Deriving Continuously Compounded Returns from Proportional Changes

# Can anyone explain to how Hull get from stock returns to continuously compounded stock returns?


# Can anyone explain to how Hull get from 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 drift 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, which is 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!

## Answer by Magic is in the chain (score 1)

https://quant.stackexchange.com/a/45096

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:

$\frac{x_t}{x_0}=e^{at}$