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ELEMENTARY ITO FORMULA EXAMPLE FOR A FRACTIONAL BROWNIAN MOTION

Consider the simple fBM SDE

$dV(t) = \sqrt{V(t)} dB_H(t)$

$d\langle V(t) \rangle = V(t) t^{2-2H} dt$

Applying the Ito formula yields

$d\sqrt{V(t)} = \frac{1}{2\sqrt{V(t)}} dV(t) - \frac{1}{4} V^{-3/2}(t) d\langle V(t) \rangle$

$= \frac{1}{2} dB_H(t) - \frac{1}{4} \frac{t^{2-2H}}{\sqrt{V(t)}} dt$

The deterministic part is

$\frac{du}{dt} = - \frac{t^{2-2H}}{4 u(t)}$

that is

$\frac{d}{dt} u^2(t) = 2 u(t) \frac{du}{dt} = - \frac{1}{2} t^{2-2H}$

This computation allows us to consider the drift term for the long memory stochastic volatility equation

$dV(t) = \kappa(\theta-V(t))dt + \nu \sqrt{V(t)} dB_H(t)$

which after applying the Ito formula to $d\sqrt{V(t)}$ produces

$\frac{d}{dt} V(t) + \frac{2\kappa}{\nu} V(t) = \frac{2\kappa\theta}{\nu} - t^{2-2H}$

Then $Z(t) = e^{\frac{2\kappa}{\nu} t}V(t)$ to solve

$Z'(t) = e^{\frac{2\kappa}{\nu} t} (\frac{2\kappa\theta}{\nu} - t^{2-2H})$

The solution of the stochastic equation is $Z(t) + B_H(t)$

Use the approximation $\int_0^t e^{As} s^B ds \sim (C/A) t^B(e^{At} - 1)$ with $A = \frac{2\kappa}{\nu}$ and $B=2-2H$ which can be obtained from termwise integration of series expansion of the exponential to conclude

$Z(t) \sim e^{\frac{2\kappa t}{\nu}} (\theta - (C/A) t^{2-2H}) + (C/A)t^{2-2H} + B_H(t)$

with $C \le \sup_{k} \frac{k+1}{k+2-2H+1}=1$, and so

$V(t) \sim (\theta - \frac{C\nu}{2\kappa}t^{2-2H}) + \frac{C\nu}{2\kappa} e^{-\frac{ \kappa t}{\nu}} t^{2-2H} + e^{-\frac{\kappa t}{\nu}}B_H(t)$

Now the fractional Brownian motion is Gaussian with mean zero and covariance $E[B_H(t) B_H(s)] = \frac{1}{2} ( t^{2H} + s^{2H} - |t-s|^{2H})$ so we can read off the covariance of volatility variance in this model.