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Is there any closed form of the Laplace transform of an integrated geometric Brownian motion ?

A geometric Brownian motion $X=(X_t)_{t \geq 0}$ satisifies $dX_t = \sigma X_t \, dW_t$ where $W=(W_t)_{t \geq 0}$ denotes a Brownian motion and the associated integrated Brownian motion is $\int_0^t X_s \, ds$. The Laplace transform of an integrated gometric Brownian motion is thus

$$ \mathcal{L}(\lambda) = \mathbb{E}\left[e^{-\lambda \int_0^t X_s ds } \right]$$

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2 Answers 2

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What is known is explained in C. Albanese, S. Lawi, Laplace transform of an integrated gometric Brownian motion, MPRF 11 (2005), 677-724, in particular in the paragraph of the Introduction beginning by A separate class of models...

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  • $\begingroup$ Google only points to this very question on MSE when typing the article/book you mentionned. $\endgroup$
    – vanna
    Commented Aug 30, 2012 at 11:30
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    $\begingroup$ Try this. $\endgroup$
    – Did
    Commented Aug 30, 2012 at 12:14
  • $\begingroup$ Very nice, thanks. $\endgroup$
    – vanna
    Commented Aug 30, 2012 at 13:28
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I would be very surprised (in a good way) if there was a closed form solution of this. Even though it is straight-forward to calculate $$E\left[\int_0^t X(s) ds\right]$$ the distribution of $$\int_0^t X(s) ds$$ is unknown. And without knowing the distribution, I believe it will be difficult to calculate the expectation. The only thing I can figure out is a lower bound $$E\left[e^{- \lambda \int_0^t X(s) ds} \right] \geq e^{-\frac{2\lambda}{\sigma^2}( e^{\frac{\sigma^2 t}{2}}-1)}$$ which is of little use.

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  • $\begingroup$ $\forall s \geq 0, \ \mathbb{E}(X(s)) = x_0$ since it is a martingale. Thus by Jensen $$ \mathbb{E}(e^{-\lambda \int_0^t X(s) ds}) \geq e^{-\lambda \mathbb{E}(\int_0^t X(s)ds)} \geq e^{-\lambda x_0 t} $$ $\endgroup$
    – vanna
    Commented Jul 18, 2012 at 12:12

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