Thanks to the work by M. Yor and colleagues, much is known about the following exponential of Brownian motion:

$X= \int_0^{\infty}{\rm d}t \ e^{-t + g \ B(t)}$

where $g$ is a real scale parameter.

A nice probabilistic way to obtain results is to use Lamperti representation to recognize $X$ as a time-changed Bessel process. Among other things, one can show that for $D\equiv g^2/2 <1$, $D X$ is a Gamma distribution of parameter $-1/D$.

For a physicist (which I am), a cheap way out to recover this would be to determine the stationary solution to the forward Chapman-Kolmogorov equation associated to the stochastic differential equation:

$dX(t) = [1+(D-1)X(t)]dt + g X(t) dB(t)$

The latter is readily integrated, and taking the $t \to \infty$ limit, one encounters the definition above for the functional $X$.

Motivated by various questions, with colleagues we have bee working hard over the last year on the following problem: what happens if $g \to \imath \ g, \ \imath \equiv \sqrt{-1}$ (in either the s.d.e. or the explicit exponential functional of BM)?

Now, of course the variable:

$Z= \int_0^{\infty}{\rm d}t \ e^{-t + \imath g \ B(t)}.$

do not live anymore on the real axis, but in the complex plane, and actually in the unit disk ($\forall g$), because of $|Z| \le \int_0^{\infty}{\rm d}t \ e^{-t}=1$.

But by whatever means, the distribution of $Z$, i.e. the joint distribution of its real and imaginary part, or of its modulus and of its angle (writing $Z=X+ \imath \ Y= R e^{\imath \theta}$), seems very hard to determine reasonably explicitely.

For instance, one can compute the integer moments:

$ E[(D Z)^m] = \frac{\Gamma(1+1+1/D)}{\Gamma(m+1+1/D)}$

but of course this is not enough to reconstruct the joint law of $(X,Y)$: one would need to solve the quadratic recurrence for the joint moments $a_{m,n} \equiv E[Z^m {\overline{Z}}^n]$ of $Z$ and of its complex conjugate $\overline{Z}$:

$ (m+n+D(m-n)^2) a_{m,n} = m a_{m-1,n}+ n a_{m,n-1}.$

The most compact characterization I have found is through the following (double) exponential generating function: $\phi(p,q) \equiv E[\exp{(p Z + q \overline{Z})}]$. The latter obeys a certain second order linear p.d.e., which can be brought to canonical form by setting $p=(\mu/2)e^{-\imath \nu}, \ q= (\mu/2)e^{\imath \nu}$, so that, with $Z=R e^{\imath \theta}$, one finds that

$\phi(p,q)=\Omega(\mu,\nu)= E[e^{\mu R \cos{(\theta-\nu)}}]$

where $\Omega$ is the unique solution with $\Omega(0,\nu)=1$ to:

$[\mu \partial_{\mu} -D \partial^2_{\nu} - \mu \cos{\nu}]\Omega(\mu,\nu)=0$

I shall post soon a paper on the arXiv on this, but I thinks it's worth drawing the attention of mathematicians to this kind of problem, which is well-posed, and for which no solution seems to have been given.

If anybody has an idea, she/he would be most welcome...


3 Answers 3


Hello Ivan

I am quite new to mathoverflow. A friend pointed this post to me as I have been doing something related to exponential functionals of BM (Brownian motion) and i have been like you reading papers from Yor, Dufresne and colleagues. Indeed, in order to just compute the measure you are interested in, there is a short method close to the one you talked about (in the real case). To my knowledge, it works for all real scale parameters g, and uses some tricks that could be valuable in treating the complex case.

  • If $ X := \int_0^\infty e^{-s +gB_s}ds $ for $g \in \mathbb{C}$, let's put $$X_t := \int_0^t e^{-s +gB_s}ds \stackrel{t\rightarrow \infty}{\rightarrow} X $$ the convergence being almost sure and in $L^1$. The process $X_t$ is definitely not Markovian, which hinders calculations considerably. A nice and useful trick is to introduce via time-inversion a process $Z_t$ that has the same fixed time marginals but that is diffusion process. For fixed t, $(B_s, 0 \leq s \leq t)$ has the same law as $(B_{t-s} - B_t, 0 \leq s \leq t)$ then $X_t$ has the same fixed time marginals than: $$ Z_t = \int_0^t e^{-s +g(B_{t-s}-B_s)}ds = e^{ -t - gB_t }\int_0^t e^{s + g.B_{s}}ds $$ A first valuable statement is that since $X_t$ converges almost surely, and $Z_t$ has the same fixed time marginals, $Z_t$ converges in law to a stationary measure. A second even more valuable one is that $Z_t$ is Markovian diffusion since if we apply Ito formula: $$ dZ_t = (1 + Z_t( \frac{g^2}{2} - 1))dt - g Z_t dB_t$$ That is exactly the SDE you exhibited. Although i believe it is not followed by $X_t$ but $Z_t$ only. It is also verified for all complex parameters $g$. Here i guess some more thorough analysis of the sde may be interesting ( Some nice change of variables? Equation for polar coordinates? Lyapounov functions?), but i guess you already tried all of that.

  • Here i turn to $g$ real parameter. That's pretty standard. I assume the process $Z_t$ has a nice smooth density $p(t,x)$ (because of the BM noise) that solves the Fokker-Planck equation starting from $Z_0 = 0$ : $$ \partial_t p = \mathcal{L}^{ * }p$$ where $\mathcal{L}$ is the infinitesimal generator of $Z$ and $\ ^{*}$ denotes the adjoint. As $Z_t$ converges in law, $p(t,.)$ converges to the measure you are interested in, $\mu(x)dx$ the law of $X$. Because everything is on the real line, we get the ODE: $$\mathcal{L}^*\mu = 0$$ Solving it easy (second order ODE that is fact a first order ODE). I find that $\frac{2}{g^2 X}$ has the law of a Gamma distribution with parameter $\frac{2}{g^2}$. It is valid for all real parameters $g$.

I guess this is just restating what you already (kind of) know. So i hope it makes it at least a bit clearer. But from this i am thinking that in the same fashion, the equation verified by the limiting density $\mu$ can be extracted. The solving will definitely be more complicated as it is going to be a 2-dimensional PDE. Maybe the use of complex analysis would help as the underlying Markov process is complex valued with special features? I am not vested in complex analysis that mixes stochastic processes. But i could give it some more thinking. In the end, what makes the real case computable is the fact we get simple ODEs...

I would definitely be interested in what you're going to post on arXiv.

Good luck.


  • $\begingroup$ Thanks Reda for the comment! (I was feeling a bit lonesome...). Indeed the complex case I am interested in seems much harder: it's really a 2-dimensional PDE one has to solve, and not some simple "product" of Yor's case. Studying the equation satisfied by $\Omega(\mu,\nu)$ (in the notations of my post), we could show for instance that the (2d) distribution of $Z(=R e^{\imath \theta}$ has an essential singularity as $R\to 1$... $\endgroup$ Sep 24, 2010 at 10:12

For those interested, we have released today an ArXiv preprint with what we were able to find: http://arxiv.org/abs/1101.1173


It turns out that the above generating function pde is related to a Lax pair of Painlevé III/Sine-Gordon, at least for the parameter value g=2. As soon as I have written this up, I'll post details


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