# Partition function for fractional Brownian motion with $H < 1/2$

Recently I was interested in computing the logarithmic derivarivative $$Z'(H)/Z(H)$$ of the following partition function: $$Z(H) = \int e^{-S_H(x)} \mathcal{D} x, \quad \text{where} \quad S_H(x) = A(H) \int \frac{\dot{x}(t) \, \dot{x}(s)}{|t-s|^{2H}} \, dt \, ds,$$ $$0 and $$A(H) = \cot ( \pi H ) / 4 \pi H$$ is the normalizing constant.

I have never calculated things like this, so I have zero experience when it comes to path integrals, but I would be very glad if some formula for this derivative may be found. Is there any hope to find some representation for this object? How do people in statistical physics and such usually treat objects like this?

Is is possible to find $$Z'(H)/Z(H)$$ at least at $$H=1/2$$? In this case the action is given by $$S_{1/2} ( x ) = \frac{1}{2} \int \dot{x}^2 ( t ) \, dt.$$ This can be shown as follows: $$\frac{1}{|t-s|^{2H}} = \frac{1}{1-2H} \frac{d}{dt} \frac{\operatorname{sign}(t-s)}{|t-s|^{2H-1}}$$ together with $$\lim_{H \to 1/2} \frac{\cot ( \pi H )}{1 - 2H} = \frac{\pi}{2}$$ gives $$\frac{A(H)}{|t - s |^{2H}} \to \frac{\pi}{2 \cdot 4 \pi \cdot 1/2} \frac{d}{dt} \operatorname{sign} ( t - s ) = \frac{1}{2} \, \delta(t-s).$$

P.S. The action $$S_H$$ corresponds to the fractional Brownian motion, I took it from the paper Path integrals for fractional Brownian motion and fractional Gaussian noise by Baruch Meerson, Olivier Bénichou, and Gleb Oshanin.

The calculation showing $$S_H \to S_{1/2}$$ is taken from the supplement to the cited paper.

• If $H=1/2$ your first expression says $S=A(1/2)\int dt ds \frac{\dot x(t)\dot x(s)}{|s-t|}$ not $S=\frac{1}{2}\int dt \dot x^2(t)$...
– hft
Commented Nov 7, 2023 at 0:17
• @hft, you are right, I should probably have said that $A(H) = \cot(\pi H) / 4 \pi H$ and it is zero at $H = 1/2$. I'll add this to the post, thanks! Commented Nov 7, 2023 at 0:20

You can use the fact that the process is zero mean Gaussian. It's easier to build intuition from finite dimensional multivariate gaussian. In the functional case, just replace the sums by integrals. Your partition function is: $$Z = \int e^{-X^T\Lambda X/2}dX$$ with $$A$$ the precision matrix, the inverse of the correlation matrix: $$C = \langle XX^T\rangle\\ \Lambda = C^{-1}$$ Using the usual Gaussian integration: $$Z = \sqrt{(2\pi)^D|C|}$$ so if $$\Lambda$$ depends on a parameter $$H$$: $$\frac{d\ln Z}{dH} = \frac{1}{2}\text{Tr}\left(\Lambda\frac{dC}{dH}\right)$$ where I used the fact that for a symmetric matrices $$A,B$$ for $$h\to0$$: $$|A+hB| = |A|(1+\text{Tr}(A^{-1}B)h)+o(h)$$
In your case, you don't need the insight from the article. For all $$H$$, $$C$$ is: $$C_H(t,s) = \frac{1}{2}(|t|^{2H}+|s|^{2H}-|t-s|^{2H})$$ and $$\Lambda$$ at $$H=1/2$$ is: $$\Lambda_{1/2}(t,s) = \delta''(t-s)$$ Thus, you just need to calculate: $$\frac{d\ln Z}{dH} = \frac{1}{2}\int \partial_t\partial_s\frac{dC_H}{dH}(u,u)du$$
Calculating the general derivative is doable in theory. Now, you need the insight from the article to get the general formula for $$\Lambda$$: $$\Lambda_H(t,s) = \frac{\cot(\pi H)}{4\pi H|t-s|^{2H}}$$
• Thanks a lot, this is very helpful! I am actually more interested in the general case, not $H=1/2$. I'll try to replicate your approach. One question: where does $\partial_t \partial_s$ come from? Should be from $\delta''(t-s)$, but I don't see how. Commented Nov 7, 2023 at 21:49
• Glad it was useful. Just integrate by parts twice, or just directly use your formula $\langle x|\Lambda_{1/2}|y\rangle=\int \dot x\dot ydt$