Quantitative Finance Stack Exchange is a question and answer site for finance professionals and academics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose I model the forward swap rate lognormal

$$dS_t = \sigma_{ln}S_tdW_t$$

On the other hand we could model it simply by a normal assumption:

$$dS_t = \sigma_{n}dW_t$$

I would like to know if there is a relationship for the volatilities $\sigma_n,\sigma_{ln}$? A friend told me, that he saw the approximation

$$\sigma_n\approx \sigma_{ln}S_t$$

However, neither my friend nor I was able to come up with a justification of this approximation. So is this a valid approximation? If so, why and if not, how else can I relate the two volatilities?

share|improve this question
up vote 2 down vote accepted

It might help to look at the solutions of the SDEs that you have there. In the first case $$ S_t/S_0 = \exp(-\sigma^2/2 t + \sigma B_t) \quad \quad (1) $$ Thus if you take the log then $\sigma$ is the volatility of the log-returns (assume that $t=1$ time step),.

In the second case $$ S_t = S_0 + \sigma B_t \rightarrow S_t - S_0 = \sigma B_t \quad \quad(2) $$ then $\sigma$ is the volatility of the absolute differences.

Coming back to your actual question, the solution should be a simple expansion of the exponential.

Take the solution of the geometric Brownian motion above (1) and we look at the time step of $\Delta t$: $$ S_{t + \Delta t} = S_t \exp(-\sigma_{ln}^2/2 \Delta t + \sigma_{ln} B_{t+\Delta t}) \approx S_t \exp( \sigma_{ln} B_{t+\Delta t}) $$ where we observe that $\sigma_{ln}^2/2$ is small. Furthermore note that $$ \exp(x) = \sum_{n=0}^\infty x^n/n! \approx 1 + x $$ where the last step is an approximation for $x$ small, thus $$ S_{t + \Delta t} \approx S_t (1+ \sigma_{ln} B_{t+\Delta t}) $$ and finally (after multiplication and rearranging terms) $$ S_{t + \Delta t} - S_t \approx S_t \sigma_{ln} B_{t+\Delta t} $$ The last equation is of the form (2) with $\sigma = \sigma_{ln} S_t$.

share|improve this answer
brilliant. many thanks! – user8 Sep 22 '14 at 18:47

For any normal variable, you have $$aX\sim N(a\mu,a^2\sigma^2).$$ So a linear transformation preserves the distribution type (note that $dW_t\sim N(0,dt)$).

When you want to approximate $dS_t$ by setting $dS_t=dS_t$, canceling the $dW_t$ you get:


share|improve this answer
I believe Pat Hagan did some work like this for his SABR model too, wilmott.com/pdfs/021118_smile.pdf formula A.64 may be relevant. – experquisite Aug 20 '14 at 23:03

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.