dstefan
07-08-2008, 01:03 PM
Chapter 5, Problem 5
In the Cox-Ross-Rubinstein parametrization for a binomial tree, the up and down factors u and d, and the risk-neutral probability p of the price going up during one time step are
\label{ex5_1}
u = A + \sqrt{A^2-1}
\label{ex5_2}
d = A - \sqrt{A^2-1}
\label{ex5_3}
p = \frac{e^{r \delta t} - d}{u-d}
where
A ~=~ \frac{1}{2} ~\left( e^{-r \delta t} + e^{(r+\sigma^2) \delta t} \right)
Use Taylor expansions to show that, for a small time step \delta t, u, d and p may be approximated by
\begin{eqnarray}
\label{ex5_4}
u = e^{\sigma \sqrt{\delta t}}
\label{ex5_5}
d = e^{- \sigma \sqrt{\delta t}}
\label{ex5_6}
p = \frac{1}{2} ~+~ \frac{1}{2} ~\left( \frac{r}{\sigma} - \frac{\sigma}{2} \right)~\sqrt{\delta t}
In other words, write the Taylor expansion for (\ref{ex5_1}-\ref{ex5_3}) and for (\ref{ex5_4}-\ref{ex5_6}) and show that they are identical if all the terms of order O(\delta t) and smaller are neglected.
Solution:
We will actually show that the Taylor expansion for (\ref{ex5_1}-\ref{ex5_3}) and for (\ref{ex5_4}-\ref{ex5_6}) are identical if all the terms of order O((\delta t)^{3/2}) and smaller are neglected.
We will use the following Taylor approximations:
e^x = 1 + x + \frac{x^2}{2} ~+~ O(x^3), ~~\mbox{as}~ x \to 0; \sqrt{1+x} = 1 + \frac{x}{2} ~+~ O(x^2), ~~\mbox{as}~ x \to 0; \sqrt{1-x} = 1 - \frac{x}{2} ~+~ O(x^2), \mbox{as}~ x \to 0
In particular, note that
\sqrt{1 + O(\delta t)} ~=~ 1 + \frac{\delta t}{2} ~+~ O((\delta t)^2)
Then,
A = \frac{1}{2} \left( e^{-r \delta t} + e^{(r+\sigma^2) \delta t} \right) = \frac{1}{2} \left( 1 - r \delta t + O((\delta t)^2) ~+~ 1 + (r+\sigma^2) \delta t + O((\delta t)^2) \right) = 1 + \frac{\sigma^2 \delta t}{2} + O((\delta t)^2)
A^2-1 = \left( 1 + \frac{\sigma^2 \delta t}{2} + O((\delta t)^2) \right)^2 - 1 = \sigma^2 \delta t + O((\delta t)^2
\sqrt{A^2-1} = \sqrt{\sigma^2 \delta t + O((\delta t)^2} ~=~ \sigma \sqrt{\delta t} ~\sqrt{1 + O(\delta t)} = \sigma \sqrt{\delta t} \cdot \left( 1 + \frac{\delta t}{2} ~+~ O((\delta t)^2) \right) = \sigma \sqrt{\delta t} + O((\delta t)^{3/2})
and
u = A + \sqrt{A^2-1} = 1 + \frac{\sigma^2 \delta t}{2} + O((\delta t)^2) ~+~ \sigma \sqrt{\delta t} + O((\delta t)^{3/2}) = 1 + \sigma \sqrt{\delta t} + \frac{\sigma^2 \delta t}{2} + O((\delta t)^{3/2})
d = A - \sqrt{A^2-1} = 1 - \sigma \sqrt{\delta t} + \frac{\sigma^2 \delta t}{2} ~+~ O((\delta t)^{3/2})
Since
e^{\sigma \sqrt{\delta t}}=1 + \sigma \sqrt{\delta t} + \frac{(\sigma \sqrt{\delta t})^2}{2} + O((\sigma \sqrt{\delta t})^3) = 1 + \sigma \sqrt{\delta t} + \frac{\sigma^2 \delta t}{2} + O((\delta t)^{3/2});
e^{-\sigma \sqrt{\delta t}} = 1 - \sigma \sqrt{\delta t} + \frac{\sigma^2 \delta t}{2} + O((\delta t)^{3/2}),
we conclude that
u = e^{ \sigma \sqrt{\delta t}} + O((\delta t)^{3/2}); d = e^{-\sigma \sqrt{\delta t}} + O((\delta t)^{3/2})
Finally,
p = \frac{e^{r \delta t} ~-~ d}{u ~-~ d} = \frac{\left( 1 + r \delta t + O((r \delta t)^2) \right) -~ \left( 1 - \sigma \sqrt{\delta t} + \frac{\sigma^2 \delta t}{2} + O((\delta t)^{3/2}) \right)}{\left( 1 + \sigma \sqrt{\delta t} + \frac{\sigma^2 \delta t}{2} + O((\delta t)^{3/2}) \right) ~-~ \left( 1 - \sigma \sqrt{\delta t} + \frac{\sigma^2 \delta t}{2} + O((\delta t)^{3/2}) \right)}
= \frac{\sigma \sqrt{\delta t} + (r-\frac{\sigma^2}{2}) \delta t + O((\delta t)^{3/2})} {2 \sigma \sqrt{\delta t} + O((\delta t)^{3/2}) } = \frac{\sigma + (r-\frac{\sigma^2}{2}) \sqrt{\delta t} + O(\delta t)}{2 \sigma + O(\delta t)} = \frac{1}{2} + \frac{1}{2 \sigma}(r-\frac{\sigma^2}{2}) \sqrt{\delta t} ~+~ O((\delta t)^{3/2}),
which is what we wanted to show; cf. (\ref{ex5_6}).
In the Cox-Ross-Rubinstein parametrization for a binomial tree, the up and down factors u and d, and the risk-neutral probability p of the price going up during one time step are
\label{ex5_1}
u = A + \sqrt{A^2-1}
\label{ex5_2}
d = A - \sqrt{A^2-1}
\label{ex5_3}
p = \frac{e^{r \delta t} - d}{u-d}
where
A ~=~ \frac{1}{2} ~\left( e^{-r \delta t} + e^{(r+\sigma^2) \delta t} \right)
Use Taylor expansions to show that, for a small time step \delta t, u, d and p may be approximated by
\begin{eqnarray}
\label{ex5_4}
u = e^{\sigma \sqrt{\delta t}}
\label{ex5_5}
d = e^{- \sigma \sqrt{\delta t}}
\label{ex5_6}
p = \frac{1}{2} ~+~ \frac{1}{2} ~\left( \frac{r}{\sigma} - \frac{\sigma}{2} \right)~\sqrt{\delta t}
In other words, write the Taylor expansion for (\ref{ex5_1}-\ref{ex5_3}) and for (\ref{ex5_4}-\ref{ex5_6}) and show that they are identical if all the terms of order O(\delta t) and smaller are neglected.
Solution:
We will actually show that the Taylor expansion for (\ref{ex5_1}-\ref{ex5_3}) and for (\ref{ex5_4}-\ref{ex5_6}) are identical if all the terms of order O((\delta t)^{3/2}) and smaller are neglected.
We will use the following Taylor approximations:
e^x = 1 + x + \frac{x^2}{2} ~+~ O(x^3), ~~\mbox{as}~ x \to 0; \sqrt{1+x} = 1 + \frac{x}{2} ~+~ O(x^2), ~~\mbox{as}~ x \to 0; \sqrt{1-x} = 1 - \frac{x}{2} ~+~ O(x^2), \mbox{as}~ x \to 0
In particular, note that
\sqrt{1 + O(\delta t)} ~=~ 1 + \frac{\delta t}{2} ~+~ O((\delta t)^2)
Then,
A = \frac{1}{2} \left( e^{-r \delta t} + e^{(r+\sigma^2) \delta t} \right) = \frac{1}{2} \left( 1 - r \delta t + O((\delta t)^2) ~+~ 1 + (r+\sigma^2) \delta t + O((\delta t)^2) \right) = 1 + \frac{\sigma^2 \delta t}{2} + O((\delta t)^2)
A^2-1 = \left( 1 + \frac{\sigma^2 \delta t}{2} + O((\delta t)^2) \right)^2 - 1 = \sigma^2 \delta t + O((\delta t)^2
\sqrt{A^2-1} = \sqrt{\sigma^2 \delta t + O((\delta t)^2} ~=~ \sigma \sqrt{\delta t} ~\sqrt{1 + O(\delta t)} = \sigma \sqrt{\delta t} \cdot \left( 1 + \frac{\delta t}{2} ~+~ O((\delta t)^2) \right) = \sigma \sqrt{\delta t} + O((\delta t)^{3/2})
and
u = A + \sqrt{A^2-1} = 1 + \frac{\sigma^2 \delta t}{2} + O((\delta t)^2) ~+~ \sigma \sqrt{\delta t} + O((\delta t)^{3/2}) = 1 + \sigma \sqrt{\delta t} + \frac{\sigma^2 \delta t}{2} + O((\delta t)^{3/2})
d = A - \sqrt{A^2-1} = 1 - \sigma \sqrt{\delta t} + \frac{\sigma^2 \delta t}{2} ~+~ O((\delta t)^{3/2})
Since
e^{\sigma \sqrt{\delta t}}=1 + \sigma \sqrt{\delta t} + \frac{(\sigma \sqrt{\delta t})^2}{2} + O((\sigma \sqrt{\delta t})^3) = 1 + \sigma \sqrt{\delta t} + \frac{\sigma^2 \delta t}{2} + O((\delta t)^{3/2});
e^{-\sigma \sqrt{\delta t}} = 1 - \sigma \sqrt{\delta t} + \frac{\sigma^2 \delta t}{2} + O((\delta t)^{3/2}),
we conclude that
u = e^{ \sigma \sqrt{\delta t}} + O((\delta t)^{3/2}); d = e^{-\sigma \sqrt{\delta t}} + O((\delta t)^{3/2})
Finally,
p = \frac{e^{r \delta t} ~-~ d}{u ~-~ d} = \frac{\left( 1 + r \delta t + O((r \delta t)^2) \right) -~ \left( 1 - \sigma \sqrt{\delta t} + \frac{\sigma^2 \delta t}{2} + O((\delta t)^{3/2}) \right)}{\left( 1 + \sigma \sqrt{\delta t} + \frac{\sigma^2 \delta t}{2} + O((\delta t)^{3/2}) \right) ~-~ \left( 1 - \sigma \sqrt{\delta t} + \frac{\sigma^2 \delta t}{2} + O((\delta t)^{3/2}) \right)}
= \frac{\sigma \sqrt{\delta t} + (r-\frac{\sigma^2}{2}) \delta t + O((\delta t)^{3/2})} {2 \sigma \sqrt{\delta t} + O((\delta t)^{3/2}) } = \frac{\sigma + (r-\frac{\sigma^2}{2}) \sqrt{\delta t} + O(\delta t)}{2 \sigma + O(\delta t)} = \frac{1}{2} + \frac{1}{2 \sigma}(r-\frac{\sigma^2}{2}) \sqrt{\delta t} ~+~ O((\delta t)^{3/2}),
which is what we wanted to show; cf. (\ref{ex5_6}).