Question about a Limit of Gaussian Integrals and how it relates to Path Integration (if at all)? 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 $t > 0$
$$(**) \qquad\lim_{n \rightarrow \infty} 2^{n} \int_{0}^{\infty} \ldots \int_{0}^{\infty} \frac{e^{\frac{-x^2}{2z_1}}}{\sqrt{2 \pi z_1}} \frac{e^{\frac{-{z_1}^2}{2z_2}}}{\sqrt{2 \pi z_2}} \ldots  \frac{e^{\frac{-{z_n}^2}{2t}}}{\sqrt{2 \pi t}} \mathrm{d}z_1 \ldots \mathrm{d}z_n = e^{-2 |x|}   $$


*

*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?

*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.)
 A: 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.
