4 Binomial Model
Binomial model in one-period and many periods. Arbitrage. Derivative pricing in the Binomial model
We now want to build a simple model for pricing derivatives. We will begin in discrete time, and imagine a world in which there is a single asset, which can only go up or down by a fixed percentage change, randomly, on which we will want to price a derivative contract. Note that, since we are now considering only one asset, we will drop the superscripts (so we write \(S_n = S_n^1\)). We will often refer to this asset as the risky asset to distinguish it from the bank account.
Suppose we have a fixed time-horizon, \(N \ge 1\), and suppose that at each time, we flip an independent, (possibly) biased coin. If we flip heads, then the asset moves up: \[S_n \mapsto S_{n+1} = uS_n,\] and if we flip tails, the asset will move down: \[S_n \mapsto S_{n+1} = dS_n,\] where \(u>d\).5 E.g. if the asset either goes up or down by 5%, then \(u = 1.05, d = 0.95\).
We can make this probabilistically rigorous by defining \((S_n)_{0 \le n \le N}\) as a sequence of r.v.’s defined on the event space \(\Omega = \{\omega = \omega_1, \omega_2, \dots, \omega_N: \omega_i \in \{H,T\}\}\). An outcome in this space is therefore a sequence \(\omega = HHTHT\dots\), which corresponds to the asset going up twice, then down, then up, then down, ….
We assign to elements of this space the usual (biased) coin probabilities, so if \(\omega\) is an event with \(k\) heads and \(N-k\) tails, then \(\mathbb{P}(\{\omega\}) = p^k (1-p)^{N-k}\). It then follows that flips are independent.
To specify a particular market, we therefore need to specify the initial value of the asset, \(S_0\), the amount the asset moves up and down, \(u,d\), the interest rate \(r\), the time-horizon \(N\) and the up probability \(p\).
Let’s consider the case where \(N=1\), a one-period Binomial model, and suppose we have a risky asset \(S_n\), and a derivative, whose value at time \(1\) depends on the price of the asset, \(S_1\). For example, a (European) call option with strike price \(K\) — that is, the option (but not the obligation6) to buy the asset at time 1, for the price \(K\). Then the call option has value \(C_1 = (S_1-K)_+ := \max\{0,S_1-K\}\) at time 1. We will refer to the final value of an option as its payoff.
Example 27. Suppose we have an asset in a one period Binomial model with \(S_0 = 10\), \(u=\frac{3}{2}, d = \frac{4}{5}, r = \frac{1}{5}\), and a call option with strike price \(K = 11\). Then we construct a (self-financing) portfolio using only the bank-account and the underlying asset which has the same payoff as the call option, and initial value \(\frac{40}{21}\).
Solution:
\(S_1(\text{H})=15\), \(C_1(\text{H})=(15-11)_+=4\)
\(S_1(\text{T})=8\), \(C_1(\text{T})=(8-11)_+=0\)
Suppose we have a portfolio \((\phi_0^0,\phi_0^1)\) at time 0.
Then at time 1, portfolio has value: \[\begin{align} V_1=(1+r)\phi_0^0+\phi_0^1 S_1= \begin{cases} \frac{6}{5}\phi_0^0+15\phi_0^1&,\ \text{if $\omega = H$}\\ \frac{6}{5}\phi_0^0+8\phi_0^1&,\ \text{if $\omega = T$} \end{cases}\end{align}\] So look for \((\phi_0^0, \phi_0^1)\) such that \[\begin{align} \frac{6}{5}\phi_0^0+15\phi_0^1=&4=C(H)\\ \frac{6}{5}\phi_0^0+8\phi_0^1=&0=C(T)\end{align}\] Then \(\phi_0^1=\frac{4}{7}\) and \(\phi_0^0=-\frac 56\cdot 8\phi_0^1=-\frac{80}{21}\). The value of the portfolio at time \(0\) is \(V_0=\phi_0^0+\phi_0^1S_0=-\frac{80}{21}+\frac 47\cdot 10=\frac{40}{21}\). \(\square\)
This really suggests that the ‘fair’ price for the call option should be \(\frac{40}{21}\). Can we make this a bit more rigorous? Consider the following definition:
Definition 28. Suppose \((S_n^1, S_n^2, \dots, S_n^K)_{n = 0,1,\dots, N}\) is a sequence of stochastic processes under some probability measure \(\mathbb{P}\), representing the prices of a set of \(K\) assets. Let \(\boldsymbol{\phi}_n\) be a self-financing portfolio where \(\phi_n^j\) depends only on the assets up to time \(n\), and whose value at time \(n\) is \(V_n\). If \(V_0 = 0\), and \[\begin{align} \tag{4.1} \mathbb{P}(V_N \ge 0) =1 \text{ and } \mathbb{P}(V_N > 0) >0\end{align}\] then \(\boldsymbol{\phi}_n\) is called an arbitrage.
Note that (4.1) says that our final portfolio is always worth something positive (we never lose money), and at least some of the time, we definitely make money.
An arbitrage means that we can make risk-free profit with strictly positive probability! Clearly, any investor would follow an arbitrage strategy, and by scaling everything up (if \(\boldsymbol{\phi}\) is an arbitrage, so too is \(\alpha \boldsymbol{\phi}\), for any \(\alpha > 0\)) we can make very large profits!
In Example27, we can construct an arbitrage if the price of the derivative is not \(\frac{40}{21}\):
if the price of the derivative is more than \(\frac{40}{21}\), we can sell the derivative, buy \(\frac{4}{7}\) units of the asset, and borrow the remaining money from the bank; at time 1, we always end up with a portfolio with a positive value;
if the price of the derivative is less than \(\frac{40}{21}\), we can buy the derivative, (short) sell \(\frac{4}{7}\) units of the asset, and invest the remaining money in the bank; at time 1, we always end up with a portfolio with a positive value.
Our key financial assumption will be that there is no arbitrage. The assumption is based on the fact that, if a situation exists that allows a trader to make money without any risk, they will take this opportunity. Therefore, should such an opportunity exist, everyone would exploit this, making massive profits without risk; since this does not happen, these opportunities must not exist! The principle of arbitrage is sometimes also expressed as: financial markets contain ‘no free lunch.’
Our first result in this direction is the following:
Theorem 29. In a one-period model, consisting only of the asset and the bank account, with \(p \in (0,1)\), there is no arbitrage if and only if \(d < (1+r) < u\).
We leave the proof of this result to an example sheet, but intuitively, the condition \(d < 1+r < u\) says that investing 1 in the risky asset is never guaranteed to be better, or guaranteed to be worse than investing 1 in the bank account.
This result is important since it means that we can use a one-period Binomial model as a (not very realistic) model for our financial market.
Our main interest is in what happens if we now include a derivative on this asset:
Theorem 30. Suppose we have a one-period Binomial model, with \(d < (1+r) < u\), and where we can also purchase a derivative contract, whose payoff at time 1, \(C_1\), depends on the value at time one of the underlying asset, so: \[C_1 = \begin{cases} c_H &, \text{ if } S_1 = uS_0\\ c_T &, \text{ if } S_1 = dS_0\\ \end{cases}.\] Then there is no arbitrage if and only if the price of the derivative at time 0, \(C_0\) is: \[\begin{align} \tag{4.2} \frac{1}{1+r}\left[ c_T \frac{u-(1+r)}{u-d} + c_H \frac{(1+r)-d}{u-d} \right].\end{align}\]
Solution: Consider portfolio \((\phi_0^0,\phi_0^1,\phi_0^2)=(\)cash, units risky asset, units derivative\()\).
Assume \(\phi_0^2\ne 0\) (otherwise we are in the “no derivative” case and Theorem3.3 says there is no arbitrage).
We want to construct an arbitrage, i.e., \(V_0=0\) and \(V_1\ge 0\) with at least one outcome for \(V_1\) being \(>0\) \(\implies\) \(V_0=0\) means \(\phi_0^0=-\phi_0^1S_0-\phi_0^2 C_0\).
Consider: outcomes \(V_1(H)\) and \(V_1(T)\)
case \(H\): \[\begin{align} V_1(H)=&\phi_0^0(1+r)+uS_0\phi_0^1+c_H\phi_0^2\\ =&-(1+r)(\phi_0^1S_0+\phi_0^2 C_0)+uS_0\phi_0^1+c_H\phi_0^2\\ =&\phi_0^1(u-(1+r))S_0+(c_H-(1+r)C_0)\phi_0^2\\ =&\left((u-(1+r))S_0\frac{\phi_0^1}{\phi_0^2}+(c_H-(1+r)C_0)\right)\phi_0^2\end{align}\]
similarly, case \(T\): \[\begin{align} V_1(T)=&\left((d-(1+r))S_0\frac{\phi_0^1}{\phi_0^2}+(c_T-(1+r)C_0)\right)\phi_0^2\end{align}\]
\(\implies\) consider \(x:=\frac{\phi_0^1}{\phi_0^2}\) variable to maximize profit per unit \(\phi_0^2\).
We have two linear functions \[\begin{align} f(x)=&(u-(1+r))S_0x+(c_H-(1+r)C_0)\\ g(x)=&(d-(1+r))S_0x+(c_T-(1+r)C_0)\end{align}\] \(V_1(H)=f(x)\phi_0^2\), \(V_1(T)=g(x)\phi_0^2\), \(f\) is increasing, \(g\) is decreasing since \(d<(1+r)<u\). Then arbitrage exists if and only if
either \(f(x)\) and \(g(x)\) are \(\ge 0\) and at least one of them is \(>0\) for some \(x\) (then take \(\phi_0^1=x\phi_0^2\) and \(\phi_0^2>0\))
or \(f(x)\) and \(g(x)\) are \(\le 0\) and at least one of them is \(<0\) for some \(x\) (then take \(\phi_0^1=x\phi_0^2\) and \(\phi_0^2<0\))
No arbitrage if and only if
either \(f(x)\ne 0\) and \(g(x)\ne 0\) have opposite signs
or \(f(x)=g(x)=0\)
No arbitrage: \(f(x_0)=0\ \iff\ g(x_0)=0\). \(f(x_0)=g(x_0)\) yields \[\begin{align} (u-(1+r))S_0x_0+(c_H-(1+r)C_0)=&(d-(1+r))S_0x_0+(c_T-(1+r)C_0)\\ \iff uS_0x_0+c_H=&dS_0x_0+c_T\\ \iff x_0=&\frac{c_T-c_H}{S_0(u-d)}\end{align}\] Putting \(x_0\) into \(f(x_0)=0\) yields \[\begin{align} 0=&(u-(1+r))S_0\frac{c_T-c_H}{S_0(u-d)}+(c_H-(1+r)C_0)\\ \iff C_0=&\frac{1}{1+r}\left((u-(1+r))\frac{c_T-c_H}{u-d}+c_H\right)\\ =&\frac{1}{1+r}\left(\frac{u-(1+r)}{u-d}c_T+c_H\frac{(1+r)-d}{u-d}\right)\end{align}\] \(\square\)
We call the price (4.2) an arbitrage-free price of the derivative. Note that (here) there is a single arbitrage-free price, so we can talk about the arbitrage-free price.
Example 31. Recall Example27. Then \(c_H = (15-11)_+ = 4\) and \(c_T = (8-11)_+ = 0\), and so the price at time 0 is: \[\begin{align} \frac{1}{1+ \frac{1}{5}} & \left[ 0 + 4 \times \frac{1+ \frac{1}{5}-\frac{4}{5}}{\frac{3}{2} - \frac{4}{5}}\right] \\ & = \frac{40}{21}\end{align}\]
We can use exactly the same idea to work out what happens in a multiple period market. If \(N>1\), we can begin by computing the price of the derivative at time \(N-1\), given e.g. we have seen \(HTHHT\dots\) up to time \(N-1\). This is now a one-period model, and we can compute the price of the derivative. Now suppose we are at time \(N-2\), now we know what the price of the derivative must be at time \(N-1\), and so we can repeat the process to compute an arbitrage-free price at time \(N-2\), etc.
Example 32. Suppose we have a Binomial model with \(N=2\), \(r = \frac{1}{4}\), \(u = \frac{4}{3}, d = \frac{5}{6}\), and \(S_0 = 2025\). Find the price of a European call option with strike \(K = 1800\).
Solution:
price at \(S_1=2700\) \[\begin{align} C_1(H)=&\frac{1}{1+r}\left(c_T\frac{u-(1+r)}{u-d}+c_H\frac{(1+r)-d}{u-d}\right)\\ =&\frac{4}{5}\left(450\frac{\frac{4}{3}-\frac{5}{4}}{\frac{4}{3}-\frac{5}{6}}+1800\frac{\frac 54-\frac 56}{\frac 43-\frac 56}\right)\\ =&1260\end{align}\]
price at \(S_1=\frac{3375}{2}\): \(C_1(T)=\frac{4}{5}\left(0\frac{\frac{4}{3}-\frac{5}{4}}{\frac{4}{3}-\frac{5}{6}}+450\frac{\frac 54-\frac 56}{\frac 43-\frac 56}\right)=300\)
price at \(S_0=2025\): \(C_0=\frac{4}{5}\left(300\frac{\frac{4}{3}-\frac{5}{4}}{\frac{4}{3}-\frac{5}{6}}+1260\frac{\frac 54-\frac 56}{\frac 43-\frac 56}\right)=880\)
\(\square\)
The European Call option in Example32 is path-independent: the payoff of the option only depends on the final value of the asset, and not on the path that the asset takes. An alternative option, which has a path-dependent payoff, could be something like a knock-in call option: this is an option with a call payoff, but which will only pay out if the asset value exceeds some barrier level some time before the maturity date.
Example 33. Suppose we have a Binomial model with \(N=3, r = \frac{1}{6}, u = \frac{3}{2}, d = \frac{5}{6}, S_0 = 1080\), and suppose we want to price a knock in call option with barrier \(B = 2400\) and strike \(K = 1700\), so the payoff of the option is: \[(S_3-K)_+\boldsymbol{1}_{\{S_n > B, \text{ for some } n = 0,1,2,3\}}.\] What is the arbitrage-free price of the derivative?
Solution:
\(C_2(HH)=\frac{6}{7}\left(325\frac{\frac 32-\frac 76}{\frac 32-\frac 56}+1945\frac{\frac 76-\frac 56}{\frac 32-\frac 56}\right)=\frac 67 \cdot 1135\)
\(C_2(XY)=0\) for \(XY\ne HH\)
\(C_1(T)=0\), \(C_1(H)=\frac{6}{7}\left(C_2(HT)\frac{\frac 32-\frac 76}{\frac 32-\frac 56}+C_2(HH)\frac{\frac 76-\frac 56}{\frac 32-\frac 56}\right)=\frac 12\left(\frac 67\right)^2\cdot 1135\)
\(C_0=\frac{6}{7}\left(C_1(T)\frac{\frac 32-\frac 76}{\frac 32-\frac 56}+C_1(H)\frac{\frac 76-\frac 56}{\frac 32-\frac 56}\right)=\frac 14\left(\frac 67\right)^3\cdot 1135\)
\(\square\)
Finally, we recall that the arbitrage-free price was determined by the fact that we could ‘replicate’ the derivative by trading in the underlying asset. We can do a similar thing with multiple time-periods, and we can interpret this as hedging. Imagine a bank who sells a client an option. The bank is then exposed to the risk associated to the final value of the product that they have sold, and may want to ‘hedge’ this risk. They can do this by trading in the underlying to ‘replicate’ the payoff of the derivative that they have sold to the client.
Given some derivative, we will call the portfolio which has the same value at maturity as the derivative the hedging portfolio, since it is how a bank who has sold a derivative would hedge their exposure to the risk associated with the contract. For a financial institution selling many derivatives, it is often as important to know the hedging portfolio as it is to know the price of the option. The number of units of the asset that we need to hold at a given time to hedge a derivative is often called the delta of the derivative.
Example 34. Consider a bank who has sold the option in Example33, and wishes to hedge their exposure. Find the portfolio that they should hold at time \(1\) if \(S_1 = 1620\).
Solution:
Need portfolio \((\phi_1^0,\phi_1^1)\) with \(\phi_1^0(1+r)+\phi_1^1S_2(\omega)=C_2(\omega)\) for \(\omega=HH,HT\) \[\begin{align} \begin{cases} \frac 76\phi_1^0+2430\phi_1^1=\frac 67\cdot 1135\\ \frac 76\phi_1^0+1350\phi_1^1=0 \end{cases}\end{align}\]
Then \(\phi_1^1=\frac{227}{252}\) “Delta” and \(\phi_1^0=-\frac{51075}{49}\)
Note \(V_1(H)=-\frac{51075}{49}+\frac{227}{252}\cdot 1620=\frac 12\left(\frac 67\right)^2\cdot 1135=C_1(H)\) as expected.
\(\square\)
Strictly, by down, we only mean that the asset does not go up as much, so we may have \(d>1\). For reasons we will see later, we will expect \(u>1+r>d\), so by down, we really mean ‘relative to the bank account’.↩︎
So, unlike a forward contract, we are not forced to buy the asset if the price goes down. ‘European’ refers to the fact that we can only make this purchase at the maturity time. This is the most common type of traded option.↩︎