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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 $\sigma>0$.

Consider $\psi$ a $C^1$ and lipschitz function from $\mathbb{R}\to\mathbb{R}$.

My aim is to compute the derivative of

$$ f(x) = \mathbb{E}\left[\psi\left(\int_{0}^{T}X_{s}^{x}ds\right)\right] $$

It seems natural to study the regularity of $f(x)$ first and given the strong assumption on $\psi$, the fact that $f$ is $C^{1}$ should be the minimum we expect from $f$, something I would like to prove first.

Finally, in order to find the derivative of $f(x)$ I have some ideas using the fact that a lipschitz function has bounded derivative (something which should be compatible with a dominated convergence theorem).

However the fact that the variable of interest is the initial condition of my SDE makes me think that before using the regularity of $\psi$ I should look to the regularity of the mapping $x\mapsto X^{x}$ something I have not seen in stochastic calculus.

If someone is able to help me in my understanding of this notion of regularity w.r.t the initial condition it would be helpful.

Thank you a lot !

Thomas Kojar
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G2MWF
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1 Answers1

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The geometric Brownian motion $$ dX_{t}^{x} = rX_{t}^{x}dt + \sigma X_{t}^{x}dB_t $$ is a linear equation with solution given by (see here generally Solution to General Linear SDE)

\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*}

Thus, for simplicity we let

$$\int_{0}^{T}X_{s}^{x}ds=x\int_{0}^{T}X_{s}^{1}ds=x\Phi_{T}^{1}.$$

So we have by chain rule

$$(\psi\left(x\Phi_{T}^{1}\right))'=\phi'(x\Phi_{T}^{1})\Phi_{T}^{1}.$$

So if $\psi$ has nice enough properties to swap derivative and expectation, we get $$f'(x)=\mathbb{E}\left[\psi'\left(x\Phi_{T}^{1}\right)\Phi_{T}^{1}\right].$$

amWhy
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Thomas Kojar
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  • Thank you a lot for your answer Thomas ! The part about the solution to this sde is very clear. However I do not see how you differentiate $\psi\left(\int_{0}^{T}X_{s}^{x}ds\right)$ with respect to $x$ that is the initial condition ? – G2MWF Feb 29 '24 at 21:58
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    @coboy see updated answer with more details. – Thomas Kojar Feb 29 '24 at 23:24
  • Thank you a lot ! I did not think of make appear explicitly $x$ in order to differentiate it, I should. Also by directly proving it is differentiable we get for free the fact it is continuous, which reduces the work to proving that the derivative is a also continuous. – G2MWF Mar 01 '24 at 09:48