Valuing a Power Payoff in the Cox-Ross-Rubinstein Model

Summary

The document derives the initial value of a claim whose payoff is the terminal stock price raised to a positive integer power in a Cox-Ross-Rubinstein binomial model. Under the risk-neutral measure, it discounts the expected terminal payoff at the per-period risk-free rate. The terminal stock price depends on the number of up and down moves, so the expectation is initially expressed as a binomial sum.

The key correction is that the binomial probabilities and combination count are not raised to the payoff power. Instead, applying the binomial theorem to the sum yields the compact valuation: the initial stock price to the power, multiplied by the risk-neutral weighted average of the powered up and down factors, raised to the number of periods, then discounted. The result relies on the stated model assumptions and equivalent martingale probability; the document gives an algebraic derivation rather than numerical validation.

Key ideas

• The claim pays the terminal stock price raised to a positive integer power.
• Risk-neutral valuation discounts the expected payoff by the accumulated risk-free return.
• The terminal expectation is a binomial sum over possible up and down moves.
• The binomial theorem compresses the sum into a powered weighted average of the up and down factors.
• The probabilities and binomial coefficients are not raised to the payoff power.

Tags

# Valuation of Cox-Ross-Rubinstein Model


# Valuation of Cox-Ross-Rubinstein Model












We have a Cox-Ross-Rubinstein model with parameters $u$ ("up"), $d$ ("down") , $r$ (interest rate) and $q$ (equivalent martingale probability) $(q=(1+r-d)(u-d)^{-1})$ . We have a contingent claim with payoff $$ X=S_1(N)^c $$ where $S_1(N)$ is the final price, and $c$ is a positive integer. I need to show that the initial valuation of a claim is: $$ \pi_X(0) = S_1(0)^c(1+r)^{-N}\left(u^cq+d^c(1-q)\right)^N $$

I know that \begin{align} \pi_X(0) & = E_Q[\frac{X}{S_0(N)}] \\ & = (1+ r)^{-N}E_Q[S_1(N)^c] \\ & = (1+ r)^{-N} \sum_{j=0}^N(S_1(0)u^jd^{N-j})^c{N \choose k}q^j(1-q)^{N-j} \end{align} and then the $S_1(0)$ can be taken out which gives me the first part, but then I'm not really sure how to proceed. I don't see how the "T choose k" bit is going to disappear, or how we can get rid of the summation sign.

## Answer by ajc3 (score 1, accepted)

https://quant.stackexchange.com/a/11187

There was an error in your expected value, which I have corrected - the probabilities and the binomial coefficient (the "N choose k") should not be raised to the power $c$. With that correction, it is a simple application of the Binomial theorem: \begin{eqnarray} \left(u^cq+d^c(1−q)\right)^N&=&\sum_{j=0}^N {N \choose j}(u^cq)^{j}(d^c(1−q))^{N-j}\\ &=&\sum_{j=0}^N {N \choose j} \left(u^j d^{N-j}\right)^c q^j (1-q)^{N-j} \end{eqnarray}

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.