# Hedging in a finite-state model (Binary Trees)

Today’s post is about hedging in a world where there are only a few outcomes. It demonstrates a lot of principles that are important in asset pricing, and it’s also a frequent topic for interview questions so almost got included as one of those!

A very basic approximation for a stock is a binary tree, as in the diagram shown below. The stock is worth 100 today, and after some time (say a year) it will be worth either 110 with probability p or 90 with probability (1-p). In this very basic model, we assume no other values are possible – the stock either does well and goes up by 10, or it does badly and goes down by 10. Furthermore, we assume that there is a risk-free yearly interest rate r at which we can deposit money in the bank and receive (1+r) times our initial deposit in a year.

We’re going to try and price options in this model. Let’s say we’ve got a call option to buy the stock in a year’s time for 100. How much is this worth? Because of the simple state of the world, we can enumerate the two possibilities – either the stock is worth 110, in which case the call option to buy it for 100 is worth 10, or the stock is worth 90, and the option is worthless. We’ll try and simulate that payoff using a combination of cash-in-the-bank and the stock itself, and then by no-arbitrage the price must be the same today.

Our portfolio consists of stocks S and cash-in-the-bank. The value of the cash will certainly be (1+r) in a year, while the stock will be 110 or 90 depending on the state, and we want the sum of these two to equal the option payoff (in either the up or the down scenario), which gives us two simultaneous equations  so  This says that a portfolio containing half of a stock and minus 45 pounds in the bank will exactly replicate the payoff of the option in either world state. The value of this portfolio today is just 0.5*100 – 45/(1+r), and I’ve plotted that as a function of r below. The price of a call option on a stock as a function of the risk-free rate in the two-state world given above

This gives meaningless prices if r > 0.1 (the price of the option must be between 0 and 10/(1+r) as these are the discounted values of the least/most that it can pay out at expiry). What does this mean intuitively? Well, if r > 0.1 then we have an arbitrage: 100 in the bank will *always* yield more than the stock at expiry, so we should short the stock as much as possible and put the proceeds in the bank. If r < -0.1 the situation is exactly reversed

The important point about all of this was that the option price DOESN’T depend on the relative chances of the stock increasing to 110 or falling to 90. As long as both of these are strictly greater than zero and less than one, then ANY set of probabilities will lead to the same price for the option. This is the discrete analogue of a result I presented before (here) – that the expected growth of the stock in the real world doesn’t matter to the option’s price, it is the risk-free rate that affects its price. Indeed, it is possible to derive the continuous result using a binary tree by letting the time period go to zero, I won’t attempt it here but the derivation can be found in many textbooks (eg. on wikipedia).

Things really get interesting when we try to extend the model in a slightly different way. Often banks will be interested in models that have not just a diffusive component, but also a ‘jump’ component, which gives a finite chance that a price will fall suddenly by a large amount (I’ll present one such model in the near future). Unlike a straight-forward Black-Scholes model, because these jumps happen suddenly and unexpectedly they are very difficult to hedge, and are meant to represent market crashes that can result in sudden sharp losses for banks.

In our simple tree model, this can be incorporated by moving to a three-branch model, shown below

We have the same two branches as before, but an additional state now exists in which the stock price crashes to 50 with probability q. In this case, again trying to price an option to buy the stock for 100 gives three simultaneous equations   Unlike before, we can’t find a single alpha and beta that will replicate the payoff in all three world states, as we have three equations and only two unknowns. Consequently, the best we will be able to to is sub-replicate or super-replicate the payoff. That is, find portfolios that either always pay equal to or greater than the option, or always pay less than or equal to the option. These will give bounds on the lowest and highest arbitrage-free option prices, but that’s as far as arbitrage-free prices will take us (in fact ANY of the prices between this limit is arbitrage-free) – choosing an actual price will require knowing the probabilities of p and q and deciding on our personal risk-premium.

In the simple three-state model, lower and upper bounds can be calculated by ignoring the second and third equation respectively above, and they give the limits shown in the graph below. Once again, as r 0.1 they converge, but note that a case where r < -0.1 is now possible, as the ‘crash’ option means that the stock can still pay out less than the bank account The limiting values given by the no-arbitrage requirement on the price of a call option as a function of the risk-free rate in the three-branch tree model given above

This is what’s called an incomplete market. The securities that exist in the market aren’t sufficient to hedge us against all future possible states of the universe. In this case, we can’t uniquely determine prices by risk-neutral pricing – sice we can’t hedge out all of the risk, risk preferences of investors will play a part in determining the market-clearing prices of securities. Most models that are used in the industry are incomplete in this way – I’ll be looking at some examples of this soon which will include models involving jump processes and stochastic volatility.