Solving for the Up State in a One-Period Binomial Call Model

Summary

The document shows how to infer the stock’s up-state price in a one-period binomial model when the current stock price, down-state price, strike, interest rate, and call price are given. It applies no-arbitrage pricing to both the stock and the call, using a risk-neutral probability to express their discounted expected payoffs. Assuming the up-state price is above the strike, the call pays in the up state and expires worthless in the down state.

The answer first rearranges the call-pricing equation to express the risk-neutral probability in terms of the unknown up price, then substitutes that expression into the stock-pricing equation and solves for the unknown. The document provides a closed-form expression for this setup. Its derivation depends on the one-period model, the stated down price, and the assumption that the call is in the money in the up state; if that payoff assumption fails, the equations must be adjusted.

Key ideas

• No-arbitrage pricing links the stock and call values through a shared risk-neutral probability.
• The call payoff depends on whether the stock price at expiry exceeds the strike.
• The risk-neutral probability can be expressed using the call price and the unknown up-state price.
• Substituting into the stock valuation equation lets the model solve for the up-state price.
• The derivation assumes the up-state price is above the strike.

Tags

# Pricing of a call option in one period binomial model


# Pricing of a call option in one period binomial model












You are given a $5\%$ call option worth $\$2.66$. The strike price $k$ is $\$41.00$. $S(0)=40$, $Sd=35$ (i.e the lower price of the stock at $t=1$) find $Su$ (i.e the high price of the stock at $t=1$).

How would this be done? I know $Cu=Su-\$41$, $C(0)=\$2.66$ and I have something written in my notes that $c(0)= B/(1+r) +ΔS(0)$, but I'm not sure what $B$ is here. I assume to find $Su$ I need to first find $Cu$, but I'm not sure how to do this from the given information. Any help is appreciated.

## Answer by user9403 (score 1)

https://quant.stackexchange.com/a/16769

By no arbitrage, $S(0)=(Su q+Sd (1-q))/(1+r)$ and $C(0)=((Su-k)^+ q+(Sd-k)^+(1-q))/(1+r)$. Simplifying and rearranging (and assuming $Su>k$), $$\left[\begin{array}{c} S(0) \\ C(0) \end{array} \right]=\frac{1}{1+r}\left[\begin{array}{cc} Su & Sd \\ (Su-k) & 0 \end{array} \right]\left[\begin{array}{c} q\\ 1-q \end{array} \right] $$ Clearly, $q=C(0)(1+r)/(Su-k) $. Substituting this back into the first equation, $$S(0)=\left(\frac{Su C(0)(1+r)}{Su-k}+Sd (1-\frac{C(0)(1+r)}{Su-k})\right)/(1+r)$$ After simplifying and solving for Su, $$Su=\frac{kS(0)(1+r)-kSd-C(0)Sd(1+r)}{S(0)(1+r)-C(0)(1+r)-Sd}$$

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