It says assuming a no-uncertainty Weiner process that models stock price: $$ \Delta S = \mu S\Delta t $$ Can be rearranged to (after taking the limit of $\Delta t \to 0$... $$ \frac{dS}{S}=\mu dt $$ Then integrating between time 0 and T to get: $$ S_T=S_0 e^{\mu T} $$

I don't understand the last step. Are they integrating with respect to t? How does the exponential come about when there was no exponential in the prior equation? Is this step a condensation of a complex calculation that they didn't show?


1 Answer 1


It's simpler than you think. Hull is just solving an ODE.

You can naively put integral signs on both sides of the equation: \begin{align*} \frac{\mathrm{d}S_t}{S_t} &=\mu \mathrm{d}t \\ \implies \int_0^T\frac{\mathrm{d}S_t}{S_t} &=\int_0^T\mu \mathrm{d}t\\ \implies \ln(S_T)-\ln(S_0) &=\mu T \\ \implies S_T&=S_0e^{\mu T}. \end{align*}

Perhaps this makes it easier: Since $S_t$ is deterministic, it is a ``normal'' function. Thus, you may want to write $y(x)=S_t$. The above equation then turns into $\frac{\mathrm{d}y}{y} =\mu \mathrm{d}x\Leftrightarrow y'=\mu y$. So, it's simply about solving a first-order ODE.

Also note that $\frac{\mathrm{d}S_t}{S_t}=\mathrm{d}\ln(S_t)$, i.e. percentage returns correspond to log-returns if time is infinitesimal. So, you shouldn't be surprised to find an exponential here.

  • 2
    $\begingroup$ Ohhhh totally didn't see it as an ODE. Thanks. (wish they had that in the wording...) $\endgroup$
    – Five9
    Apr 13, 2020 at 1:54

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