# 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?

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 $$S_0 \exp(r T) >= S_0$$.
• 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. Commented May 28, 2020 at 0:16
• And another question. In my problem, $T=2$ so wouldn't $S_T$ be $=p^2S_0u^2+2p(1-p)S_0ud+(1-p)^2S_0d^2$? Commented May 28, 2020 at 0:20