I'm reading Chapter 13 of Hull's book and am stuck on how he got from stock returns to continuously compounded stock returns. As a recap, he built the generalized Wiener Process, which describes a change in some variable x is a function of a drift term and stochastic term.

dx = a * dt + b * ∈ * sqrt(dt)

where a * dt is the drif term, b * ∈ * sqrt(dt) is the stochastic term, and ∈ is a standard normal distribution N(0,1)

I get that part. However, that only describes the change in x not a return of x. So we need to make a modification. Under 13.3, he says we just multiply the drift and stochastic terms by x itself so that we can re-arrange the equation to be:

dx = a * x * dt + b * x * ∈ * sqrt(dt)

dx/x = a * dt + b * ∈ * sqrt(dt)

Which allows us to describe the returns of x. I get all of that. What I don't get is the case where b is 0, aka when there is no stochastic process. He says the above equation will just become:

dx/x = a * dt

Which when you take the integral with bounds 0 and T, you get:

xt = x0 * exp (a * t)

How do you get to the last equation, which is the continuously compounded return? Isn't the derivative of exp(x) equal to itself? And thus if the anti-derivative has exp(x) in it, that means the equation itself has exp(x) in it? I don't get how exp(x) appears out of thin air.


  • 1
    $\begingroup$ hi: integrate both sides of the first equation. $ln(x) = at + c$ where $c$ is a constant. then take exp of both sides. c is determined by initial conditions which leads to $x_{0}$. $\endgroup$ – mark leeds Apr 11 at 15:42
  • $\begingroup$ thanks! that helps! $\endgroup$ – vt_og Apr 11 at 18:24

The key is on the left hand side. Recall that the differential of log of x is:

$d \ln x =\frac{1}{x}dx$

So you get:

$\ln x_t-\ln x_0=at$

Which you will need to exponentiate to get rid of the log:



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