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 $t > 0$
$$(**) \qquad\lim_{n \rightarrow \infty} 2^{n} \int_{0}^{\infty} \ldots \int_{0}^{\infty} \frac{e^{\frac{-x^2}{2z_1}}}{\sqrt{2 \pi z_1}} \frac{e^{\frac{-{z_1}^2}{2z_2}}}{\sqrt{2 \pi z_2}} \ldots \frac{e^{\frac{-{z_n}^2}{2t}}}{\sqrt{2 \pi t}} \mathrm{d}z_1 \ldots \mathrm{d}z_n = e^{-2 |x|} $$
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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
https://physics.stackexchange.com/q/7552/2757 -
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?
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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.)
Best Answer
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.