Comparing a Vanilla Call with an Average-Price Call in a Binomial Model

Summary

The document asks whether a standard call option or a contract paying on the average of the initial and terminal stock prices must have the higher arbitrage-free value in a two-period binomial model. The answer introduces a risk-neutral probability for a one-step model and relates the expected terminal stock price to the risk-free growth of the initial price. It then argues that, with a positive risk-free rate, the terminal-price call should be more valuable than the average-price contract.

This is a useful starting point for comparing contingent payoffs and connecting option valuation to risk-neutral expectations. The reasoning is brief and does not fully analyze the two-period tree or compare the payoffs state by state. Its conclusion relies on an unstated or assumed nonnegative rate and on conventions for the up and down factors; the stated conditions alone do not establish all those assumptions. A rigorous comparison would price both payoffs across the tree using the model’s specified discounting and probabilities.

Key ideas

• The two contracts depend on different stock-price references: terminal price alone versus an average of initial and terminal prices.
• Risk-neutral valuation uses probabilities consistent with the risk-free growth of the stock price.
• The answer claims the ordinary call is more valuable when the risk-free rate is positive.
• The brief argument does not fully establish the result for the stated two-period model or all possible rates.

Tags

# call vs average of prices


# call vs average of prices












> Consider a two-period binomial model, with one risky asset. The are two types of options: call option with strike price $K$, i.e., the payoff is given by $g(S_T)=(S_T-K)^{+}$ option with payoff given by the average of prices, i.e., $g(S_T)=(\frac12(S_0+S_T)-K)^{+}$ where $X^{+}=\max\{X,0\}$. Assume that $u>1$ and $ud>1$. Is it possible to know which option has higher arbitrage free price?

What I've tried:

I plotted the payoff functions for both contracts and realized that the second option is better than the second if the price of the stock at maturity, $S_T$ lies in $(K-\frac12S_0,K+S_0)$ and the first option is better if the $S_T$ lies in $(S_T+S_0,+\infty)$

Intuitively, I would say that the call option is better and then it would cost more. But I don't know if I'm correct or how can I determine which option should be more expensive.

Any ideas?

## Answer by stackoverblown (score 2)

https://quant.stackexchange.com/a/54472

The call is worth more unless the risk free rate is zero. Let $p$ be the probability of $S_0$ going up, $r$ be risk free rate, $T$ is the one time step. Then no arbitrage means $$ S_T = S_0 \exp(r T) = p S_0 u + (1-p) S_0 d$$. I am assuming $u$ and $d$, which you did not say, are the up and down factors. Then obviously $$ S_0 \exp(r T) >= (S_0 \exp(r T) + S_0)/2 $$ because $$ S_0 \exp(r T) >= S_0$$.

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