Stochastic Processes and Financial Mathematics
(part two)
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13.2 Geometric Brownian motion
In this section we focus on a particular SDE, namely
\(\seteqnumber{0}{13.}{7}\)
\begin{equation}
\label {eq:gbm} dX_t=\alpha X_t\, dt+\sigma X_t\, dB_t
\end{equation}
where \(\alpha \in \R \) and \(\sigma \geq 0\) are deterministic constants. The parameter \(\alpha \) is known as the drift, and \(\sigma \) is known as the volatility. Equation (13.8) will be important to us because it will be our next step in establishing better models for stock prices (to be continued,
in Section 15.1).
The solution to equation (13.8), which we will shortly show exists, is known as geometric Brownian motion.
The key step to solving (13.8) is to work with the logarithm of \(X\). With this aim in mind, we assume (for
now) that there is a solution \(X\) that is strictly positive, and consider the process
\[Z_t=\log X_t.\]
Of course our assumption may not be true - but if such a solution does exist we hope to (do some calculations and with \(Z\) and) find an explicit formula for it, at which point we can go back and check we really
do have a solution.
-
Taking logarithms is a natural idea to try. To see this, consider the special case
\(\sigma =0\), where we find ourselves back in the world of differential equations: \(x(t)=\int _0^t \alpha x(u)\,du\). The fundamental theorem of calculus gives \(\frac {dx}{dt}=\alpha x\). We can
solve this equation by considering \(z=\log x\), obtaining
\[\frac {dz}{dt}=\frac {dz}{dx}\frac {dx}{dt}=\frac {1}{x}\alpha x=\alpha \]
Thus \(z(t)=\alpha t+C\) and \(x(t)=e^{z(t)}=C'e^{\alpha t}\).
Using Ito’s formula, with \(Z_t=\log X_t\) (i.e. \(f(t,x)=\log x\)) we obtain
\(\seteqnumber{0}{13.}{8}\)
\begin{align*}
dZ_t&=\l (0+\alpha X_t\frac {1}{X_t}+\frac 12(\sigma X_t)^2\frac {-1}{X_t^2}\r )\,dt + \sigma X_t \frac {1}{X_t}\,dB_t\\ &=\l (\alpha -\frac 12\sigma ^2\r )\,dt+\sigma
\,dB_t.
\end{align*}
In integral form this gives us
\(\seteqnumber{0}{13.}{8}\)
\begin{align*}
Z_t&=Z_0+\int _0^t\alpha -\frac {1}{2}\sigma ^2\,du+\int _0^t\sigma \,dB_u\\ &=Z_0+\l (\alpha -\frac 12\sigma ^2\r )t+\sigma B_t.
\end{align*}
Since \(Z_t=\log X_t\), raising both sides to the power \(e\) gives us
\(\seteqnumber{0}{13.}{8}\)
\begin{equation}
\label {eq:gbm_soln} X_t=X_0\exp \l ((\alpha -\tfrac 12\sigma ^2)t+\sigma B_t\r ).
\end{equation}
As we said, this was all based on the assumption that a (strictly positive) solution exists. But, now we have found the formula (13.9), we can go back and check that it does really give us a solution. This part is left for you, see exercise 13.12.
For future use, applying (13.9) at times \(t\) and \(T\), we obtain that
\(\seteqnumber{0}{13.}{9}\)
\begin{equation}
\label {eq:gbm_soln_tT} X_T=X_t\exp \l ((\alpha -\tfrac 12\sigma ^2)(T-t)+\sigma (B_T-B_t)\r ).
\end{equation}