# Fractional Brownian motion

In Mandelbrot(1968)'s paper, the fractional brownian motion, denoted by $B_{H}(t,\omega)$,(t>0) is defined by $$B_{H}(0,\omega)=b_{0}$$ $$B_{H}(t,\omega)-B_{H}(0,\omega)=\frac{1}{\Gamma(H+\frac{1}{2})}\{\int^{0}_{-\infty}[(t-s)^{H-1/2}-(-s)^{H-1/2}]dB(s,\omega)+\int^{t}_{0}(t-s)^{H-1/2}dB(s,\omega)\}$$ I have difficulty understanding fractional brownian motion by self study.Is there an intuitive interpretation of this definition? Why time s can have negative value? Thanks!

• Why should $s$ be negative?
– Ric
May 18 '15 at 10:23
• I mean that the lower bound of the first integral on the right hand side is negative infinity. May 18 '15 at 13:31

Only the pure Brownian motion regime doesn't depend on the past. You don't define any variables, but I'd guess Mandelbrot is using the Hurst coefficient in the equation you give, where pure Brownian motion is H = 0.5. In this case the first integral gives 1 - 1, i.e. no dependence on the past (times leading up to time 0, i.e. "negative" time). For other cases (0 <= H <= 1, H != 0.5), the process leading up to time zero, and continuing on until time $t$, each incremental step in the process does depend on the previous steps.
For a Brownian motion, if you wait $dt$, the variance will grow linearly with (proportionally to) $dt$.
For a fractional Brownian motion, it will grow with a power law of $dt$, in fact in $dt^{H}$, where $H$ is the Hurst exponent. See wikipedia for more details.
It means the fBM will somehow keep memory of the past. When $H$ is lower than 1/2, it will mean revert, when it is greater than 1/2, it will be superdiffusive. At $H$, it is a Brownian motion.