Can anyone explain to how Hull get's from the stock returns to continuously compounded stock returns?

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.

Thanks!

• 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 $x_{0}$. – mark leeds Apr 11 at 15:42
• thanks! that helps! – vt_og Apr 11 at 18:24

$$d \ln x =\frac{1}{x}dx$$
$$\ln x_t-\ln x_0=at$$
$$\frac{x_t}{x_0}=e^{at}$$