Question: 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.

Answer: 
To solve the expectation directly, you need to remember that a density function is not the same as the probability of the event.

We have, $\frac{S\_1}{S\_0} \sim \ln \mathcal{N} \left(-\frac{\sigma^2}{2},\sigma\right)$, therefore,

\begin{eqnarray}
\mathbb{E}\left(\frac{S\_1}{S\_0}\right) &=& \int\_{-\infty}^\infty x\, f\_{\frac{S\_1}{S\_0}}(x)dx\\
&=&\int\_{-\infty}^\infty x \, \frac{1}{x\sqrt{2\pi}\sigma}\exp\left[-\frac{\left(\ln x+\frac{\sigma^2}{2}\right)^2}{2\sigma^2}\right]dx.
\end{eqnarray}
The usual substitution method aims to manipulate the expression into looking like a standard normal. To do this we use $z=\frac{\ln x + \frac{\sigma^2}{2}}{\sigma}-\sigma$ and we get the substitutions $dx=x \sigma dz$ and $x=\exp(\sigma z + \frac{\sigma^2}{2})$.

\begin{eqnarray}
\mathbb{E}\left(\frac{S\_1}{S\_0}\right) &=& \int\_{-\infty}^\infty \frac{1}{\sqrt{2\pi} \sigma} \sigma \exp\left(\sigma z + \frac{\sigma^2}{2}\right) \exp\left( \frac{-z^2 - 2\sigma z -\sigma^2}{2} \right)dz\\
&=&\int\_{-\infty}^\infty \frac{1}{\sqrt{2\pi}}\exp\left(-\frac{z^2}{2}\right) dz\\
&=&1.
\end{eqnarray}
If $S\_t = S\_0 e^{(\mu-\sigma^2/2)t + \sigma W\_t}$, we can compute
$$\mathbb{E}^Q\left[S\_T\middle\vert \mathcal{F}\_0\right] = S\_0 e^{r T} = \text{forward price of } S\_T \text { at time } 0. $$
To show the details, 
$\mathbb{E}^Q\left[S\_T\middle\vert \mathcal{F}\_0\right] = S\_0 e^{(r-\sigma^2/2) T} \mathbb{E}^Q\left[e^{\sigma W\_T} \middle\vert \mathcal{F}\_0\right]$. To compute $\mathbb{E}^Q\left[e^{\sigma W\_T} \middle\vert \mathcal{F}\_0 \right]$, note that, for $X \sim \mathcal{N}(\mu,\sigma^2)$, $\mathbb{E}\left[e^X \right]=e^{\mu+\sigma^2/2}$.

In your case, since $r=0$ and $T = 1$, 
$$\mathbb{E}^Q\left[\frac{S\_1}{S\_0}\middle\vert \mathcal{F}\_0\right] = 1. $$