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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?

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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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  • $\begingroup$ Why did you write $S_T=p S_0 u + (1-p) S_0 d$? I understand why this is true but I didn't understand why you had to write it. $\endgroup$ – Babado May 28 at 0:16
  • $\begingroup$ And another question. In my problem, $T=2$ so wouldn't $S_T$ be $ =p^2S_0u^2+2p(1-p)S_0ud+(1-p)^2S_0d^2$? $\endgroup$ – Babado May 28 at 0:20
  • $\begingroup$ @Babado Sure but it won't change the conclusion. You can split T into 2, 3, 10, 10000, or as many time steps as you want, it won't change the conclusion. $\endgroup$ – stackoverblown May 28 at 20:06

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