Risk Neutral Pricing of a Call Option with a Two-State Tree

Risk Neutral Pricing of a Call Option with a Two-State Tree

In our last article on Hedging the sale of a Call Option with a Two-State Tree we showed that there was one unique price for a call option on an underlying stock, in a world with two-future states. This was guaranteed by the principle of no arbitrage. The most surprising consequence of the argument was that the probability of the stock going up or down did not factor into the discussion. We will now utilise a probability argument and show that the value $C$ of the call-option is achieved. Note: It will be necessary to read the prior articles on the Binomial Trees in order to familiarise yourself with the example of the stock and option before proceeding. Click for Part 1 and Part 2.

Consider the same world as before, which has a stock valued today at $S$ equal to 100, with the possibility of a rise in price to 110 or a fall in price to 90. Our task as an insurance firm is to price a call option struck at $K = 100$ such that all risk is eliminated from the sale of this option to a purchaser. We will use a probability argument for this particular technique, which is known as risk neutral pricing.

Let us assume that the probability of the stock going up to 110 is given by $p$ and that the probability of it falling is given by $1-p$. Since these are the only two probabilities, we can see that $p + (1-p) = 1$, i.e. that both probabilities sum to unity and thus one of the events must occur. Thus, the expected value of our stock $S$ tomorrow, is given by:

\begin{eqnarray*} \mathbb{E}(S_2) = 110p + 90(1-p) \end{eqnarray*}

This leads to the expected value of the option price $C$ to be:

\begin{eqnarray*} \mathbb{E}(C ) = 10p + 0(1-p) = 10p \end{eqnarray*}

The only value of $p$ which causes the option value $C$ to agree with the price obtained from the hedging argument is $p=0.5$. How does this affect the expected value of the stock in tomorrow's world? Well, $\mathbb{E}(S_2) = 110p + 90(1-p) = 110\cdot 0.5 + 90\cdot (1-0.5) = 100$. Thus the expected value of $S$ is today's price. Note that this is a risk free price because we are still setting interest rates to zero and can synthesise the stock using zero-coupon bonds worth 100, as before.

It is very important to realise that we have assumed any purchasers of this stock are risk neutral and do not need to be compensated for taking on the extra risk associated with a stock that can take on two differing values. In reality, this is not likely to be the case. These purchasers will require compensation for taking on this uncertainty, which will cause our probability $p$ to be larger than 0.5 and thus $\mathbb{E}(C )$ will be larger than the no-arbitrage value. The fact that we can hedge the entire portfolio has removed the diversifiable risk and thus eliminated the premium usually associated with holding this risk.

We haven't yet considered the possibility that $p=1$, i.e. that the stock is guaranteed to go up. This would in fact lead to an arbitrage opportunity. In order to see this, we could borrow money at the risk free rate (which is currently zero!) and then purchase some stock today. Tomorrow when the bond matures we could sell the stock, which is guaranteed to increase in value, and then pay back the bond, leaving us with a risk-free profit. Thus the probabilities in this argument are only present in order to allow both world states to occur.

So what is actually happening here and why does this method work? Once the value of the option has been specified to be its risk neutral value, which is determined by the probability $p$, we can conclude that every instrument in the market is valued today by the its risk neutral expected value tomorrow. Thus, $\mathbb{E}(C )=C_1$ implies that $\mathbb{E}(S)=S_1$.

Next we will consider a third method of pricing an option, that of replication.