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We consider a continuous stochastic stochastic process $X_t$ with the following dynamic on $[0,T]$ :

$$ dX_{t}^{x} = rX_{t}^{x}dt + \sigma X_{t}^{x}dB_t $$

Where $X_{0}^{x}=x$ is the initial condition, $r>0$, $B_t$ is a standard Brownian motion and $\sigma>0$.

A solution is given by

\begin{align*} X_t^x =& x \exp\left( t\left( r - \frac{1}{2}\sigma^2(s) \right) + \sigma B_t\right) =xX_{t}^{1}. \end{align*}

(See the post Solution to General Linear SDE)

I am interesting on the computation of the following

$$ \int_{0}^{T} D_s I_{n}ds $$

Where $D_s$ is the Malliavin derivative and

$$ I_{n} = \int_{0}^{T} t^n X_{t}^{x} dt $$

I have no clue on how to start this problem. I thought about a classical integration by part for Riemann integral but I do not see how to take an anti derivative or a derivative at the usual sense of such a thing.

If you have some hints to provide, I would appreciate. Thank you

G2MWF
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1 Answers1

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I am basing this on Nualart's book The Malliavin Calculus and Related Topics.

First we use that the Malliavin-derivative is linear and so

$$D_{s}I_{n}=\int_{0}^{T}t^{n}D_{s}(X_{t}^{x})dt.$$

Then we use Malliavin-chain rule

$$D_{s}(X_{t}^{x})=x \exp\left( t\left( r - \frac{1}{2}\sigma^2(s) \right) + \sigma B_t\right)\sigma D_{s}(B_{t})$$

and then we use that $ D_{s}(B_{t})=1_{[0,t]}(s)$ to get

$$D_{s}I_{n}=\int_{0}^{T}t^{n}x \exp\left( t\left( r - \frac{1}{2}\sigma^2(s) \right) + \sigma B_t\right)\sigma 1_{[0,t]}(s)dt$$

$$=\int_{0}^{min(T,s)}t^{n}x \exp\left( t\left( r - \frac{1}{2}\sigma^2(s) \right) + \sigma B_t\right)\sigma dt.$$

Thomas Kojar
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