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I'm viewing the following derivation of a Call Option price using the CRR model. There is one piece of the derivation which I cannot understand.

\begin{align} C_0 &= e^{-rT} \sum_{i=0}^{N} (S_{0}\,u^{N-i}\,d^{i} - K)^{+} \binom {N}{i} q^{N-i}(1-q)^{i}\\ &= e^{-rN \Delta t} \sum_{i=a}^{N} (S_{0}\,u^{N-i}\,d^{i} - K) \binom {N}{i} q^{N-i}(1-q)^{i}\\ &= S_0 \sum_{i=a}^{N} \binom {N}{i} (u\,q\,e^{-r \Delta t})^{N-i}\, (d\,e^{-r \Delta t}\,(1- q))^{i} - Ke^{-rT} \sum_{i=a}^{N} \binom {N}{i} q^{N-i} (1-q)^{i}\\ &= S_0 \sum_{i=a}^{N} \binom {N}{i} \overline{q}^{N-i}\, (1 - \overline{q})^{i} - Ke^{-rT} \sum_{i=a}^{N} \binom {N}{i} q^{N-i} (1-q)^{i}\\ &= S_0 \mathcal{Q}_1 - K e^{-rT} \mathcal{Q}_2 \end{align}

where $\overline{q} = uqe^{-r\Delta t}$.

$\textbf{Question}$

If $\overline{q} = uqe^{-r\Delta t}$, then I'm assuming $(1-\overline{q}) = d\,e^{-r \Delta t}\,(1- q)$, however I cannot seem to derive this equality.

Appreciate any help understanding why.

Many thanks,

John

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1 Answer 1

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Note that \begin{align*} q= \frac{e^{r\Delta t} -d}{u-d}. \end{align*} Then, \begin{align*} u = \frac{e^{r\Delta t} -d}{q} + d. \end{align*} Therefore, \begin{align*} 1-\bar{q} &= 1-uqe^{-r\Delta t}\\ &=1- \big(e^{r\Delta t} -d\big)e^{-r\Delta t}-dqe^{-r\Delta t}\\ &=de^{-r\Delta t} -dqe^{-r\Delta t}\\ &=de^{-r\Delta t}(1-q). \end{align*}

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  • $\begingroup$ Of course it is, how embarrassing. In my defence it's been a long day. Many thanks Gordon, appreciated. $\endgroup$
    – John Smith
    Mar 31, 2015 at 17:20

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