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.
Stochastic Processes – Stochastic Integration by Parts of Geometric Brownian Motion
stochastic-differential-equationsstochastic-integralsstochastic-processes
Best Answer
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