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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?

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2 Answers 2

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 $\sigma = \sigma_{ln} S_t$.

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brilliant. many thanks! –  user8 Sep 22 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:

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

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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 at 23:03

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