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Find the analytical solution of the stochastic integral equation $$\tag1 X(t) = -\frac18-\int_0^t\frac14sX(s)ds-\int_0^t\frac{1}{40}X(s)dB_H(s) $$ where $(B_H(t))_{t\ge0}$ is a fractional Brownian motion, i.e. a continuous Gaussian process with $B_H(0)=0$, $\mathbb E[B_H(t)]=0$ $\forall t$, covariance $\mathbb E[B_H(t)B_H(s)] = \frac12(|t|^{2H}+|s|^{2H}-|t-s|^{2H})$ and $H\in(0,1)$.


SOLUTION $$\tag2 X(t) = -\frac18\exp\Big(-\frac{1}{40}B_H(t)-\frac{t^2}{8}-\frac{t^{2H}}{3200}\Big) $$ P.S. I have not to verify that this is the solution, but rather to find it from $(1)$.


I tried to apply the Ito formula for fractional integrals enter image description here

where $f(s,x) : \mathbb R \times \mathbb R \to \mathbb R$ is a function of $C^{1,2}(\mathbb R \times \mathbb R)$.

Using $f(t,X(t)) = \log X(t)$ I got

$$ \log X(t) = \log X(0)+\int_0^t\frac{1}{X(s)}dX(s)+H\int_0^t-\frac{1}{X^2(s)}s^{2H-1}ds+\underbrace{\int_0^t\frac{\partial\log X(s)}{\partial s}ds}_{=0\ ?} $$

and by substituting $dX(s) = -\frac14sX(s)ds-\frac{1}{40}X(s)dB_H(s)$ (obtained from $(1)$ by differentiating)

$$\begin{align*} \log\frac{X(t)}{X(0)} &= \int_0^t \Big(-\frac14sds-\frac{1}{40}dB_H(s)\Big)+\underbrace{H\int_0^t-\frac{1}{X^2(s)}s^{2H-1}ds}_{=\ ?} \\ &=-\frac{t^2}{8}-\frac{1}{40}B_H(t)+? \end{align*}$$

Then by taking exponential both sides

$$ X(t) = X(0)\exp\Big(-\frac{t^2}{8}-\frac{1}{40}B_H(t)+?\Big) $$

which partially is equal to $(2)$. Moreover I notice that $H\int_0^ts^{2H-1}ds = t^{2H}/2$, which is also contained in $(2)$. But I don't know how to deal with $\frac{1}{X^2(s)}$.


Maybe another way is to assume that the solution is of the form $$ X(t) = a\exp(bB_H(t)+c(t)) $$ but how to find $a, b\in\mathbb R$ and the real function $c$?

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  • $\begingroup$ What do you mean by integration against fBm for $H<1/2$? $\endgroup$
    – user658409
    Feb 23, 2020 at 17:36
  • $\begingroup$ i think the answer is on yuliya mishura's stochastic calculus for fBm $\endgroup$
    – sound wave
    Feb 23, 2020 at 22:55
  • $\begingroup$ Defining integration against fBm with bad regularity is highly nontrivial and nonunique. $\endgroup$
    – user658409
    Feb 23, 2020 at 23:32
  • $\begingroup$ what about $H > 1/2$? is it possible to find the solution in this case? $\endgroup$
    – sound wave
    Feb 24, 2020 at 8:30

1 Answer 1

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You didn't apply Itô's formula correctly. The Itô formula which you stated at the beginning of your calculation applies for expressions of the form $f(t,B_H(t))$ - but in your computation you need to compute the stochastic differential of $f(t,X_t)$. Compare this with the situation in classical stochastic calculus (i.e. for Brownian motion): Itô's formula for Brownian motion looks different than Itô's formula for general Itô processes.

Here is the version of Itô's formula which you need to solve the problem: If $$X_t = \int_0^t b(s) \,ds + \int_0^t \sigma(s) \, dB_H(s)$$

then (assuming everything is well-defined)

\begin{align*} f(t,X_t) -f(0,X_0) &= \int_0^t \partial_s f(s,X_s) \, ds + \int_0^t \partial_x f(s,X_s) \, d\color{red}{X_s} \\ &+ H \int_0^t \partial_x^2 f(s,X_s) \color{red}{\sigma(s)^2} s^{2H-1} \, ds. \tag{1} \end{align*}

Note that for $X_t = B_H(t)$ we get, as a particular case, the Itô formula which you stated.

Applying Itô's formula $(1)$ for $f(t,x) = \log x$, it follows that

\begin{align*} \frac{\log X_t}{\log X_0} &= \int_0^t -\frac{1}{4} s \, ds - \frac{1}{40} \int_0^t dB_H(s) + H \int_0^t - \frac{1}{X_s^2} \color{red}{\left( \frac{1}{40} X_s\right)^2} s^{2H-1} \, ds \\ &= - \frac{t^2}{8} - \frac{1}{40} B_H(t) - \frac{1}{3200} t^{2H}.\end{align*}

(The term in red is the one which is missing in your calculation.) Hence,

$$X_t = X_0 \exp \left(-\frac{t^2}{8} - \frac{1}{40} B_H(t) - \frac{1}{3200} t^{2H} \right).$$

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  • $\begingroup$ Thank you very much! While I study your answer, could you tell me if considering the logarithm of $X_t$ (equation (1) of my question) is completely correct? I'm asking since all terms in $X_t$ are negative, so it seems that log is not defined. But maybe the results of the two integrals in $X_t$ are negative, hence the two terms turn positive. $\endgroup$
    – sound wave
    Feb 29, 2020 at 14:41
  • $\begingroup$ @soundwave Since $X_t$ is (strictly) negative, you can apply itô's formula for $-X_t$ to compute the stochastic differential of $\log(-X_t)$. This doesn't really change much about the calculation $\endgroup$
    – saz
    Feb 29, 2020 at 16:13
  • $\begingroup$ Do you know a source for the Ito formula that you used? I found several papers and books but there are several Ito formulas w.r.t. the fBm, and I cannot find the one you used. $\endgroup$
    – sound wave
    Mar 12, 2020 at 1:09
  • $\begingroup$ p.s. the books I was talking about are Mishura "Stochastic Calculus for Fractional Brownian Motion and Related Processes" and Biagini, Hu, Oksendal, Zhang "Stochastic Calculus for Fractional Brownian Motion and Applications". The closest Ito formula to yours which I could find is i.imgur.com/09yPMAA.png which is in the Mishura book. $\endgroup$
    – sound wave
    Mar 12, 2020 at 5:54
  • $\begingroup$ Moreover, is the Ito formula that you stated based on the pathwise approach or on the Malliavin/Skorohod approach? $\endgroup$
    – sound wave
    Mar 12, 2020 at 8:02

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