User jzadeh - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T01:45:16Z http://mathoverflow.net/feeds/user/10632 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/87098/does-the-existence-of-an-asymtpotic-density-imply-the-existence-of-a-measure-on-i Does the existence of an asymtpotic density imply the existence of a measure on infinite dimensional (path) space? jzadeh 2012-01-31T01:36:21Z 2012-01-31T02:05:08Z <p>This question is related to the following question <a href="http://mathoverflow.net/questions/59513/question-about-a-limit-of-gaussian-integrals-and-how-it-relates-to-path-integrati" rel="nofollow">http://mathoverflow.net/questions/59513/question-about-a-limit-of-gaussian-integrals-and-how-it-relates-to-path-integrati</a></p> <p>A couple of authors have observed that composing a random walk an infinite number of times gives an asymptotic time invariant density. The original reference is "Fractional diffusion equations and processes with randomly varying time" Enzo Orsingher, Luisa Beghin <a href="http://arxiv.org/abs/1102.4729" rel="nofollow">http://arxiv.org/abs/1102.4729</a>. Roughly speaking I am curious if this notion of iterating a random walk infinitely often and that fact that this iteration converges to some fixed density imply the existence of an infinite dimensional measure.</p> <p>The line (3.14) of Orsingher and Beghins paper reads for $t > 0$ and $x \in \mathbb{R}$ $$(*) \qquad\lim_{n \rightarrow \infty} 2^{n} \int_{0}^{\infty} \ldots \int_{0}^{\infty} \frac{e^{\frac{-x^2}{2z_1}}}{\sqrt{2 \pi z_1}} \frac{e^{\frac{-{z_1}^2}{2z_2}}}{\sqrt{2 \pi z_2}} \ldots \frac{e^{\frac{-{z_n}^2}{2t}}}{\sqrt{2 \pi t}} \mathrm{d}z_1 \ldots \mathrm{d}z_n = e^{-2 |x|}$$</p> <p>Since (*) is very similar to normalizations carried out in computing the propagator in quantum mechanics or just the formulations of path integrals in general I was curious about how rigorous we could make the following statements. Also the way I have seen these type of constructions carried out is either via the the standard definition of Wiener measure on finite dimensional "cylinder sets" or some application of Bochner-Milnos combined with a normalization of Gaussian measure on $\mathbb{R}^n$. So I am wondering if this is something contained within the construction of wiener measure or other infinite dimensional measures on Banach spaces.</p> <p>1) Does (*) imply the existence of a measure on the space of continuous functions with finite support (paths)?</p> <p>2) If such a measure does exist is it equivalent to Wiener measure?</p> http://mathoverflow.net/questions/59513/question-about-a-limit-of-gaussian-integrals-and-how-it-relates-to-path-integrati Question about a Limit of Gaussian Integrals and how it relates to Path Integration (if at all)? jzadeh 2011-03-25T03:20:53Z 2011-04-05T14:34:50Z <p>I have come across a limit of Gaussian integrals in the literature and am wondering if this is a well known result. </p> <p>The background for this problem comes from the composition of Brownian motion and studying the densities of the composed process. So if we have a two sided Brownian motion $B_1(t)$ we replace t by an independent Brownian motion $B_2(t)$ and study the density of $B_1(B_2(t))$. If we iterate this composition n times we get the iterated integral in (**) below as an expression for the density of the n times iterated Brownian motion. The result I am interested in is derived in the following paper:</p> <p>The original reference is "Fractional diffusion equations and processes with randomly varying time" Enzo Orsingher, Luisa Beghin <a href="http://arxiv.org/abs/1102.4729" rel="nofollow">http://arxiv.org/abs/1102.4729</a></p> <p>Line (3.14) of Orsingher and Beghins paper reads for $t > 0$</p> <p>$$(**) \qquad\lim_{n \rightarrow \infty} 2^{n} \int_{0}^{\infty} \ldots \int_{0}^{\infty} \frac{e^{\frac{-x^2}{2z_1}}}{\sqrt{2 \pi z_1}} \frac{e^{\frac{-{z_1}^2}{2z_2}}}{\sqrt{2 \pi z_2}} \ldots \frac{e^{\frac{-{z_n}^2}{2t}}}{\sqrt{2 \pi t}} \mathrm{d}z_1 \ldots \mathrm{d}z_n = e^{-2 |x|}$$</p> <ol> <li><p>How do you prove this result without using probability? Edit: there has been a solution posted to 1) using saddlepoint approximation but I am still not clear on how to make the argument rigorous <a href="http://physics.stackexchange.com/q/7552/2757" rel="nofollow">http://physics.stackexchange.com/q/7552/2757</a></p></li> <li><p>I have been studying a slight generalization of ** from the probability side of things and have been trying to use dominated convergence to show the LHS of ** is finite but I am having problems finding a dominating function over the interval $[1,\infty)^n$. Is dominated convergence the best way to just show the LHS of (**) is finite?</p></li> <li><p>Is this a type of path integral (functional integral)? Or is this integrand some kind of kinetic plus potential term arsing in quantum mechanics? Do expressions like (**) ever come up in physics literature?</p></li> </ol> <p>(I tried using the change of variable theorem for Wiener measure to transform (**) into a Wiener integral with respect a specific integrand and have had some success with this.. I think this shows how to compute a Wiener integral with respect to a function depending on a path and not just a finite number of variables but did not see how to take this any further - The change of variable theorem for Wiener Measure was taken from "The Feynman Integral and Feynman's Operational Calculus" by G. W. Johnson and M. L. Lapidus.)</p> http://mathoverflow.net/questions/59748/what-structure-is-needed-to-define-a-gaussian-distribution-on-a-given-space/59752#59752 Answer by jzadeh for What structure is needed to define a Gaussian distribution on a given space? jzadeh 2011-03-27T19:08:50Z 2011-03-27T19:26:20Z <p>One general construction can be found in Revuz and Yor "Continuous Martingales and Brownian Motion" for instance:</p> <p>proposition (1.3)</p> <p>Let $H$ be a separable real Hilbert space. There exist a probability space $\left( \Omega, \mathcal{F}, \mathbb{P} \right)$ and a family $X(h)$, $h \in H$ of random variables on the space such that</p> <p>i) the map $h \rightarrow X(h)$ is linear</p> <p>ii) for each $h$ the random variable $X(h)$ is Gaussian centered and $\mathbb{E} [ X(h)^2] = ||h||_{H}^{2}$</p> <p>Nualarts book "Malliavian Calculus" also starts with the notion of isonormal Gaussian process which is general as well as Adler's books on Gaussian processes.</p> <p>Alternatively you could look at one of T. Hida's books on White noise analysis for a construction based on the bochner-milnos theorem and Nuclear spaces.</p> <p>Sorry none of these have a geometric perspective that I am aware of... </p> http://mathoverflow.net/questions/52448/time-integrals-of-diffusion-processes/52542#52542 Answer by jzadeh for Time integrals of diffusion processes jzadeh 2011-01-19T18:54:40Z 2011-01-20T11:38:06Z <p>My Stochastic calculus professor always used to say "When in doubt use Ito"</p> <p>So let $f(t,x) = t x$ and compute $\partial_t f(t,x) = x$, $\partial_x f(t,x) = t$ and $\partial_{xx} f(t,x) = 0$</p> <p>Now the Ito lemma says for $f$ twice differentiable with respect to $x$ and once differentiable with respect to $t$ then the following formula holds for any Ito process:</p> <p>$f(t, X_t) = f(0,X_0) + \int_{0}^{t}\partial_s f(s,X_s) ds + \int_{0}^{t}\partial_x f(s,X_s) dX_s$ $+ \int_{0}^{t}\partial_{xx} f(s,X_s) d \left&lt; X,X \right>_s$</p> <p>So applying the above fact to the function $f(t,x) = tx$ gives:</p> <p>$t X_t = 0 + \int_{0}^{t}X_s ds + \int_{0}^{t} s dX_s + 0$ or to give you a starting answer $\int_{0}^{t}X_s ds = t X_t - \int_{0}^{t} s dX_s$</p> <p>Edit 3: The following only holds now if $X_t$ is a Gaussian process which is not true in general... So in vague words we have that The (Riemann) integral of an ito process is equal to the difference of two Gaussian processes which should again be Gaussian .... (if all this logic is correct it should then suffice to characterize the processes covariance structure in order to have a complete understanding of the law of the process.)</p> <p>For example one can compute the variance if $X_t$ is standard Brownian motion:</p> <p>$\mathbb{E}[(\int_{0}^{t}X_s ds)^2] = t^2 \mathbb{E}[(X_t)^2] -2t \mathbb{E}[X_t \int_{0}^{t} s dX_s] + \mathbb{E}[(\int_{0}^{t}sdX_s)^2]$</p> <p>By the Ito isometry we have $\mathbb{E}[(\int_{0}^{t}sdX_s)^2] = \int_{0}^{t}s^2ds = t^{3}/3$.</p> <p>To compute $\mathbb{E}[X_t \int_{0}^{t} s dX_s]$ notice first that the Ito integral of a deterministic function is always a Gaussian process. EDIT: Shavi has given that $\mathbb{E}[X_t \int_{0}^{t} s dX_s] = t^2/2$</p> <p>$\mathbb{E}[(\int_{0}^{t}X_s ds)^2] = t^3 - t^3 + \frac{t^{3}}{3} = \frac{t^3}{3}$</p> <p>Computing the covariance $\mathbb{E}[\int_{0}^{t}X_s ds \int_{0}^{u}X_s ds]$ involves dealing with terms $\mathbb{E}[\int_{0}^{t}s dX_s \int_{0}^{u}s dX_s]$ and $\mathbb{E}[X_t X_u]$ which are again probably well known in certain cases (the second term is obviously equal to $min(t,u)$ when $X$ is b.m.) but may be difficult to handle in your general case. </p> <p>Edit 2: To give an approach to answer the question "Is it possible that the integrated processes have equivalent laws?"</p> <p>Since $\int_{0}^{t}X_{s}^{(1)}ds$ and $\int_{0}^{t}X_{s}^{(2)}ds$ are Gaussian processes (we proved this using ito) it suffices to check if there covariance functions $g_{1}(t,u)=\mathbb{E}[\int_{0}^{t}X_{s}^{(1)}ds \int_{0}^{u}X_{s}^{(1)}ds]$ and $g_{2}(t,u) = \mathbb{E}[\int_{0}^{t}X_{s}^{(2)}ds \int_{0}^{u}X_{s}^{(2)}ds]$ are equal for all $t,u >0$ to show that the two processes have equivalent laws.</p> <p>Now applying the result we got above from the ito calculation lets us start computing the covariance:</p> <p>$\mathbb{E}[\int_{0}^{t}X_s ds \int_{0}^{u}X_s ds]$ = $\mathbb{E}[( t X_t - \int_{0}^{t} s dX_s)( u X_u - \int_{0}^{u} s dX_s)]$ $= t u \mathbb{E}[ X_t X_u ] - t \mathbb{E}[X_t \int_{0}^{u} s dX_s ] -u \mathbb{E}[X_u \int_{0}^{t} s dX_s ]$ $+\mathbb{E}[\int_{0}^{t}s dX_s \int_{0}^{u}s dX_s]$</p> <p>I refer to my above example on ways to deal with the terms in this expression given certain assumptions on $\mu$ and $\sigma$.Edit 3: Again this is just a way to start and obviously the calculations involving standard Brownian motion are trivial but the point is that the laws of $Y^{(1)}$ and $Y^{(2)}$ are equivalent (as opposed to equal) as soon as you show $g_1(t,u) = g_2(t,u)$ for all $t,u>0$.</p> http://mathoverflow.net/questions/48771/proofs-that-require-fundamentally-new-ways-of-thinking/52395#52395 Answer by jzadeh for Proofs that require fundamentally new ways of thinking jzadeh 2011-01-18T09:50:30Z 2011-01-18T09:50:30Z <p>Malliavin's proof of Hormander's theorem is very interesting in the sense that one of the basic ingredients in the language of the proof is a derivative operator with respect to a Gaussian process acting on a Hilbert space. The adjoint of the derivative operator is known as the divergence operator and with these two definitions one can establish the so called "Malliavin Calculus" which has been used to recover classical probabilistic results as well as give new insight into current research in stochastic processes such as developing a stochastic calculus with respect to fractional Brownian motion. What makes his proof more interesting is that Malliavin was trained in geometry and only used the language of probability in a somewhat marginal sense at times - alot of his ideas are very geometric in nature which can be seen for example in his very dense book: P. Malliavin: Stochastic Analysis. Grundlehren der Mathematischen Wissenschaften, 313. Springer-Verlag, Berlin, 1997.</p> http://mathoverflow.net/questions/50329/generalizations-of-a-product-formula-for-the-gamma-function Generalizations of a product formula for the gamma function jzadeh 2010-12-25T07:44:45Z 2010-12-26T07:59:29Z <p>Hello and Happy holidays. </p> <p>I am interested in generalizations of the following product formula for the gamma function $\Gamma(z)= \int_{0}^{\infty} t^{z-1}e^{-t}dt$ when $n \geq 2$:</p> <p>\begin{align} \displaystyle\prod_{k = 1}^{n} \frac{\Gamma(\frac{z}{2^k}+\frac{1}{2})}{\Gamma(\frac{1}{2})} = &amp; \frac{\Gamma(z+1)}{2^{2z(1-\frac{1}{2^n})} \Gamma(\frac{z}{2^n}+\frac{1}{2})} \end{align}</p> <p>Let $H_1,H_2,...H_n \in (0,1)$ and $z \in \mathbb{R^+}$.<br> 1) Then is it true that the following formula holds for $n \geq 2$?</p> <p>\begin{align} \frac{\Gamma(zH_1 + \frac{1}{2})\Gamma(zH_1H_2 + \frac{1}{2}) \dotsb \Gamma(zH_1H_2 \dotsb H_n + \frac{1}{2})}{\prod_{k=1}^{n} \Gamma(\frac{1}{2})} =<br> \end{align}</p> <p>$\frac{\Gamma(z+1)}{2^{2z(1-H_1H_2 \dotsb H_n)} \Gamma( z H_1 H_2 \dotsb H_n + \frac{1}{2})}$ </p> <p>2) As $n$ tends to $\infty$ is the LHS of the last expression finite?</p> <p>3) Does question 1) hold if $H_1 = 1$?</p> <p>(In the context of my research the $H_i$'s are Hurst parameters from n+1 independent fractional Brownian motions)</p> http://mathoverflow.net/questions/44528/a-simple-decomposition-for-fractional-brownian-motion-with-parameter-h1-2/45156#45156 Answer by jzadeh for A simple decomposition for fractional Brownian motion with parameter $H<1/2$ jzadeh 2010-11-07T10:23:34Z 2010-11-07T10:30:09Z <p>Sorry I don't have time to write a better answer. I would be willing to bet Nualart has thought about this problem at least and his answer could very well be encompassed in this paper: (In particular your problem might be a special case described in section 3)</p> <p>P. Lei and D. Nualart: A decomposition of the bifractional Brownian motion and some applications. Statistics and Probability Letters 79, 619-624, 2009. </p> <p><a href="http://arxiv.org/PS_cache/arxiv/pdf/0803/0803.2227v1.pdf" rel="nofollow">http://arxiv.org/PS_cache/arxiv/pdf/0803/0803.2227v1.pdf</a></p> http://mathoverflow.net/questions/87098/does-the-existence-of-an-asymtpotic-density-imply-the-existence-of-a-measure-on-i Comment by jzadeh jzadeh 2012-02-03T02:56:53Z 2012-02-03T02:56:53Z @AlexanderChervov. Thanks for your ideas but I am still left with the feeling that the RHS of * can be usesd to a come up with a measure that concentrates on something different than Holder continuous paths with modulus 1/2. Furthermore * is an expression for the probability density of Iterating Brownian motion and the density is not Gaussian and its transition probabilities do not satisfy Kolmogorov-Chapman so one is led to believe the induced measure is not a so called &quot;Gaussian Measure&quot;. I wonder if * can give some way to study the induced measure of the IBM process itself. http://mathoverflow.net/questions/87098/does-the-existence-of-an-asymtpotic-density-imply-the-existence-of-a-measure-on-i Comment by jzadeh jzadeh 2012-01-31T18:57:17Z 2012-01-31T18:57:17Z @AlexanderChervov Thanks for your comment. I see your point and so to make things a little more clear how about this: Using equation * can we construct a measure on the space of continuous functions? Equation * has generalizations given by considering iterating fractional Brownian motion and so I am curious to see what type of (if any) measures on function spaces are induced by considering iterating certain classes of random walks an infinite number of times. http://mathoverflow.net/questions/59513/question-about-a-limit-of-gaussian-integrals-and-how-it-relates-to-path-integrati/60601#60601 Comment by jzadeh jzadeh 2011-04-05T00:12:52Z 2011-04-05T00:12:52Z Is is clear that $T$ has a fixed point because of your comments that $K$ is Hilbert-Schmidt? http://mathoverflow.net/questions/59513/question-about-a-limit-of-gaussian-integrals-and-how-it-relates-to-path-integrati/60601#60601 Comment by jzadeh jzadeh 2011-04-05T00:11:52Z 2011-04-05T00:11:52Z Thanks for your help. I am am not sure I completely understand the iteration process in terms of $T$. Dont you want to show $\lim_{n\rightarrow \infty } T^{n} \phi = \phi$? http://mathoverflow.net/questions/59748/what-structure-is-needed-to-define-a-gaussian-distribution-on-a-given-space/59752#59752 Comment by jzadeh jzadeh 2011-03-27T20:15:03Z 2011-03-27T20:15:03Z sorry is it appropriate to remove this as answer then? One last thing I can add is the construction of &quot;brownian motion&quot; in the free potability setting where you are working over von neumann algebras of operators as in section 1.1 of <a href="http://www.iecn.u-nancy.fr/~nourdin/4th-moment-Wigner-KNPS.pdf" rel="nofollow">iecn.u-nancy.fr/~nourdin/&hellip;</a>. Just in case you were unaware of an example of a type of Brownian motion taking values in a non commutative space. http://mathoverflow.net/questions/59513/question-about-a-limit-of-gaussian-integrals-and-how-it-relates-to-path-integrati/59628#59628 Comment by jzadeh jzadeh 2011-03-26T02:24:19Z 2011-03-26T02:24:19Z Can a moderator remove this if I flag it? http://mathoverflow.net/questions/59513/question-about-a-limit-of-gaussian-integrals-and-how-it-relates-to-path-integrati Comment by jzadeh jzadeh 2011-03-25T21:09:18Z 2011-03-25T21:09:18Z Concerning 2) and and the comment following 3) I have been moving back and forth between analyzing the iterated density of $X_n$ and analyzing the behavior of its moment generating function and these comments really apply to the mgf. http://mathoverflow.net/questions/59513/question-about-a-limit-of-gaussian-integrals-and-how-it-relates-to-path-integrati Comment by jzadeh jzadeh 2011-03-25T06:56:58Z 2011-03-25T06:56:58Z The density of $X_n(t)$ is given by the iterated integral in ** http://mathoverflow.net/questions/59513/question-about-a-limit-of-gaussian-integrals-and-how-it-relates-to-path-integrati Comment by jzadeh jzadeh 2011-03-25T06:56:03Z 2011-03-25T06:56:03Z Yes this is true if you look at this from a probabilistic perspective you can argue by self similarity. So set $X_n(t)=B_n(B_{n-1}(...(B1(t))...))$ where $B_i$ is two-sided Brownian motion. Then the following equalities hold in distribution: $X_n(t)=t^{\frac{1}{2^n}} X_n(1)$ taking limits on both sides we see that the random variable $\lim_{n\rightarrow \infty} X_n(t)$ depends only on $X_n(1)$ (i.e. is time invariant). Other authors have made this more rigorous (there is a proof that the asymptotic density is time invariant based on method of moments) http://mathoverflow.net/questions/52448/time-integrals-of-diffusion-processes/52542#52542 Comment by jzadeh jzadeh 2011-01-20T11:42:46Z 2011-01-20T11:42:46Z And here is my mistake. I was trying to figure out why $Y$ would be Gaussian in general but it is not. My argument breaks down I should have said that $tX_t - \int_{0}^{t}X_sds = \int_{0}^{t}sdX_s$ is a Gaussian process. http://mathoverflow.net/questions/52448/time-integrals-of-diffusion-processes/52542#52542 Comment by jzadeh jzadeh 2011-01-20T11:02:58Z 2011-01-20T11:02:58Z Since the processes are Gaussian they will have equivalent laws (as opposed to equal) for all time if the covariance functions equal. That is if $g_1(t,u) = g_2(t,u)$ for all $t,u &gt;0$ the laws of the processes will be equivalent (as opposed to equal). Do you disagree with this fact Didier? http://mathoverflow.net/questions/52448/time-integrals-of-diffusion-processes/52542#52542 Comment by jzadeh jzadeh 2011-01-20T10:16:19Z 2011-01-20T10:16:19Z Thanks for the downvote The Bridge.... I refer you to my above passage &quot;it should then suffice to characterize the processes covariance structure in order to have a complete understanding of the law of the processes&quot;. Since that is obviously to vague I have elaborated a little more in edit 2 and I refer you to one of the excellent texts by Robert Adler for the theorems I am citing on Gaussian processes. # R.J. Adler, (1990), , An Introduction to Continuity, Extrema, and Related Topics for General Gaussian Processes, IMS Lecture Notes-Monograph Series, Vol 12, vii + 160 – jzadeh 0 secs ago http://mathoverflow.net/questions/52448/time-integrals-of-diffusion-processes/52542#52542 Comment by jzadeh jzadeh 2011-01-19T22:48:43Z 2011-01-19T22:48:43Z Thank you for the help I will make the appropriate edits. http://mathoverflow.net/questions/50329/generalizations-of-a-product-formula-for-the-gamma-function Comment by jzadeh jzadeh 2010-12-26T07:30:14Z 2010-12-26T07:30:14Z Yes in the context of exact covering systems here is the reference. Exact Covering Systems and the Gauss-Legendre Multiplication Formula for the Gamma Function John Beebee Proceedings of the American Mathematical Society Vol. 120, No. 4 (Apr., 1994), pp. 1061-1065 (article consists of 5 pages) <a href="http://jbeebee.net/math%20web%20pages/gauss_legendre.pdf" rel="nofollow">jbeebee.net/math%20web%20pages/gauss_legendre.pdf</a> http://mathoverflow.net/questions/50329/generalizations-of-a-product-formula-for-the-gamma-function/50332#50332 Comment by jzadeh jzadeh 2010-12-26T00:06:36Z 2010-12-26T00:06:36Z Thank you very much for your time and nice one line answer. We also have been considering the case $H_1 = 1$. For fixed $z \in \mathbb{R}$ and $H_1 = 1$ is there a way to multiply by an appropriate constant to still make the equality hold?