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I have come across a limit of Gaussian integrals in the literature and am wondering if this is a well known result.

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:

The original reference is "Fractional diffusion equations and processes with randomly varying time" Enzo Orsingher, Luisa Beghin http://arxiv.org/abs/1102.4729

Line (3.14) of Orsingher and Beghins paper reads for $t > 0$

$$(**) \qquad\lim_{n \rightarrow \infty} 2^{n} \int_{0}^{\infty} \ldots \int_{0}^{\infty} \frac{e^{\frac{-x^2}{2z_1}}}{\sqrt{2 \pi z_1}} \frac{e^{\frac{-{z_1}^2}{2z_2}}}{\sqrt{2 \pi z_2}} \ldots \frac{e^{\frac{-{z_n}^2}{2t}}}{\sqrt{2 \pi t}} \mathrm{d}z_1 \ldots \mathrm{d}z_n = e^{-2 |x|} $$

  1. 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 http://physics.stackexchange.com/q/7552/2757

  2. 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?

  3. 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?

(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.)

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Looks like the left hand side of (**) depends on $t$, but the right hand side does not? –  Bjørn Kjos-Hanssen Mar 25 '11 at 5:37
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) –  jzadeh Mar 25 '11 at 6:56
The density of $X_n(t)$ is given by the iterated integral in ** –  jzadeh Mar 25 '11 at 6:56
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. –  jzadeh Mar 25 '11 at 21:09

1 Answer 1

up vote 2 down vote accepted

The expression you are interested in is of the form $\lim_n T^n\psi_t$ where $T$ is the integral operator $$Tf(x)=2\int_0^\infty \frac{e^{-\frac{x^2}{2y}}}{\sqrt{2\pi y}} f(y)dy$$ and $\psi_t(x)= \frac{e^{-\frac{x^2}{2t}}}{\sqrt{2\pi t}}$. Note that $T$ is an operator with a positive kernel $T(x,y)=2 I[y>0] \frac{e^{-\frac{x^2}{2y}}}{\sqrt{2\pi y}}$ which satisfies $\int_{-\infty}^\infty T(x,y)dx =1.$ That is, $T$ is much like a stochastic matrix and the limit you wish to obtain could only hold if $e^{-2|x|}$ is the principle eigenvector (with eigenvalue $1$).

To realize $T$ as something like a stochastic matrix, we should present it as a compact operator. It is not compact on $L^2(\mathbb{R})$, but if we think of it as an integral operator on the space $L^2(e^x dx)$ of functions with $$\int_{-\infty}^\infty |f(x)|^2 dx <\infty,$$ i.e. $$Tf(x)=\int_{-\infty}^\infty K(x,y) f(y) e^y dy$$ with $K(x,y)= 2I[y>0] \frac{e^{-\frac{x^2}{2y} -y}}{\sqrt{2\pi y}}$, then $\int\int K(x,y)^2 e^{x+y}dxdy <\infty$ so $T$ is Hilbert-Schmidt on $L^2(e^x dx)$, hence compact. Because $K(x,y)>0$ for all $x,y$ the Perron-Frobenius theorem (suitably generalized to compact operators of this type) shows that $T$ has a unique positive eigenvalue $\lambda_0$ with a positive eigenfunction and all other eigenvalues $\lambda$ are of modulus $|\lambda|<\lambda_0$. I claim that $T\phi=\phi$ where $\phi(x)=e^{-2|x|}$, so the unique positive eigenvalue is one! (Certainly this has to do with the origins of the problem in probability theory.) To see that $T\phi=\phi$ it is most convenient to take a Fourier transform in $x$ to get $$ \widehat{T\phi}(k)=2 \int_0^\infty e^{-\frac{k^2}{2}z}e^{-2z}dz=\frac{4}{k^2 +4}=\widehat{\phi}(k).$$

The other thing we need is the left principle eigenvector -- the eigenvector of $T^\dagger$ with eigenvalue $1$. Here the adjoint must be on $L^2(e^x dx)$ so

$$ T^\dagger f(x) =\int_{-\infty}^\infty K(y,x)f(y)e^y dy= 2 e^{-x} I[x>0] \int_{-\infty}^\infty \frac{e^{-\frac{y^2}{2x}}}{\sqrt{2\pi x}} e^y f(y).$$

Observe that $T^\dagger \widetilde{\phi}(x)=\widetilde{\phi}(x)$ where $\widetilde{\phi}(x)=2 e^{-x}I[x>0]$. The factor of $2$ enforces the normalization $\langle \phi,\widetilde \phi \rangle =1$ (with the inner product in $L^2(e^xdx)$). It now follows that

$$T^n f =\phi \langle \widetilde{\phi},f\rangle +o(1)$$

for any function $f\in L^2(e^xdx)$. Since

$$\langle \widetilde{\phi},\psi_t \rangle =\int_{0}^\infty \widetilde{\phi}(x)\psi_t(x)e^x dx=2 \int_0^\infty \psi_t(x)= $$

your identity follows.

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Thanks for your help. I am am not sure I completely understand the iteration process in terms of $T$. Dont you want to show $\lim_{n\rightarrow \infty } T^{n} \phi = \phi$? –  jzadeh Apr 5 '11 at 0:11
Is is clear that $T$ has a fixed point because of your comments that $K$ is Hilbert-Schmidt? –  jzadeh Apr 5 '11 at 0:12
I expanded the explanation to answer your questions. The observation that $T$ is Hilbert-Schmidt is just an easy way to prove compactness. –  Jeff Schenker Apr 5 '11 at 14:36

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