I am reading Rebonato's Volatility and Correlation (2nd Edition) and I think it's a great book. I'm having difficulty trying to derive a formula he used that he described as the expression for standard deviation in a simple binomial replication example:

\begin{eqnarray}\sigma_S\sqrt{\Delta t}=\frac{\ln S_2-\ln S_1}{2}\end{eqnarray}

This expression is equation (2.48) on page 45. You can read that page and get some context from Google Books: http://goo.gl/uDgYg3

I understand continuous compounding is used in the example, if that helps any. It's a little confusing because the equations he listed a few pages above (pg.43; not available in Google Books) use a discrete rate of return, not continuous compounding. But in any case, this discrepancy does not seem to provide any hint as to how the standard deviation is obtained.

Any help is much appreciated.


2 Answers 2


Consider a one-period, two-state market model. The known asset price at time $t=0$ is $S_0.$ At time $\Delta t$, the asset price can be $S_1$ with probability $\pi_1$ or $S_2$ with probability $\pi_2$, where $\pi_1+\pi_2=1$.

In this example the probabilities are fixed as $\pi_1=\pi_2= \frac{1}{2}$.

Define log-returns over this period as

$$R_1 = \ln(S_1/S_0), \\R_2 = \ln(S_2/S_0).$$

The expected return is

$$E(R)= \pi_1R_1+\pi_2R_2=\frac{1}{2}[\ln(S_1)+\ln(S_2)]-\ln(S_0).$$

The deviations of returns around the mean are

$$R_1-E(R)= [\ln(S_1)-\ln(S_0)]-\{\frac{1}{2}[\ln(S_1)+\ln(S_2)]-\ln(S_0)\}=\frac{1}{2}[\ln(S_1)-\ln(S_2)]$$


$$R_2-E(R)= [\ln(S_2)-\ln(S_0)]-\{\frac{1}{2}[\ln(S_1)+\ln(S_2)]-\ln(S_0)\}=\frac{1}{2}[\ln(S_2)-\ln(S_1)]$$

and the variance of return is

$$var(R)= E[(R-E(R))^2]=\pi_1(R_1-E(R))^2+\pi_2(R_2-E(R))^2=\frac{1}{2}\frac{[\ln(S_1)-\ln(S_2)]^2}{4}+\frac{1}{2}\frac{[\ln(S_2)-\ln(S_1)]^2}{4}=\frac{[\ln(S_2)-\ln(S_1)]^2}{4}.$$

The one-period volatility is the standard deviation of return and is given by

$$\sigma_P=\sqrt{var(R)}= \frac{1}{2}[\ln(S_2)-\ln(S_1)].$$

If $\sigma$ denotes the annualized volatility then

$$\sigma\sqrt{\Delta t}= \sigma_P= \frac{1}{2}[\ln(S_2)-\ln(S_1)].$$

  • $\begingroup$ Excellent thank you! $\endgroup$
    – JohnC
    Commented Jun 4, 2014 at 21:56
  • $\begingroup$ You're welcome. I should have read some of the preceding pages first. $\endgroup$
    – RRL
    Commented Jun 4, 2014 at 22:05

The binomial tree is a discrete approximation of the continuous lognormal stochastic process for the underlying asset price.

The tree specifies a price $S_{ij}$ at node $(i,j)$ where $j$ is the time index and $i$ indexes the asset price at a fixed time. Potentially, a very large number of parameters could be used to construct a tree, but these are usually restricted in number in the more computationally efficient recombining tree -- in order to match the first two moments of the continuous distribution. These parameters are the time step size $\Delta t$, the up/down move parameters $u$ and $d=1/u$ and branching probability $p$. In this way the price at nodes $(i+1,j+1)$ and $(i,j+1)$ are given by,

$$S_{i+1,j+1} = S_{ij}u \\\ S_{i,j+1} = S_{ij}d$$

The parameters u and d are specified as

$$u = e^{\sigma \sqrt{\Delta t}} \\d= 1/u = e^{-\sigma \sqrt{\Delta t}}$$

where $\sigma$ is the annualized volatility of the asset

The remaining parameter $p$ is set to match the risk-neutral drift.

We can back out the volatility from these equations as follows

$$\sigma \sqrt{\Delta t}= \frac{1}{2}[\ln(u)-\ln(d)]=\frac{1}{2}[\ln(S_{i+1,j+1}/S_{ij})-\ln(S_{i,j+1}/S_{ij})]=\frac{1}{2}[\ln(S_{i+1,j+1})-\ln(S_{i,j+1})]$$

and using the notation in the reference

$$\sigma \sqrt{\Delta t}= \frac{1}{2}[\ln(S_{2})-\ln(S_{1})].$$

  • $\begingroup$ I'm afraid this answer makes too many assumptions outside of the framework set up in the context of the Worked-Out Example in question. In the example, the pricing is approached from a "market price of risk" angle. I am trying to figure out how he came to the formula for standard deviation without all the implicit assumptions and framework of a tree. $\endgroup$
    – JohnC
    Commented Jun 4, 2014 at 19:43
  • $\begingroup$ @JohnC: There is an equivalence. Nevetheless I posted again with a derivation in the framework of your example. $\endgroup$
    – RRL
    Commented Jun 4, 2014 at 21:19

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