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\title{Derivatives Lab}
\author{Session 10}
\date{October 24, 2011}
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\def\e{\mathrm e}

\begin{enumerate}

\item Recall that for a binomial tree with probability for upward
movement $p$, the expected single period return is
\[
  \hat \mu = p \, \ln \frac{u}d + \ln d
\]
with single period variance
\[
  \hat \sigma^2 = p \, (1-p) \, \ln^2 \frac{u}d \,.
\]
Recall further that, to obtain a limit as $n \to \infty$, we must have
\[
  \hat \mu \sim \mu \, \frac{T}n
\]
and
\[
  \hat \sigma^2 \sim \sigma^2 \, \frac{T}n
\]
where $T$ is the total duration in years, $\mu$ the annual
continuously compounded rate of return, and $\sigma$ the so-called
annualized volatility.

So far, we have satisfied these relationships choosing
\begin{gather*}
  u = \e^{\sigma \sqrt{T/n}} \,, \\
  d = \e^{-\sigma \sqrt{T/n}} \,, \\
  p = \frac12 + \frac12 \, \frac\mu\sigma \, \sqrt{\frac{T}n} \,.
\end{gather*}

What expressions do you obtain by fixing $p=1/2$ (rather than
$u=1/d$)?  Does the binomial tree model compute the same option
prices?

\begin{enumerate}
\item Modify your binomial tree program to try this out.
\item Why does it work?
\end{enumerate}
  
\item Compare your call option prices against those computed by the
Black--Scholes formula.  Does the error scale like a power of $n$?
(Plot the logarithm of the error vs.\ $n$.  Do you obtain a straight
line?)



\end{enumerate}


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