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


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,



1 Answer 1


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*}

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.