Deriving the Adjusted Probability in CRR Call Option Pricing

Summary

The document addresses a step in the Cox–Ross–Rubinstein binomial model derivation for a European call. The option value is written as a discounted sum over terminal stock prices above the strike, then rearranged into two binomial tail sums: one weighted by an adjusted probability for the stock-price component and another by the risk-neutral probability for the strike component.

The key identity is that if the adjusted probability is defined as the up factor times the risk-neutral probability discounted over one time step, its complement equals the down factor times the discounted probability of a down move. This follows by substituting the CRR risk-neutral probability, which is determined by the risk-free growth factor and the up and down factors. The explanation verifies the algebra, but assumes the standard CRR setup and does not discuss alternative lattices or implementation details.

Key ideas

• The CRR call value can be rearranged into two binomial tail sums for the stock and strike terms.
• The adjusted probability is the up factor multiplied by the risk-neutral probability and discounted over one time step.
• Using the CRR risk-neutral probability shows that the complement of the adjusted probability equals the discounted down-move term.
• The derivation depends on the standard CRR relationship between the risk-free growth factor and the up and down factors.

Tags

# Call option pricing using CCR model - derivation problem


# Call option pricing using CCR model - derivation problem












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

## Answer by Gordon (score 1, accepted)

https://quant.stackexchange.com/a/17180

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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