# Fair Value Of a Call Option

I am dealing with the following question/information (which may not be complete):

The price of the underlying asset is $100$. European call option with exercise price $150$ in a year. The price of the underlying asset in a year is $200$ with probability $p$ and $50$ with probability $1-p$. Calculate the fair value of the option, given that there is no arbitrage opportunity. Risk-free interest rate is $i$. Hint: Calculate the probability so that there can't be arbitrage opportunity first.

There are several problems I have with this question.

1. The hint confuses me. Given the information, I wouldn't have even guessed one can calculate $p$
2. It seems to me that the first sentence is incomplete? I suppose it is the price at the current date.
3. General confusion as I've sadly never dealt with this kind of question. (I do however know and understand the put-call-parity argument)

Thanks for any help!

This problem assumes there are two possible states of the market in a single time period.

At time $t=0$, the asset price is $S_0 = 100$ and the call option price is $C_0$, to be determined. At time $t=1$, there are two possible asset prices $S_{1u} = S_0(1+u)=200$ with probability $p$ and $S_{1d} = S_0(1+d)=50$ with probability $1-p$. In this case, $u=1$ and $d = -0.5$ Let $K=150$ denote the option strike. The payoff of the option at expiration is

$$C_1 = \max(S_1-K,0),$$

where the random variable $S_1$ is $S_{1u}$ or $S_{1d}$ depending on the future state of the market:

$$C_{1u} = \max(S_{1u}-K,0)\\\ C_{1d} = \max(S_{1d}-K,0)$$

The value of the option at time $t=0$ is the discounted expected value of the payoff. However this expected value is not calculated with real probability $p$ -- but rather the risk-neutral probability $\hat{p}$ that preclude arbitrage opportunities.

We can determine the no-arbitrage risk-neutral probability $\hat{p}$ by showing it is possible to construct a hedged portfolio of the option and the asset that is risk free -- it has the same value in both future states. Hence the value of the portfolio grows in time at the risk-free rate of interest $i$.

Suppose the portfolio is long $1$ call option and short $\Delta$ shares of the asset. The value at time $t=0,1$ is

$$V_t= C_t-\Delta S_t.$$ We can solve for the hedge ratio $\Delta$ so that the value of the portfolio at time $t=1$ is independent of the state of the market:

$$C_{1u}-\Delta S_{1u}=C_{1d}-\Delta S_{1d},$$

or

$$C_1-\Delta S_1=C_{1u}-\Delta S_0(1+u)=C_{1d}-\Delta S_0(1+d).$$

This value of the hedge ratio is independent of the probabilities for the value of the asset at time $t=1$:

$$\Delta = \frac{C_{1u}-C_{1d}}{S_0(u-d)}$$

Consequently, in the absence of arbitrage, the portfolio grows at the risk-free rate:

$$C_1-\Delta S_1 = (C_0 - \Delta S_0)(1+i)$$

and we can solve for the value of the call option at time $t=0:$

$$C_0 -\Delta S_0= \frac{C_{1u}-\Delta S_0 (1+u)}{1+i},$$

$$C_0 = \frac{\Delta S_0 (1+i) +C_{1u}-\Delta S_0 (1+u)}{1+i},$$

$$C_0 = \frac{-\Delta S_0 (u-i) + C_{1u}}{1+i}.$$

Substituting for $\Delta$, we obtain

$$C_0 = \frac{\hat{p}C_{1u}+(1-\hat{p})C_{1d}}{1+i}=\frac{1}{1+i}\{\hat{p}\max[S_0(1+u)-K,0)]+(1-\hat{p})\max[S_0(1+d)-K,0)]\},$$

where

$$\hat{p} = \frac{i-d}{u-d}.$$

This has the form of an expected value with a different probability -- the risk-neutral probability. The fair value of the call option is the discounted expected value under the risk-neutral probability measure.

To understand the no-arbitrage argument, consider the fact that the hedged option portfolio assumes the same value in the future regardless of the state of the market. If this portfolio did not grow at the risk-free rate, then we could make a sure profit. For example if the portfolio grows at a greater rate, we can borrow cash at rate $i$, buy the portfolio, earn the sure payoff and repay the loan with a sure profit.

In general, for any contingent claim on an asset, the no-arbitrage fair value is the discounted expected value of the payoff under the risk-neutral measure. For this simple binomial model, we can find the risk-neutral probabilities in a more straigtforward way. The present value of the underlying asset itself must equal the discounted risk-neutral expected value of the future price.

Hence,

$$S_0 = \frac{\hat{p}S_0(1+u) + (1-\hat{p})S_0(1+d)}{1+i},$$

and

$$\frac{\hat{p}(1+u) + (1-\hat{p})(1+d)}{1+i} = 1.$$

Solving for the risk-neutral probability $\hat{p}$ we again obtain

$$\hat{p} = \frac{i-d}{u-d}.$$

• Thank you very much. This question is a question from a past exam from a lecture and your answer seems to go beyond the scope of the lecture (don't get me wrong, it is perfectly understandable, I just cannot imagine a person that'd answer this, having only done the course, as we've never done anything like that and one would have to pull it out of thin air using a great deal of intuition). Thus I am wondering, are there perhaps any hidden assumptions or simplifications (even if they change the nature of the question a bit) that could've been made in a more rudimentary course?
– Dahn
Commented Jun 3, 2014 at 7:24
• You're welcome. That's what I was thinking as well There is a quicker way to the result which I will post for you
– RRL
Commented Jun 3, 2014 at 10:49
• @Dahn Jahn: I added a shorter derivation of the risk-neutral (no-arbitrage) probabilities. The point of the hedging argument (the discrete analog of Black-Scholes) was to provide some explanation for the connection between risk-neutral probability and no-arbitrage.
– RRL
Commented Jun 3, 2014 at 14:23