be $U(n-1)$ acting on $V = \{z_1 + \ldots + z_n = 0\}$. The construction of explicit basis by Adjoint action remains. </p> http://mathoverflow.net/questions/92118/identify-a-curious-subgroup-in-un/92186#92186 Answer by John Jiang for identify a curious subgroup in $U(n)$ John Jiang 2012-03-25T18:31:47Z 2012-03-25T19:26:02Z <p>It is clear from the structure of the generators $S A S^{-1}$ that the resulting group lies in the group $U(n-1) \subset U(n) \subset SO(2n)$, where the last inclusion is through the classical identification of $I$ with $1$ and $J$ with $i$, and $U(n-1)$ acts on the complex vector space $V = \{z_1 + \ldots + z_n = 0\}$. </p> <p>So to show the generated group $G$ is isomorphic to $U(n-1)$, it suffices to show that the Lie algebra has at least $(n-1)^2$ linearly independent elements. One set of linear independent elements are $Ad(S, A'):= S A' S^{-1}$, for $S$ corresponding to $\sigma \in S_n$. These provide $\binom{n}{2}$ elements. </p> <p>So we still need $\binom{n-1}{2}$ other elements. Those are provided by $Ad(Ad(S_{ij},A'),Ad(S_{jn}, A'))$, $1\le i &lt; j \le n-1$, where $S_{ij}$ stands for the permutation matrix corresponding to the transposition $(ij)$. So we have identified a basis of $\mathfrak{g}$ using only Adjoint action on $A'$ by the generator elements.</p> http://mathoverflow.net/questions/92126/an-innocent-looking-subgroup-of-un An innocent looking subgroup of $U(n)$ John Jiang 2012-03-25T03:18:52Z 2012-03-25T04:19:06Z <p>Consider the Lie subalgebra of $\mathfrak{u}(n)$ given by $L = \{A \in \mathfrak{u}(n): \sum_{j=1}^n A_{ij} = 0 \text{ for all } i \in [n]\}$. What is its dimension? What does the corresponding Lie subgroup of $U(n)$ look like? </p> <p>Edit: I believe the dimension is $(n-1)^2$, by explicitly determining the number of independent constraints. I highly suspect the subgroup is $U(n-1)$, acting on the subspace $\{z_1 + \ldots + z_n = 0\}$.</p> http://mathoverflow.net/questions/91581/probability-of-a-set-of-random-vectors-over-finite-field-being-a-spanning-set Probability of a set of random vectors over finite field being a spanning set John Jiang 2012-03-19T00:15:24Z 2012-03-19T06:28:01Z <p>Suppose I have a set of random vectors $f(a_1, \ldots, a_\ell) := (v_1, \ldots, v_m) \subset F_p^n$, $m \ge n$, given by a matrix valued polynomial function $f$, where the $a_i$'s are independent, uniformly distributed in $F_p \setminus {0}$, and each component of $f$ consists of some polynomial in the $a_i$'s whose coefficients are all in ${0,1}$, and such that the degree of each variable is at most $1$. For example, the following set of random vectors fits the discription:</p> <p>$$v_1 = (a_1, a_2 a_1, a_3); v_2 = (a_2, a_1 a_2 a_3 + a_2, 0); v_3 = (a_1 + a_2, a_2 + a_3, a_3 a_1)$$.</p> <p>Now consider the quantity $\pi_p = \mathbb{P}(f(a_1, \ldots, a_\ell) \text{ spans } F_p^n)$. My question is, is $\pi_p$ monotone non-decreasing in $p$? If not can one give a counterexample? The motivation comes from a recent result of Yuval Peres and Allan Sly (Arxiv preprint arXiv:1105.4402, 201) giving the right order of mixing time for the most natural random walk on uni-upper triangular matrices over $F_p$. Knowing the above will extend their result of $\mathcal{O}(n^2)$ to $p$ that grows with $n$, which is highly anticipated. </p> <p>Edit: Will Sawin below essentially solved an earlier version of this problem, where I forgot to state the degree condition on the $a_i$'s.</p> http://mathoverflow.net/questions/90862/what-are-some-examples-of-ingenious-unexpected-constructions/91221#91221 Answer by John Jiang for What are some examples of ingenious, unexpected constructions? John Jiang 2012-03-14T21:07:21Z 2012-03-14T21:07:21Z <p>The discovery and construction of Schramm-Loewner evolutions by Oded Schramm, and the subsequent proofs that many random discrete curves (self-avoiding walk (still open), interface of critical percolation, contour curves of uniform spanning trees, interface of critical Ising model, loop-erased random walk) converge to them with various parameters, is perhaps the most celebrated result in probability theory today. </p> http://mathoverflow.net/questions/90160/root-system-generalizations-of-sekiguchi-debiard-aka-laplace-beltrami-operators root system generalizations of Sekiguchi-Debiard (aka Laplace-Beltrami) operators John Jiang 2012-03-03T23:23:06Z 2012-03-04T01:13:45Z <p>For the root system $A_n$, taking the limit $q = t^\alpha$ and $t \to 1$, and letting $Y = (t-1) X -1$ one obtains from the Macdonald operator the so-called Sekiguchi-Debiard operator: $$D_\alpha(X) = a_\delta(x)^{-1} \sum_{w \in S_n} \epsilon(w) x^{w \delta} \prod_{j=1}^n (X + (w \delta)_j + \alpha x_j \partial_j).$$</p> <p>The eigenfunctions of this family of operators indexed by powers of $X$ are called symmetric Jack polynomials with parameter $\alpha$. It is well-known that Heckman-Opdam polynomials generalize Jack symmetric polynomials to other root systems. My question is, are there generalizations of the Laplace-Beltrami operators to root systems as well? The operator is defined in $A_{n+1}$ up to affine transformation as $$D_\alpha^2 f = \frac{\alpha}{2} \sum_{i=1}^n (x_i \partial_i)^2 + \sum_{i &lt; j} \frac{x_i^2 \partial_i - x_j^2 \partial_j}{x_i - x_j}.$$</p> <p>Applying this operator to power sum polynomial $p_\lambda$ has a natural random walk interpretation, see the paper "a probabilistic interpretation of Macdonald polynomials" for detail.</p> <p>Edit: I figured out that the notation $\partial_\alpha$ with respect to a weight $\alpha$ is indeed the same as $x_i \partial_i$ since the latter is differentiation with respect to maximal torus element, i.e., $\partial_\alpha e^{k \alpha} = k e^{k \alpha} = \partial_{x_i} x_i^k$ if $e^\alpha = x_i$. Thus Heckman-Opdam operators do provide the Laplace Beltrami operators in the other root systems. </p> http://mathoverflow.net/questions/80619/some-weird-relations-among-beta-random-variables some weird relations among beta random variables John Jiang 2011-11-10T18:55:13Z 2012-03-02T00:59:14Z <p>Let $X_i, Y_i, Z_i$ be three iid family of standard normal random variables. Then the following random variable is distributed the same as the first coordinate of a uniform $2$-sphere:</p> <p>$$\frac{X_1}{\sqrt{X_1^2+X_2^2}} \frac{Y_1}{\sqrt{Y_1^2+Y_2^2}} + \frac{X_2}{\sqrt{X_1^2+X_2^2}} \frac{Y_2}{\sqrt{Y_1^2+Y_2^2}} \frac{Z_1}{\sqrt{Z_1^2+Z_2^2+Z_3^2}}$$.</p> <p>Of course, $\frac{X_1}{\sqrt{X_1^2+X_2^2}}$ squared has $\beta(1/2,1/2)$ distribution. </p> <p>I know this is true by looking at the matrix product $\gamma_{1,2}(\theta_1) \gamma_{2,3}(\theta_2) \gamma_{1,3}(\theta_3)$ where $\gamma_{i,j}(\theta)$ is the rotation matrix by $\theta$ in the $i \wedge j$ plane in $\mathbb{R}^3$. But is there a probabilistic way to prove it using beta-gamma algebra or otherwise?</p> http://mathoverflow.net/questions/76859/what-is-known-about-zero-sets-of-schur-polynomials/89226#89226 Answer by John Jiang for What is known about zero-sets of Schur polynomials? John Jiang 2012-02-22T20:37:02Z 2012-02-22T20:37:02Z <p>For $k :=|\lambda| \ge r$, the statement that all $s_\lambda(x_1, \ldots, x_r)$ vanish is equivalent to all the elementary polynomials $e_j(x_1, \ldots, x_r) := \sum_{i_1 &lt; \ldots &lt; i_j} x_{i_1} \ldots x_{i_j}$, $j \le r$, vanish, since the latter form a basis of the algebra $\Lambda_r$. But this is true if and only if all the $x_1, \ldots, x_r$ are zero, since $e_j$ is the $y^{r-j}$ coefficient of a degree $r$ polynomial, i.e., $$\prod_{i=1}^r (y - x_i) = \sum_{j=0}^r (-1)^j e_j(x_1, \ldots, x_r) y^{r-j}.$$ If all the $e_j$'s are zero, except $e_0 \equiv 1$, then it must be $y^r$, hence all the $x_i$'s vanish.</p> <p>For $k &lt; r$, the zero set consists of roots of polynomials of the form $y^r + c_{k+1} y^{r-k-1} + c_{k+2} y^{r-k-2} + \ldots + c_r$.</p> http://mathoverflow.net/questions/88488/meaning-of-connected-part-cumulant-expectation-in-physics-algebraic-geometry-a Meaning of connected part / cumulant expectation in physics, algebraic geometry and random matrix literature John Jiang 2012-02-15T05:03:32Z 2012-02-15T05:03:32Z <p>I have seen more than once the following notation in algebraic geometry or physics papers: $$\langle tr \frac{1}{M-x_1} \ldots tr \frac{1}{M-x_k} \rangle_c$$ where the angle brackets stand for expectation with respect to some measure (typically given by a Hamiltonian) on the space of matrices $M$, and the author also mentions that the subscript $c$ stands for the connected part of cumulant.I am always puzzled by this last remark. Can anyone point me to the relevant place that explains carefully what the subscript $c$ means in actual computation? I sorta know what cumulant means, but why does it have to do with connected part? </p> http://mathoverflow.net/questions/88474/simplify-an-expression-of-symmetric-polynomials-related-to-macdonald-operators Simplify an expression of symmetric polynomials related to Macdonald operators John Jiang 2012-02-15T00:15:39Z 2012-02-15T00:27:38Z <p>I am trying to understand the action of the second Macdonald operator $D_{q,t}^2$ on power sum polynomials $p_\lambda$, i.e., express $D_{q,t}^2 p_\lambda$ in terms of other $p_\mu$. After many steps of simplification, I reduce the problem to calculating the following expression: $$B_n(u,v) = \frac{1}{4} \sum_{s=0}^{n-2} \sum_{r=0}^s \sum_{p=0}^s (-1)^{s+p} (t^2-1)^{n-2-s} (t-1)^s \binom{s}{r}$$ $$(s_{(r+p+u-s,v-r)} + s_{(r+p+v-s,u-r)} + s_{(p+u-r,r+v-s)} + s_{(p + v-r,u+r-s)}) e_{s-p}$$ Here $s_{a,b}$ denotes the Schur polynomial indexed by the integer vector $(a,b)$ and $e_a$ denotes the elementary symmetric polynomial of degree $a$. When $u=v=0$, the answer is pretty nice: $$B_n(0,0) = \frac{(t^2-1)^{n-1} - (t-1)^{n-1}}{t^2 -t}$$.</p> <p>I am wondering if someone knows how to deal with convolution product of Schur and elementary polynomials as stated. One thing I tried was to use the Jacobi-Trudy identity to write $s_{(a,b)} = h_a h_b - h_{a+1} h_{b-1}$, where $h_\lambda$ is the complete symmetric polynomial indexed by the partition $\lambda$.</p> <p>Then use the relation between the generating functions of elementary and complete symmetric polynomials $E(-t) H(t) =1$. Unfortunately the expression above as $p$ ranges from $0$ to $s$ is only a partial convolution. Maybe I am missing something, but I am eager to learn from experts. The similar computation for $D_{q,t}^1$ can be found in the paper by Diaconis and Ram: stat.stanford.edu/~cgates/PERSI/papers/100726macdonaldpoly.pdf</p> http://mathoverflow.net/questions/87877/jacobis-equality-between-complementary-minors-of-inverse-matrices Jacobi's equality between complementary minors of inverse matrices John Jiang 2012-02-08T09:30:41Z 2012-02-08T10:04:49Z <p>What's a quick way to prove the following fact about adjugates of an invertible matrix $A$ and its inverse?</p> <p>Let $A[I,J]$ denote the submatrix of an $n \times n$ matrix $A$ obtained by keeping only the rows indexed by $I$ and columns indexed by $J$. Then </p> <p>$$|\det A[I,J]| = | (\det A)^{-1} \det A^{-1}[I^c,J^c]|,$$ where $I^c$ stands for $[n] \setminus I$, for $|I| = |J|$. It is trivial when $|I| = |J| = 1$ or $n-1$. This is apparently proved by Jacobi, but I couldn't find a proof anywhere in books or online. Horn and Johnson listed this as one of the advanced formulas in their preliminary chapter, but didn't give a proof. In general what's a reliable source to find proofs of all these little facts? I ran into this question while reading Macdonald's book on symmetric functions and Hall polynomials, in particular page 22 where he is explaining the determinantal relation between the elementary symmetric functions $e_\lambda$ and the complete symmetric functions $h_\lambda$. </p> <p>I also spent 3 hours trying to crack this nut, but can only show it for diagonal matrices :(</p> <p>Edit: It looks like Ferrar's book on Algebra subtitled determinant, matrices and algebraic forms, might carry a proof of this in chapter 5. Though the book seems to have a sexist bias.</p> http://mathoverflow.net/questions/87770/analogues-of-power-sum-polynomials-for-symmetric-laurent-polynomials analogues of power sum polynomials for symmetric Laurent polynomials John Jiang 2012-02-07T06:48:35Z 2012-02-07T07:41:56Z <p>To deal with root systems of type B C D, one needs to understand symmetric Laurent polynomials $\Lambda$. I am wondering if the naive definition of power sum symmetric Laurent polynomials form a basis in $\Lambda$. More specifically, define $$p_r = \sum_{i=1}^N x_i^r + x_i^{-r},$$ and define power sum symmetric Laurent polynomial indexed by a partition $\lambda \vdash n$ as $$p_\lambda = \prod_{j=1}^n p_{\lambda_i}.$$ What are some standard references that talk about the connection of $\Lambda$ with root systems, as well as Macdonald operators? Thanks.</p> <p>edit: I guess it is obvious that $p_\lambda$ does form a basis, if I restrict to symmetric Laurent polynomials that are also symmetric with respect to $x_i \mapsto x_i^{-1}$. So I would like to change my question to just asking for a reference relating $\Lambda$ with Macdonald operators. </p> http://mathoverflow.net/questions/87629/schur-weyl-duality-for-real-orthogonal-groups-and-relation-to-hyperoctahedral-gro schur weyl duality for real orthogonal groups and relation to hyperoctahedral groups John Jiang 2012-02-06T02:03:18Z 2012-02-06T02:30:52Z <p>I am wondering whether the Lie groups $SO(n)$ and the hyperoctahedral groups $H_n$ form some sort of duality. I am mainly interested in how to parametrize the conjugacy classes of $H_n$ in terms of the dominant weights of root systems of type $B_n$ or $C_n$. So far I haven't found an accessible reference that draws complete analogy with the case of $U(n)$ and $S_n$. I am also interested in the analogue of symmetric functions in the case of $H_n$. Any pointer to literature would be greatly appreciated.</p> http://mathoverflow.net/questions/84853/l2-basis-of-class-functions-on-a-compact-lie-group-that-are-point-wise-small L^2 basis of class functions on a compact Lie group that are point-wise small John Jiang 2012-01-04T03:17:37Z 2012-01-05T13:38:25Z <p>Consider first the torus group $\mathbb{T}^k$. A natural $L^2$ basis is given by the 1-dimensional complex representations: $(\theta_1, \ldots, \theta_k) \mapsto e^{i \sum_j c_j \theta_j}$ for integral $c_j$'s. This basis has also the nice property that each element in it is point-wise bounded by $1$ in absolute value. In general, an $L^2$ function takes the form $g = \sum_n q_n f_n$ where $f_n$ are the basis functions described above. $g$ can have arbitrary bad $L^\infty$ bound, since the only requirement is that $\sum q_n^2 = 1$, but $\sum_n q_n$ can be $\infty$. </p> <p>Now my question is, for the unitary or orthogonal group, what is the smallest uniform $L^\infty$ bound one can achieve on an orthogonal basis on the space of class functions? For the unitary groups, any orthogonal basis of class functions forms an orthogonal basis in the space of symmetric functions of the eigenvalues under the inner product such that the Schur polynomials are orthonormal. So the question can be phrased in terms of symmetric functions. For the orthogonal groups, one orthogonal (not normalized) basis of class functions comes from $\prod_{i=1}^n (Tr X^i)^{\alpha_i}$ where for some odd $i$, $\alpha_i$ is odd (I am not entirely sure about this, but see <a href="http://stat.stanford.edu/~cgates/PERSI/papers/random_matrices.pdf" rel="nofollow">this paper</a> theorem 4).</p> <p>Edit: what about all $L^2$ functions in general? Can one get a uniform $L^\infty$ bound on the basis of size $e^{\mathcal{O}(n)}$ for $SO(n)$ or $U(n)$?</p> http://mathoverflow.net/questions/84915/rigidity-of-eigenvalues-of-circular-ensemble rigidity of eigenvalues of circular ensemble John Jiang 2012-01-04T23:05:54Z 2012-01-04T23:05:54Z <p>Given a circular unitary ensemble, with the following joint density:</p> <p>$p(\theta_1,\ldots, \theta_n) = Z_n \prod_{j &lt; k} |e^{i \theta_j} - e^{i \theta_k}|^2$, </p> <p>is the following statement true? With high probability the eigenvalues are within distance $\mathcal{O}(1)$ from the evenly spaced set of $n$ points $(0,2\pi/n, 4\pi/n, \ldots, 2(n-1)\pi/n)$, rotated by some angle $\theta$. More precisely, is it true that</p> <p>$P[\int_\alpha d(\{\theta_1, \ldots, \theta_n\}, \{0, 2\pi/n, \ldots, 2(n-1)\pi/n\} + \alpha (\text{mod }2\pi)) &lt; C] \to 1$ for some sufficiently large constant $C$?</p> <p>Here the distance is the induced Riemannian distance on $\mathbb{T}^n/ S_n$, where the action of $S_n$ on $\mathbb{T}^n$ is permutation of the coordinates. </p> <p>I know Erdos Schlein Yau have proved a rigidity theorem for Wigner ensembles, but their result is slightly weaker than what I need. It seems natural to investigate this question for the exactly solvable case.</p> http://mathoverflow.net/questions/84848/volume-of-compact-simple-lie-groups-under-the-natural-euclidean-embedding volume of compact simple Lie groups under the natural Euclidean embedding John Jiang 2012-01-04T01:01:12Z 2012-01-04T08:48:17Z <p>I am looking for a quick reference for the volume formula for all the compact simple Lie groups embedded as matrix groups in the natural way. The one I care most for are the real orthogonal groups. I don't see how to deduce them from the Weyl integration formula. I believe they are of the order diameter raised to the power of the dimension. So for instance for $SO(n)$ the volume should be of order $n^{\Theta(n^2)}$. </p> http://mathoverflow.net/questions/84858/riemannian-hausdorff-distance-between-two-conjugacy-classes-in-a-compact-lie-grou Riemannian Hausdorff distance between two conjugacy classes in a compact Lie group John Jiang 2012-01-04T03:49:15Z 2012-01-04T06:14:32Z <p>I am interested in the distance between two conjugacy classes in a group like $SO(n)$. However let's consider $U(n)$ for simplicity. My conjecture is that the Hausdorff distance between the conjugacy classes of $\Lambda_1$ and $\Lambda_2$, both of which are diagonal matrices, is given by the distance of the eigenvalues $(\lambda_1, \ldots, \lambda_n)$ and $(\lambda^{(2)}_1, \ldots, \lambda^{(2)}_n)$ as elements in the homogeneous space $\mathbb{T}^n / S_n$, where the action of the symmetric group $S_n$ on $\mathbb{T}^n$ is by permutation of the coordinates. Thus by translation invariance, the closest element to $X$ in the conjugacy class of $Y$ is any one that can be simultaneously diagonalized with $X$. However I don't know how to prove this plausible statement. </p> http://mathoverflow.net/questions/84603/rotation-along-a-coordinate-axis-and-kac-random-walk rotation along a coordinate axis and Kac random walk John Jiang 2011-12-30T19:10:56Z 2011-12-31T00:56:55Z <p>With due respect, I am profoundly puzzled by an amazing <a href="http://www.springerlink.com/index/1y9bkugfdldk6wek.pdf" rel="nofollow">paper</a> of David K. Maslen on the eigenvalues of the Kac random walk on $SO(n)$. The Kac walk is essentially given by choosing a random pair of coordinates $i \neq j \in [n]$ and rotating by a uniformly random angle about the 2-plane $i \wedge j$. In section 3.1, he defined an auxilliary random walk by "the following procedure of choosing random elements in $SO(n)$: first choose a random coordinate axis in $\mathbb{R}^n$, then choose a random angle $\theta$ uniformly in $[0,2\pi)$. then the rotation by $\theta$ about the chosen axis is the chosen random element of $SO(n)$."</p> <p>To me a random rotation always occurs along a 2-dimensional plane. So I have no idea how to rotate along a coordinate axis.</p> <p>He then went on to define the Fourier transform of this random rotation in an irreducible representation space. " Let $P_{\hat{i}}$ be the self-adjoint projection from $V_\rho$ onto the subspace of $\sigma SO(n-1) \sigma^{-1}$-invariant vectors, where $\sigma$ is the transposition matrix for $(i,n)$. Then $\frac{1}{n}(P_{\hat{1}} + \ldots + P_{\hat{n}})$ is the Fourier transform of the measure we defined for a uniform random rotation in a random coordinate hyperplane. " If someone can make sense of the description of the random rotation defined in the first paragraph, please also enlighten me why the Fourier transform is of the above form. I assume he is talking about real representation here?</p> <p>In particular, I see that if $\rho$ is the natural representation, then the Fourier transform would simply be $1/n$ times the identity matrix. </p> <p>As a final remark, the Kac walk is not constant on conjugacy classes, hence standard representation theory isn't useful for its analysis. I must say that Maslen's paper contains such a rich amount of new discoveries about representation theory and semi-standard Young tableau stuff that I am eager to find out its connection with standard representation theory on compact simple Lie groups. Also if someone thinks he or she understands this paper quite thoroughly, please help me out by getting in touch with me.</p> http://mathoverflow.net/questions/84471/bounding-the-probability-that-a-polynomial-is-near-0 bounding the probability that a polynomial is near 0 John Jiang 2011-12-28T19:54:37Z 2011-12-28T20:47:59Z <p>Given a polynomial $p(x_1,\ldots, x_k)$ in $k$ variables with maximum degree $n$, and $x_1,\ldots, x_k \in [0,1]$. Suppose $\max_{x \in [0,1]^k} p(x) = 1$, can we get an upper bound on the probability $\mathbb{P}(|p(x)| &lt; \epsilon)$ where we treat the product space $[0,1]^k$ as a natural probability space? Using techniques from <a href="http://www.its.caltech.edu/~kaloshin/research/dioph-SO3.pdf" rel="nofollow">this</a> paper: one can show that the above probability is small provided $\epsilon$ is of tower exponential order $e^{-e^n}$. But I need $\epsilon$ to be of poly-exponential order, i.e., $e^{-(nk)^c}$. Looking at the polynomial $x_1^n x_2^n \ldots x_k^n$, one deduce by central limit theorem that $\mathbb{P}(|p(x)| &lt; \epsilon)$ is small provided $\epsilon = \mathcal{O}(e^{-kn})$. But I can't prove that this polynomial has the the highest probability of staying near zero. </p> <p>If needed, one can also bound $\epsilon$ in terms of the total degree, i.e., the sum of degrees of all the monomial terms in $p$. </p> http://mathoverflow.net/questions/81728/probability-of-overlapping-of-repetitive-events/81743#81743 Answer by John Jiang for Probability of overlapping of repetitive events John Jiang 2011-11-23T19:17:38Z 2011-11-23T19:17:38Z <p>Consider a circle of length $t+\ell$. Then I think your problem is asking if I drop $N$ points uniformly at random onto that circle, what is the probability that at least $m$ of them are in an interval of length $\ell$. When $m/(N-m) >> \ell / t$, so that the tail event that there are $m$ points in a fixed interval of size $\ell$ is exponentially small, you can use the union bound</p> <p>$$N \sum_{j\ge m} (t/(t+\ell))^{N-j} (\ell/(t+\ell))^j \binom{N}{j}$$. </p> <p>You could also consider using the Brownian bridge (from $(0,0)$ to $(1,1))$ approximation of the partial sum $\sum_{j=1}^k X_j$ where $X_j$ is the distance between the $j$th and $j+1$st points in say the clockwise direction, with a particular chosen first point $p$. Then the question roughly becomes what is the chance that there is some $s \in [0,1]$ such that $B_{s + \ell/(\ell + t)} - B_s$ exceeds $m/N$. So for instance if $m/N \le \ell / (t + \ell) + o(\sqrt{\ell / (t + \ell)})$ then this probability should be very close to $1$. </p> http://mathoverflow.net/questions/127327/extension-of-the-jacobi-triple-product-identity Comment by John Jiang John Jiang 2013-04-13T05:23:51Z 2013-04-13T05:23:51Z if you typed all this up yourself i can see you do love math. http://mathoverflow.net/questions/126739/sum-k0-infty-frac1km-k-equiv-1-for-m2/126744#126744 Comment by John Jiang John Jiang 2013-04-07T22:03:04Z 2013-04-07T22:03:04Z @quid: to be honest I didn't check the comments above when I wrote the answer either:) http://mathoverflow.net/questions/126789/does-the-prime-number-theorem-prove-that-the-primes-cannot-be-exactly-identified Comment by John Jiang John Jiang 2013-04-07T21:44:01Z 2013-04-07T21:44:01Z The question is a natural one to ask for someone who first saw the statement of the prime number theorem and perhaps didn't have the background to understand its significance. http://mathoverflow.net/questions/126743/what-is-the-solution-to-min-k-k-fracbk-n-lfloor-bk-n-rfloor Comment by John Jiang John Jiang 2013-04-07T20:28:37Z 2013-04-07T20:28:37Z I interpret it as b,n are given. All I can say is that the solution is between $\log n / \log b$ and $2 \log n/ \log b$. http://mathoverflow.net/questions/126677/linear-recurrence-relations-in-2-variables-with-variable-coefficients/126685#126685 Comment by John Jiang John Jiang 2013-04-07T01:03:32Z 2013-04-07T01:03:32Z @mkreisel: yes you are right. The lower half of the plane is the more interesting part. There is more structure to the problem than I thought. http://mathoverflow.net/questions/125865/a-curious-definition-in-measure-theory Comment by John Jiang John Jiang 2013-03-30T13:53:40Z 2013-03-30T13:53:40Z How can we talk about uniform convergence of a sequence of functions at a \emph{single} point? http://mathoverflow.net/questions/125805/a-question-on-a-variant-of-hardys-inequality Comment by John Jiang John Jiang 2013-03-30T13:48:26Z 2013-03-30T13:48:26Z @Higgs: I think the idea is arithmetic geometric inequality: $\sqrt{ab} \le (a+b)/2$, which implies $\frac{|u|^{\alpha r_1 + (1-\alpha) r_2}}{|x|^{\alpha s_1 + (1- \alpha) s_2}} \le \frac{|u|^{r_1}}{|x|^{s_1}} + \frac{|u|^{r_2}}{|x|^{s_2}}$. http://mathoverflow.net/questions/126003/does-partial-overline-partialf0-imply-f-equiv-c-for-particular-kind-of-f/126011#126011 Comment by John Jiang John Jiang 2013-03-30T13:40:52Z 2013-03-30T13:40:52Z This is a nice problem to learn about Hartog's theorem! http://mathoverflow.net/questions/96866/random-permutation-with-fixed-cycle-length Comment by John Jiang John Jiang 2012-05-14T05:25:18Z 2012-05-14T05:25:18Z @gmath: It would also be nice to have the references of the two results you mentioned. http://mathoverflow.net/questions/7834/undergraduate-differential-geometry-texts/8048#8048 Comment by John Jiang John Jiang 2012-05-11T05:57:45Z 2012-05-11T05:57:45Z Just like calculus on manifolds should be read before Spivak 5 volume, perhaps Ted's multivariable mathematics text serves as a great prerequisite builder for his differential geometry text. http://mathoverflow.net/questions/96109/analogues-of-the-knuth-and-forgotten-equivalences-on-permutations-have-they-been Comment by John Jiang John Jiang 2012-05-06T03:55:08Z 2012-05-06T03:55:08Z Wow you are 23 and already mentoring students!? http://mathoverflow.net/questions/84858/riemannian-hausdorff-distance-between-two-conjugacy-classes-in-a-compact-lie-grou/84862#84862 Comment by John Jiang John Jiang 2012-05-05T02:59:58Z 2012-05-05T02:59:58Z @Anton, I guess it depends on how you put a Riemannian metric structure on the orbit space right? http://mathoverflow.net/questions/94807/are-the-real-components-of-s-roots-subharmonic Comment by John Jiang John Jiang 2012-04-22T05:03:26Z 2012-04-22T05:03:26Z Why is $u(z)$ a well-defined function? You can define it in regions away from the zeros of $f(z)$, but in any neighborhood of $a zero of$f(z)\$ there is no way to define it in a continuous way in general, so I wouldn't expect it to be subharmonic, however it is defined. http://mathoverflow.net/questions/36816/representation-theoretic-interpretation-of-jack-polynomials Comment by John Jiang John Jiang 2012-04-04T23:34:11Z 2012-04-04T23:34:11Z @Alexander and Steven: the Jack polynomials here are symmetric ones, defined by certain upper triangularity condition and orthogonality with respect to appropriate inner product.The Calogero-Moser operators might be the same as the Laplace-Beltrami operators in a different disguise, but I haven't checked. http://mathoverflow.net/questions/92195/orbit-of-the-identity-matrix-under-lie-group-algebra-actions Comment by John Jiang John Jiang 2012-03-26T01:40:20Z 2012-03-26T01:40:20Z @Jim: No problem. I am glad ppl here helped me understand the actions on matrices, which is crucial for my study of the Kac random walk. See for instance: <a href="http://arxiv.org/abs/0905.1539" rel="nofollow">arxiv.org/abs/0905.1539</a>