Stochastic integration by parts of geometric brownian motion

Question

My objective is to calculate the integral of a geometric brownian motion $Y=e^{\alpha t+\beta W_t}$, i.e. $$\int_{0}^T e^{\alpha s+\beta W_s}ds$$ and to characterize the moments of resulting random variable. If I make the transformation $g_t=x_ty_t$ with $x_t=e^{\alpha t}$ and $y_t=e^{\beta W_t}$ (where $\alpha,\beta>0$) and consider Ito integration by part, i.e. \begin{equation} d(x_ty_t)=x_tdy_t+y_tdx_t+dx_tdy_t \end{equation} I end up with: \begin{eqnarray} e^{\alpha T+\beta W_T} &=& 1+\int_{0}^T x_tdy+\int_{0}^T y_tdx_t+\int_{0}^Tdx_tdy_t\\ &=& 1+\int_{0}^T e^{\alpha s}e^{\beta W_s} \beta W_s dW_s +\int_{0}^T \alpha e^{\alpha s}e^{\beta W_s} ds+\int_{0}^T \alpha\beta e^{\alpha s}e^{\beta W_s}ds dW_t \end{eqnarray} Here, I made use of $dx_t=\alpha x_tdt$, $dy_t=\beta y_t dW_t$ and $W_0=0$. After rearranging I get: \begin{eqnarray} \alpha \int_{0}^T e^{\alpha s}e^{\beta W_s} ds &=& e^{\alpha T+\beta W_T}-1-\beta e^{\alpha s}\int_{0}^T e^{\beta W_s} W_s dW_s -\int_{0}^T \alpha\beta e^{\alpha s}e^{\beta W_s}dsdW_t \end{eqnarray} Now I am stuck since I do not know how to cope with the last two integrals. Any help and hints are very much appreciated and many thanks in advance.

Answer 1

You can't go any further than $\int e^{B_{t}+at}dt$. This is called the integrated Geometric Brownian motion. For references on its law see

• "The Integral of Geometric Brownian Motion"
• "On Some Exponential Functionals of Brownian Motion"
• "On the moments of the integrated geometric Brownian motion"

Let $B$ stand for one-dimensional standard Brownian motion starting at $0$, and define the 'integral-exponential functional' of Brownian motion to be $$ A_t^{(\mu)} = \int_0^te^{2\mu\tau + 2B_\tau} d\tau, \quad t \ge0, \mu \in \mathbb{R}. \tag{1.1} $$ The main goal of this paper is to derive a new expression for the probability density function of $A_t^{(\mu)}$. Some expressions have previously been obtained for this density. Yor (1992a, Proposition 2) states that $$ \begin{align} P(A_t^{(\mu)} \in du \mid B_t + \mu t = x) &= a_t(x,u) d u \\ &= \frac{\sqrt{2 \pi t}}{u} \exp\left( \frac{x^2}{2t} - \frac{1}{2u} (1 + e^{2x})\right) \theta _ {e^x / u} (t) du, \tag{1.2} \end{align} $$ where $$ \theta_r(t) = \frac{r}{\sqrt{2 \pi^3 t}} \exp \left( \frac{\pi^2}{2t} \right) \int_0^\infty \exp \left( - \frac{y^2}{2t} \right) \exp (-r \cosh y) (\sinh y) \sin \left( \frac{\pi y}{t} \right) dy. $$