One-Period Binomial Model

An option is a financial contract that gives the buyer the right but not obligation to buy or sell an underlying asset. Because the payoff of an option is a nonlinear function of another asset, banks and dealers have historically had difficulty pricing options at a fair value, where “fair” means that both the seller and buyer have zero expected profit from the option. Without a good model of this fair value, selling an option would be risky: sell too high, and no one buys; sell to low, and the seller risks significant losses. Thus, while options have traded for centuries, they were obscure and illiquid contracts until relatively recently (Cox & Rubinstein, 1985; Shreve, 2005).

What changed was that in the 1970s, economists Fischer Black, Myron Scholes, and Robert Merton published their work on what has come to be known as the Black–Scholes or Black–Scholes–Merton options-pricing model (Black & Scholes, 1973; Merton, 1973). Black–Scholes was the first mathematical model for fairly valuing an option. By providing a theoretical fair price, Black–Scholes legitimized options and increased their liquidity in financial markets (Cox & Rubinstein, 1985).

While Black–Scholes is a useful model, it relies on an understanding of stochastic calculus. Much of its mathematical complexity arises because Black–Scholes assumes (correctly) that time is continuous. However, the model can be simplified by discretizing time. This is the essential idea behind the binomial options-pricing model (Cox et al., 1979; Rendleman, 1979), a discrete-time analog to Black–Scholes.

The goal of this post is to understand the simplest version of this already simplified model: the one-period binomial options-pricing model. Here, we’ll just explore the binomial model over a single discrete time period. In future posts, I’ll explore the multi-period binomial model and then ultimately Black–Scholes. For now, this simple framing will help us explore the basic setup of the binomial options-pricing model, as well as two important ideas for determining a fair price: the no-arbitrage condition and risk-neutral pricing. These ideas will be relevant for Black–Scholes as well.

Finally, this post relies heavily on an excellent introduction to this material, Stochastic Calculus for Finance I (Shreve, 2005). Some of the proofs and examples are inspired by exercises from that text.

Options

Let’s start by first describing options. If you already understand them, you should have no issue jumping to the next section.

A European option is a financial contract that gives the buyer the right but not obligation to buy or sell an underlying asset¹ at a fixed price, called the strike price, on a fixed date, called the exercise date. An American option differs in that it can be exercised any time up to and including the exercise date. Throughout this post, I’ll always use “option” to refer to a European option². A call option gives the buyer the right to buy the underlying asset, while a put option gives the buyer the right to sell (short) the underlying asset. Finally, the parity value of an option is the value of the option when exercised (i.e. parity with the underlying), while the premium over parity is the price of the option minus its parity value (i.e. the market price for optionality).

To see why pricing options is tricky and non-obvious, let’s look at the parity value of an option as a function of the underlying’s price. Consider a call or put option for a stock whose price today is $S_0$. Let $K$ denote the strike price. Now imagine that the exercise date is one discrete-time point into the future, i.e. when the stock is valued at $S_1$. Then the parity value of a call option at time one, denoted $V_1$, is

$V_1 = \max\{0, S_1 - K\}. \tag{1}$

In words, if the value of the stock is greater than the strike price, then we want to exercise our option (pay $K$), since we can immediately sell the stock back on the open market for $S_1$, pocketing $S_1 - K$ less the premium over parity, which we paid when we bought the option. If the stock is worth less than the strike price, then the option is worthless (Figure $1$, left).

Figure 1. Diagrams of option parity value and profit for call (left) and put (right) options. The difference between these two lines is the premium over parity.

In this setting, the parity value of a put option is

$V_1 = \max\{0, K - S_1\}. \tag{2}$

Here, if the strike price is greater than the stock, we want to exercise the put option (sell the stock short) for $K$ and then immediately buy it back on the open market for $S_1$, pocketing $K - S_1$ less the premium over parity. If the stock is worth more than the strike price, then the option is worthless (Figure $1$, right).

A key point is here that options are tricky to price because their value is a nonlinear function of the underlying’s price. A question that both Black–Scholes and the binomial options-pricing model answer is: as a dealer, how can one sell options without substantial risks?

One-period binomial model

Our goal is to find the fair value for an option. We’re going to come at this problem seemingly indirectly. Let’s instead talk about a simple mathematical model for the dynamics of a single stock, the one-period binomial model³.

At time zero, we have a stock whose price we denote with $S_0 \gt 0$. This is non-random. The stock price at time one, $S_1 \gt 0$, is a random variable that is a function of a possibly biased coin toss taking the value heads ($H$) or tails ($T$). We denote the two possible prices as $S_1(H)$ or $S_1(T)$ (Figure $2$).

Figure 2. One-period binomial model. At time zero, a stock has price $S_0 \gt 0$. At time one, it has price $S_1 \gt 0$. The new price $S_1$ depends on the outcome of a possibly-biased coin toss taking values $H$ or $T$. So $S_0$ either increases by up factor $u$ or decreases by down factor $d$ (Equation $3$).

Here’s our first big assumption. Let’s assume the relationship between $S_0$ and $S_1$ is defined by an up factor $u$ and a down factor $d$ such that

$S_1(H) = u S_0, \qquad S_1(T) = d S_0, \qquad 0 \lt d \lt u. \tag{3}$

For example, let $u=2$ and $d=1/2$. This means that the stock price either doubles on heads or halves on tails. Notice that the down factor can be greater than one, meaning that the stock price can still go up in the “down branch”. However, the up factor is strictly greater than the down factor.

No-arbitrage condition

Now that we understand the dynamics of the stock, let’s state an important assumption that will let us eventually price the option: the no-arbitrage condition. Here, we’re defining an arbitrage as an opportunity such that, beginning with zero wealth, we have zero probability of losing money and positive probability of making money. The layman’s version of the no-arbitrage condition is that there is “no free lunch”.

In the binomial options-pricing model, the no-arbitrage condition is that

$0 \lt d \lt 1+r \lt u, \tag{4}$

where $r$ is the risk-free interest rate, which yields $1+r$ dollars at time one. For example, imagine that $r = 0.02$, and we have one million USD at time zero. Then we time one, we would have $20,000$ extra USD in risk-free interest.

Why is Equation $4$ the no-arbitrage condition? Because if either inequality did not hold, then we could perform an arbitrage. Let’s see why. (See A1 for a proof.)

The down factor $d$ is strictly positive because stock prices are strictly positive. We already assumed this when we assumed that $S_0, S_1 \gt 0$. Now if $d \geq 1+r$, then the stock, in the worst case, would do better than what we could earn with risk-free assets like US Treasury bonds. So we would have an arbitrage. Starting with zero wealth, we would borrow money and simply buy the stock. In the worst case, the value of our stock would increase faster than our debt increases. So either way, we can repay our debt while pocketing the difference⁴.

Alternatively, if $u \leq 1+r$, we have a different arbitrage. Starting with zero wealth, we borrow money, sell the stock short, and put the proceeds into the money market. Even in the worst case for us (best case for the stock), the stock will only go up to $u S_0$, which is less than what we would make on interest on our loan.

In summary, we need to assume the no-arbitrage condition. Otherwise, the problem is un-interesting, as market participants will simply arbitrage prices until the condition does hold.

Delta hedging

So far, we have introduced options and the one-period binomal model for the dynamics of stock prices. But what does this have to do with the prices of options? The key insight for the one-period binomial model is that we can replicate the dynamics of an option by holding a particular position in the underlying stock. If we are a dealer, this would allow us to hedge the risk of selling a call (put) option by taking a short (long) position in the underlying stock. Thus, we will see that the fair price of an option is actually just the cost to hedge it with a stock position that has the same dynamics!

To start, let’s just look at an example. Assume the price of a stock today (time zero) is $S_0 = 100$ USD, and let $u = 11/10$ and $d = 9/10$. According to the one-period binomial model, at time one, the stock’s value will either be

$\begin{aligned} S_1(H) &= d S_0 = \frac{11 \cdot 100}{10} = 110, \\ S_1(T) &= u S_0 = \frac{9 \cdot 100}{10} = 90. \end{aligned} \tag{5}$

Next, imagine that the risk-free rate is $r = 1/50$. You should verify that we’ve picked numbers such that the no-arbitrage condition (Equation $4$) is satisfied. Now what is the value of a call option at time one? It takes one of two values:

$\begin{aligned} V_1(H) &= \max(0, S_1(H) - K) \\ &= \max(0, 110-105) \\ &= 5, \\\\ V_1(T) &= \max(0, S_1(T) - K) \\ &= \max(0, 90-105) \\ &= 0. \end{aligned} \tag{6}$

We want to take a particular position in the underlying stock such that our wealth at time one, denoted $W_1$, is equal to $V_1$. If we buy $\Delta_0$ of the stock at price $S_0$, then our wealth at time one is

$W_1 = \Delta_0 S_1 + (1+r) (W_0 - \Delta_0 S_0). \tag{7}$

$\Delta_0 S_1$ is simply the value of our stock position at one, while the remaining terms are our cash position after buying the stock and investing in a risk-free money-market asset like a US bond. Note that $\Delta_0$ need not be positive. If it is negative, then we are short the stock. Furthermore, our wealth need not be greater than $\Delta_0 S_0$. If $W_0 \lt \Delta_0 S_0$, then we are borrowing money to take the position, and $r$ is the interest rate on our debt.

Now here are some magic numbers. You are not supposed to understand how we picked these yet. Imagine that our initial wealth is $W_0 = 2.94$, and that we buy $\Delta_0 = 1/4$ shares of the stock. (Let’s assume that we can buy fractional shares of the stock.) Then our wealth at time one takes one of two values:

$\begin{aligned} W_1(H) &= \Delta_0 S_1(H) + (1+r) (W_0 - \Delta_0 S_0) \\ &= (1/4)(110) + (1+1/50)(-22.06) \\ &= 5, \\\\ W_1(T) &= \Delta_0 S_1(T) + (1+r) (W_0 - \Delta_0 S_0) \\ &= (1/4)(90) + (1+1/50)(-22.06) \\ &= 0. \end{aligned} \tag{8}$

As we can see, $W_1 = V_1$. In words, our wealth given our stock position replicates the value of the stock’s option. The reason this example worked is because I intentionally picked values $W_0$ and $\Delta_0$ to replicate the value of the option.

This example suggests that for a given stock’s dynamics, we may be able to sell an option and then hedge it with a particular stock position. At least in this example, the natural price for the option is thus $W_0$! We can sell a call (put) option for $W_0$ and then use that money to take a short (long) position of $\Delta_0$ of the stock at price $S_0$. Then the money we lose (make) in the option is offset by the money we make (lose) in the stock. Our expected profit from the parity value of the option is zero, and the only money we make is the premium over parity, which for our purposes can be loosely thought of as a transaction fee. So $W_0$ seems like a fair price for the option.

No-arbitrage condition, redux

Before we solve for the exact values of $W_0$ and $\Delta_0$, let’s ask an important question: is the idea above robust to arbitrage? As an exercise, imagine we are an options dealer, and we have an adversary, who wants to arbitrage us if we misprice the option. Will they succeed?

At time zero, this adversary begins with zero wealth and then buys $\delta$ of a stock at price $S_0$ and $\gamma$ of an option at price $V_0$. This leaves the adversary with a cash position of $-S_0 \delta - V_0 \gamma$. This positive (negative) value can be lent (borrowed) in the money market with interest rate $r$. Thus, at time one, the wealth of our adversary, denoted $A_1$, is

$A_1 = \delta S_1 + \gamma V_1 - (1+r) \left( S_0 \delta + V_0 \gamma \right). \tag{9}$

What we want to show is that if we price our option such that we can replicate its price with a stock, then our adversary has no arbitrage. This means that if there is a positive probability that $A_1$ is positive, then there is a positive probability that $A_1$ is negative. Formally,

$\mathbb{P}(A_1 \gt 0) \gt 0 \iff \mathbb{P}(A_1 \lt 0) \gt 0. \tag{10}$

This should make intuitive sense, since we’ve priced the option to perfectly offset the value of a stock. However, let’s prove it.

Imagine that we, the dealer, buy $\Delta_0$ of the stock at price $S_0$, to offset the risk of selling an option to this adversary. By assumption, we buy $\Delta_0$ in precisely the amount such that the value of our stock position matches the value of the option at time one. And we showed in the previous section that $W_0$ is the cost to replicate the stock and thus the fair price for the option ($V_0 = W_0$). So by construction, we have

$V_1 = \Delta_0 S_1 + (1+r)(W_0 - \Delta_0 S_0). \tag{11}$

Now let’s compute our adversary’s wealth, assuming they buy our option whose value at time one is $V_1$ (i.e. plug Equation $11$ into Equation $9$):

$\begin{aligned} A_1 &= \delta S_1 + \gamma V_1 - (1+r) ( S_0 \delta + V_0 \gamma ) \\ &= \delta S_1 + \gamma \left[ \Delta_0 S_1 + (1+r)(W_0 - \Delta_0 S_0)\right] - (1+r) (\delta S_0 + \gamma V_0) \\ &= (\delta + \gamma \Delta_0) S_1 - (\delta + \gamma \Delta_0) (1+r) S_0 - (1+r) V_0 \gamma + \gamma (1+r) W_0 \\ &= (\delta + \gamma \Delta_0) S_1 - (\delta + \gamma \Delta_0) (1+r) S_0 \\ &= \left[ \delta + \gamma \Delta_0 \right] (S_1 - (1+r) S_0). \end{aligned} \tag{12}$

So the adversary’s wealth takes one of two values, depending on the outcome of a coin toss:

$\begin{aligned} A_1(H) &= \left[ \delta + \gamma \Delta_0 \right] (u - (1+r)) S_0, \\ A_1(T) &= \left[ \delta + \gamma \Delta_0 \right] (d - (1+r)) S_0. \end{aligned} \tag{13}$

Now the no-arbitrage condition (Equation $4$) implies that $A_1(H)$ and $A_1(T)$ have opposite signs, since

$\begin{aligned} d \lt 1+r &\implies d - (1+r) \lt 0, \\ 1+r \lt u &\implies u-(1+r) \gt 0. \end{aligned} \tag{14}$

So if $A_1(H)$ is positive (negative), then $A_1(T)$ is negative (positive). So our adversary has no arbitrage if we price our option using $W_0$ and hedge our position with $\Delta_0$ of the stock.

Solving for $W_0$ and $\Delta_0$

Now let’s see how to find $W_0$ and $\Delta_0$.

As we just saw, we want our wealth to mimic the dynamics of an option; and, we want $W_1 = V_1$. Thus, we want the following two equations to hold:

$\begin{aligned} V_1(H) &= \Delta_0 S_1(H) + (1 + r)(W_0 - \Delta_0 S_0), \\ V_1(T) &= \Delta_0 S_1(T) + (1 + r)(W_0 - \Delta_0 S_0), \end{aligned} \tag{15}$

which we can rewrite as

$\begin{aligned} \frac{1}{1+r} V_1(H) &= \Delta_0 \left( \frac{S_1(H)}{1+r} - S_0 \right) + W_0, \\ \frac{1}{1+r} V_1(T) &= \Delta_0 \left( \frac{S_1(T)}{1+r} - S_0 \right) + W_0. \end{aligned} \tag{16}$

We can then multiply the top equation by a number $\tilde{p}$ and the bottom equation by $\tilde{q} = 1 - \tilde{p}$, and then add the two equations together, letting terms cancel. This gives us

$\frac{1}{1+r} \left[ \tilde{p} V_1(H) + \tilde{q} V_1(T) \right] = \Delta_0 \left( \frac{1}{1+r} \left[ \tilde{p} S_1(H) + \tilde{q} S_1(T) \right] - S_0 \right) + W_0. \tag{17}$

Now let’s chose $\tilde{p}$ cleverly such that

$S_0 = \frac{1}{1+r} \left[ \tilde{p} S_1(H) + \tilde{q} S_1(T) \right]. \tag{18}$

Why? This makes the terms in parentheses in Equation $17$ cancel. We can see from Equation $18$ that $\tilde{p}$ and $\tilde{q}$ do not depend on $\Delta_0$, so this should work. For now, don’t worry about the exact values of $\tilde{p}$ and $\tilde{q}$. We’ll discuss them more in a moment. So this gives us

$W_0 = \frac{1}{1+r} \left[ \tilde{p} V_1(H) + \tilde{q} V_1(T) \right], \tag{19}$

which is the initial wealth $W_0$ that we need in order to replicate the value of an option with its underlying stock.

Now let’s solve for $\Delta_0$. We can do this by subtracting the first line of Equation $16$ by the second:

$\Delta_0 = \frac{V_1(H) - V_1(T)}{S_1(H) - S_1(T)}. \tag{20}$

This is the amount of the stock we need to buy in order to replicate the value of an option with its underlying stock. Equation $20$ is sometimes called the delta-hedging formula.

So we have shown that if we buy $\Delta_0$ stock (Equation $20$) using initial wealth $W_0$ (Equation $19$), then the value of our wealth at time one will match the value of the option ($V_1 = W_1$). The fair price of the option at time zero is $W_0$, since this is the cash we need in order to delta hedge.

If we’re an options dealer, we can’t celebrate just yet. Notice that $\Delta_0$ depends explicitly on $S_1$, while $W_0$ depends implicitly on $S_1$ through $\tilde{p}$ and $\tilde{q}$. Since $S_1$ is random, we can only fairly price the option by having a good estimate of $u$ and $d$! This is why an options trader is actually a volatility trader!

Risk-neutral probabilities

Now let’s return to the numbers $\tilde{p}$ and $\tilde{q}$. We can solve for them using Equation $18$:

$\tilde{p} = \frac{1+r-d}{u-d}, \qquad \tilde{q} = \frac{u-1-r}{u-d}. \tag{21}$

See A2 for a derivation. Because of the no-arbitrage condition (Equation $4$), it should be clear that both the numerators and denominators in Equation $15$ are positive, so both $\tilde{p}$ and $\tilde{q}$ are positive. Furthermore, since $\tilde{p} + \tilde{q} = 1$, it’s clear these numbers are in the range $(0, 1)$. Therefore, we can think of $\tilde{p}$ and $\tilde{q}$ as the probabilities of heads and tails respectively. But they aren’t the “real” probabilities in the sense that they are not the true probabilities of the stock increasing or decreasing in value. Instead, they are the probabilities under which an investor is risk neutral. Why? A different way to write Equation $19$ is

$(1+r) W_0 = \tilde{p} V_1(H) + \tilde{q} V_1(T). \tag{22}$

If $\tilde{p}$ and $\tilde{q}$ are probabilities, then the right-hand side is the expected value of the option. This equality, then, states that the expected value of the option is equal to the value of our initial wealth in the money market, earning risk-free interest $r$. If this were true, if $\tilde{p}$ were the actual probability for heads, then no one should invest in the stock market, as one might as well get the same money risk free via lending. So the true probabilities are $p$ and $q$ is such that Equation $22$ becomes

$(1+r) W_0 \lt p V_1(H) + q V_1(T). \tag{23}$

Here, an investor is rewarded in expectation for risking their money in the stock market.

Because of this, we can think of $\tilde{p}$ and $\tilde{q}$ as the risk-neutral probabilities. They are abstract probabilities that depend on the price movements of the underlying stock, and they are useful for understanding the dynamics of the option. The actual probabilities for heads and tails, $p$ and $q$, do not appear in Equation $19$ precisely because we have priced the option at $V_0=W_0$ to ensure that we can delta hedge our exposure with the underlying valued at $S_0$ and position size $\Delta_0$. In other words, we are agnostic to whether the price goes up or down, and thus the probabilities are irrelevant to us. This idea is essential to pricing options and is called risk-neutral pricing.

Conclusion

In this post, we have explored one discrete time period of the binomial options-pricing model. The goal of going through this simple model was to clarify some key ideas, such as the no-arbitrage condition and risk-neutral pricing. The one-period binomial model also makes some of the same assumptions as the more advanced Black–Scholes model, such as the existence of a risk-free rate, the ability to buy fractional shares, and the behavior of the underlying asset as a random walk. In future posts, I’ll explore the multi-period binomial model and eventually Black–Scholes.

Appendix

A1. Proving the no-arbitrage condition

Assume that Equation $4$ holds. Then we want prove that there is no arbitrage. Consider the scenario in which at time zero, we buy $\Delta_0$ of a stock at price $S_0$. ($\Delta_0$ may be negative, in which case we are taking a short position.) Then at time one, our wealth $W_1$ is

$W_1 = \Delta_0 S_1 + (1 + r)(W_0 - \Delta_0 S_0). \tag{A1.1}$

This is just the value of our position, $\Delta_0 S_1$, plus our initial wealth minus the cost to buy the stock, $W_0 - \Delta_0 S_0$, propagated into the future with the risk-free interest rate.

We want to show that if $W_1$ is positive with positive probability, then $W_1$ is negative with positive probability as well. $W_1$ is a random variable that is a function of a coin toss with outcomes $H$ and $T$. So formally, we want to show that

$\mathbb{P}(W_1 \gt 0) \gt 0 \iff \mathbb{P}(W_1 \lt 0) \gt 0. \tag{A1.2}$

We can write $W_1(H)$ as

$\begin{aligned} W_1(H) &= \Delta_0 S_1(H) + (1+r)(W_0 - \Delta_0 S_0) \\ &= \Delta_0 u S_0 + (1+r)(- \Delta_0 S_0) \\ &= \Delta_0 u S_0 - \Delta_0 S_0 - \Delta_0 S_0 r \\ &= (u - 1 - r) \Delta_0 S_0. \end{aligned} \tag{A1.3}$

$W_0$ disappears because we assume an arbitrageur begins with zero wealth. Our assumption implies that $(u - 1 - r)$ above is positive, since

$\begin{aligned} 1+r &\lt u \\ 0 &\lt u - (1+r) \\ 0 &\lt (u - 1 - r). \end{aligned} \tag{A1.4}$

Furthermore, $S_0$ is the stock price and is assumed to be positive. So we can say that $W_1(H)$ has the same sign as $\Delta_0$ and is zero if $\Delta_0$ is zero.

And of course, we can write $S_1(T)$ in the same form as above, where $u$ is replaced with $d$:

$W_1(T) = (d - 1 - r) \Delta_0 S_0. \tag{A1.5}$

By assumption, we can see that $d - (1+r) \lt 0$. Again, $S_0$ is positive. So we can say that $W_1(T)$ has the opposite sign of $\Delta_0$ and is zero if $\Delta_0$ is zero. So we have shown that no arbitrage exists.

A2. Solving for the risk-neutral probabilities

Let’s solve for $\tilde{p}$ using Equation $18$:

$\begin{aligned} S_0 &= \frac{1}{1+r} \left[ \tilde{p} S_1(H) + (1 - \tilde{p}) S_1(T) \right] \\ -S_1(T) + (1+r) S_0 &= \tilde{p} S_1(H) - \tilde{p} S_1(T) \\ &\Downarrow \\ \tilde{p} &= \frac{(1+r) S_0 - S_1(T)}{S_1(H) - S_1(T)} \\ &= \frac{(1+r) S_0 - d S_0}{u S_0 - d S_0} \\ &= \frac{1 + r - d}{u - d}. \end{aligned} \tag{A2.1}$

To solve for $\tilde{q}$, just compute $1 - \tilde{p}$.