8. Binomial Pricing

Put-call parity says that a European call is the same as a European put and a long forward contract on the stock. In the absence of arbitrage opportunities, this relationship is exact and quite robust to market frictions. It would be nice, however, to derive the price of the European call option independently from the price of the European put.

One of the most important results in option pricing is the fact that, under certain conditions, we can replicate the price evolution of the call option from some more basic assets, namely the stock itself and a risk-free bond. In order to do this, though, we need a stochastic model of the evolution of the stock price over time.

8.1. Binomial Trees

One of the easiest ways to describe the evolution over time of a stock price is to use what in finance we call a binomial tree. At each point in time, there are only two possibilities for the future stock price occurring with probability \(p\) and \(1 - p\), respectively. Specifically, if the current stock price is \(S\), then over the next period the stock price can take the values \(S_{u} > S\) and \(S_{d} < S.\)

In the following we will usually compute \(S_{u} = S \times u\) and \(S_{d} = S \times d,\) where \(u\) and \(d\) are the gross percentage increase and decrease of the stock price over the next period, respectively. We will later relate these quantities to the volatility of stock returns in order to make the tree consistent with observed data.

Example 8.1 (One-Period Binomial Tree)

The current stock price is $100. Next period, the asset can go up or down by 10% with probability \(p\) and \(1 - p\), respectively. In this example \(u = 1.10\) and \(d = 0.90.\)

The power of binomial trees becomes apparent when we combine one-period trees to build larger trees, as the next example suggests.

Example 8.2 (Two-Period Binomial Tree)

The current stock price is $100. Each period, the asset can go up or down by 5% with probability \(p\) and \(1 - p\), respectively.

Sometimes we will avoid explicitly writing down the probabilities to alleviate the notation.

Example 8.3 (Four-Period Binomial Tree)

The current stock price is $100. Each period the asset can go up or down by 2.5%.

An interesting fact in the previous examples is that going up and then down yields the same as going down and then up. When this happens we say that the tree recombines. These types of binomial trees are very useful in modeling the stochastic behavior of financial assets because the number of nodes increases linearly with the number of periods, i.e after \(n\) periods there are \(n+1\) possible nodes. If the tree does not recombine then the number of nodes increases exponentially, i.e. after \(n\) periods there are \(2^{n+1}\) possible nodes.

A sufficient condition for the tree to recombine is that the percentage increase and decrease of the stock price at each node are kept constant throughout the tree. As the following example shows, in a recombinant tree \(u\) need not be equal to \(d\).

Example 8.4

Is it the same for an asset to go up by 80% and then down by 30%, compared to first go down by 30% and then go up by 80%? Consider an asset whose current price is $100.

The tree recombines because of the associative property of multiplication:

\[\begin{equation*} S_{ud} = 100 \times 1.80 \times 0.70 = 100 \times 0.70 \times 1.80 = S_{du} \end{equation*}\]

8.2. The Replicating Portfolio Approach

The idea of pricing derivatives such as call and put options by replication is quite straightforward. If we were able to manufacture an option written on a stock by trading in the stock and risk-free bonds, the price of the option would have to be the price of the replicating portfolio.

If the option premium was more expensive than the price of its replicating portfolio, then it would make sense to buy the replicating portfolio, which provides us with a synthetic option, and sell the option for a higher price. If on the contrary the price of the option was lower than its replicating portfolio, if would make sense to buy the option and sell the synthetic option obtained through the replicating portfolio.

In financial markets, these types of trading strategies constitute an arbitrage opportunity, that is, an easy way to make free money. There are both theoretical and empirical arguments as to why arbitrage opportunities cannot exist for long.

Theoretically, a strategy that requires zero investment and generates positive payoffs in the future would incentivize anyone who finds it to exploit it without bounds. This behavior would clearly violate any economic equilibrium model of security prices. The demand for the option or the replicating portfolio, whichever is cheapest, would make the price to quickly adjust until the arbitrage opportunity disappears.

Empirically, we do observe that when markets are functioning well arbitrage opportunities do not last for long. It is important to note that in periods of market stress, or when trading constraints are significant, arbitrage opportunities can last longer. Nevertheless, most of the time we observe that derivative prices prevent obvious arbitrage opportunities.

8.2.1. Replicating a Put Option

In order to show how the method works, we will use the replicating portfolio approach to price a European put option. Consider a non-dividend paying stock that currently trades for $100. Over the next 6-months the stock can go up or down by 10%.

The interest rate is 6% per year with continuous compounding. What should be the price of a European put option with maturity 6 months and strike price $100?

In order to replicate the put, we will need another financial instrument that pays differently from the stock. Consider then a risk-free bond with maturity 6 months and face value equal to the strike price of the put, i.e., $100. The price of the bond is:

\[\begin{equation*} B = 100 e^{-0.06 \times 6/12} = \$97.04. \end{equation*}\]

Note that the bond pays $100 at maturity no matter what happens to the stock price in six months. Therefore, the corresponding binomial tree for the bond is:

The put pays nothing if the stock price is $110 and pays $10 if the stock price is $90. The idea now is to use the the stock and the risk-free bond to replicate the payoffs of the put. The figure below depicts the payoffs of the bond and the put for the two different values of the stock.

Say we purchase \(N_{S}\) units of the stock and \(N_{B}\) units of the bond. Such a portfolio would pay

\[\begin{equation*} \text{Payoff} = \begin{cases} 110 N_S + 100 N_B & \text{if $S = 110$} \\ 90N_S + 100 N_B & \text{if $S = 90$} \end{cases} \end{equation*}\]

Furthermore, say we choose \(N_S\) and \(N_B\) such that payoff of the portfolio matches the payoff of the put, i.e.

\[\begin{align*} 110 N_S + 100 N_B & = 0 \\ 90N_S + 100 N_B & = 10 \end{align*}\]

We can solve for \(N_S\) and \(N_B\) to find:

\[\begin{align*} N_S & = \frac{0 - 10}{110 - 90} = -0.50 \\ N_B & = - \frac{110}{100} N_S = (-1.1)(-0.5) = 0.55 \end{align*}\]

Therefore, by shorting \(0.50\) units of the stock and going long \(0.55\) units of the bond we can exactly match the payoffs of the put. It is common to call the number of shares needed to replicate the option the delta (\(\Delta\)) of the put.

The price of the put must then match the price of the replicating portfolio, otherwise there would be an arbitrage opportunity:

\[\begin{equation*} P = -0.5 \times 100 + 0.55 \times 97.04 = \$3.38. \end{equation*}\]

Example 8.5

Let us analyze what would happen in the previous analysis if the put was trading for $3. Then it would make sense buy the put and sell a synthetic put, that is, purchasing 0.5 shares of the stock and selling 0.55 units of the risk-free bond.

By selling the synthetic put, we collect a cash flow of $3.38, which is more than enough to buy the put for $3, leaving a free profit of $0.38 per transaction. The fact that we are long a put and short a synthetic put guarantees that we are hedged in six months.

The figure below shows the cash flow of these transactions. We can see that this would be an obvious arbitrage opportunity since we collect $0.38 per transaction with no risk afterwards.

The replication analysis shows that the put can be seen as an investment in the risk-free bond that is financed in part by shorting stocks. We could see the put as the equity position of a limited liability entity that has a risk-free investment worth $53.37 that is financed with a short stock position worth $50 and $3.38 of equity.

Since the risk-free position does not depend on the value of the stock, a change in the stock price causes an opposite change in the value the put. For example, an increase of 1% in the stock reduces the put value approximately by \(0.5/3.38 = 14.8\%.\) Therefore, the put carries more risk than the underlying stock and correlates inversely with it.

8.2.2. Replicating A Call Option

Consider now a European call option with the same maturity and strike price as the put. The call pays $10 is the stock price is $110 and $0 otherwise.

As before, we replicate the payoffs of the call option by trading the stock and the bond:

\[\begin{align*} 110 N_{S} + 100 N_{B} & = 10 \\ 90N_{S} + 100 N_{B} & = 0 \end{align*}\]

We can solve for \(N_{S}\) and \(N_{B}\) to find:

\[\begin{align*} N_{S} & = \frac{10 - 0}{110 - 90} = 0.50 \\ N_{B} & = - \frac{90}{100} N_{S} = (-0.90)(0.5) = -0.45 \end{align*}\]

Therefore, by buying 0.50 units of the stock and shorting 0.45 units of the bond we can exactly match the payoffs of the call. The price of the call must then match the price of the portfolio, otherwise there would be an arbitrage opportunity:

\[\begin{equation*} C = 0.5 \times 100 - 0.45 \times 97.04 = \$6.33. \end{equation*}\]

The replication analysis reveals that the call can be seen as a levered position on the stock. We could see the call as the equity position of a limited liability entity that has a stock position worth $50 that is financed with $43.67 of debt and $6.33 of equity.

The risk of the call is amplified by the implicit leverage in the position. A 1% decrease in the stock price reduces the value of the call approximately by \(0.5/6.33 = 7.9\%.\) Thus, the risk of a call option is many times larger than the risk of the underlying asset.

8.2.3. Replicating A Generic Derivative

The analysis so far suggests that we can generalize the replicating approach to price any derivative. By a derivative we mean a security that pays a certain amount \(X_{u}\) is the stock goes up and \(X_{d}\) is the stock price goes down at time \(T.\)

We replicate the derivative by trading the stock and a zero-coupon risk-free bond with face-value \(F\) and expiring at \(T\) as well. If the risk-free rate with continuous compounding is \(r\), the value of the bond today is \(B = F e^{-r T}.\)

As before, we start by matching the payoffs of the derivative using \(N_{S}\) shares of the stock and \(N_{B}\) units of the bond:

\[\begin{align*} S_{u} N_{S} + F N_{B} & = X_{u} \\ S_{d} N_{S} + F N_{B} & = X_{d} \end{align*}\]

Solving for \(N_{S}\) and \(N_{B}\) shows

\[\begin{align*} N_{S} & = \frac{X_{u} - X_{d}}{S_{u}- S_{d}}, \\ N_{B} & = \frac{X_{u} - S_{u} N_S}{F} = \frac{X_{d} - S_{d} N_S}{F}. \end{align*}\]

The price of the derivative must then match the price of the portfolio, otherwise there would be an arbitrage opportunity:

\[\begin{equation*} X = N_{S} S + N_{B} B. \end{equation*}\]

The number of shares \(N_S\) needed to replicate the derivative is called the delta of the instrument. Note that the specific face-value of the risk-free bond is not really relevant since if we choose a different value the number of bonds bought or sold would get adjusted accordingly.

Example 8.6

A non-dividend paying stock currently trades at $50 and can either increase to $60 or decrease to $40 over the next 6 months. An investment bank is offering to its clients a product that pays in 6 months $1000 if the stock goes up and $200 if the stock goes down. The risk-free rate is 5% per year with continuous compounding. What is the no-arbitrage price of the derivative? How many shares of the stock does the bank needs to buy or sell in order to hedge the derivative?

As we said before, we can choose any face value for the risk-free bond. If we use $100 then the value of the bond is \(B = 100 e^{-0.05 \times 6/12}.\) The number of shares needed to hedge the derivative is

\[\begin{equation*} N_{S} = \frac{1000 - 200}{60 - 40} = 40, \end{equation*}\]

whereas the number of bonds is

\[\begin{equation*} N_{B} = \frac{1000 - 60 \times 40}{100} = -14. \end{equation*}\]

Finally, the no-arbitrage price of the derivative is

\[\begin{equation*} 40 \times 50 + (-14) \times 100 e^{-0.05 \times 6/12} = 634.57. \end{equation*}\]

Therefore, the bank needs to buy 40 shares of the stock to hedge the derivative and sell it for $634.57.

8.3. The Risk-Neutral Approach

In replicating the payoffs of the option, we never used the actual probabilities. As a matter of fact, these probabilities might even change based on whose thinking about the asset. Since the previous reasoning is silent about the probabilities and the type of investor pricing the asset, we can assume in our reasoning that all investors are risk neutral. Even if this is not true in real markets, such assumption would not affect \textbf{the logic} of the replicating-portfolio argument.

The real probabilities are thus irrelevant. In a world populated by risk-neutral investors, all expected payoffs should be discounted at the risk-free rate, regardless of their riskiness. Therefore, the price of any asset \(X\) is equal to its expected payoffs discounted at the risk-free rate:

\[\begin{align*} X = (X_{u} p + X_{d} (1 - p)) e^{-r T}. \end{align*}\]

In practice, we can compute \(p\) by using the price of the stock as:

\[\begin{align*} S = (S_{u} p + S_{d} (1 - p)) e^{-r T} \end{align*}\]

which implies that

(8.1)\[\begin{align} p = \frac{S e^{r T} - S_{d}}{S_{u} - S_{d}}. \label{eq:rn_prob} \end{align}\]

We can then use the risk-neutral probabilities to compute the price of a call or put option.

An alternative way to think about the risk-neutral probabilities is the following. In a risk-neutral world, investors are indifferent between receiving an expected cash flow \(p S_{u} + (1 - p) S_{d}\) or selling the stock forward for a fixed forward price \(F = S e^{r T}.\) This means that in a risk-neutral world the forward price is the best forecast of the price of the stock, that is:

(8.2)\[\begin{align} \underbrace{S e^{r T}}_{\text{Forward Price}} = \underbrace{p S_{u} + (1 - p) S_{d}}_{\text{Forecast}}. \label{eq:rn_forward_price} \end{align}\]

Equation \(\eqref{eq:rn_forward_price}\) implies that we can also write \(\eqref{eq:rn_prob}\) as

\[\begin{align*} p = \frac{F - S_{d}}{S_{u} - S_{d}} \end{align*}\]

where \(F\) denotes the forward price of the stock.

8.3.1. Pricing the Call and Put Again

Continuing with our example, the price of the stock should be equal to the expected payoff discounted at the risk-free rate:

\[\begin{equation*} 100 = (110 p + 90 (1 - p)) e^{-0.06 \times 6/12}. \end{equation*}\]

We can reverse-engineer the probability of the stock going up that makes consistent valuations in this world:

\[\begin{equation*} p = \frac{100 e^{0.06 \times 6/12} - 90}{110 - 90} = 0.6522. \end{equation*}\]

The price of the call is also equal to the expected payoff under this risk-neutral probability, discounted at the risk-free rate:

\[\begin{equation*} C = (10 p + 0 (1 - p)) e^{-0.06 \times 6/12} = \$6.33. \end{equation*}\]

Similarly, for the put we have that:

\[\begin{equation*} P = (0 p + 10 (1 - p)) e^{-0.06 \times 6/12} = \$3.38. \end{equation*}\]

Of course, the prices are the same as before since both approaches are consistent with each other.

Example 8.7

A non-dividend paying stock trades at $50 and over the next 6-months can go up to $60 or down $40. The risk-free rate is 6% per year with continuous compounding. Let’s compute the price of a European call option expiring in 6 months with strike price $48.

We start by computing the risk-neutral probability of the stock moving up:

\[\begin{equation*} p = \frac{50 e^{0.06 \times 6/12} - 40}{60 - 40} \end{equation*}\]

Therefore, the price of the call is:

\[\begin{equation*} C = (12 p + 0 (1 - p)) e^{-0.06 \times 6/12} = 6.71. \end{equation*}\]

Example 8.8

A non-dividend paying stock trades at $120 and over the next 3-months can increase or decrease by 10%. The risk-free rate is 5% per year with continuous compounding. What is the price of an asset that pays in 3 months $100 if the stock increases in price and $200 otherwise?

We have that the stock can move up to \(120 \times 1.10 = 132\) or down to \(120 \times 0.90 = 108.\)

Therefore, the risk-neutral probability of the stock moving up is:

\[\begin{equation*} p = \frac{120 e^{0.05 \times 3/12} - 108}{132 - 108} \end{equation*}\]

The price of the asset is:

\[\begin{equation*} X = (100 p + 200 (1 - p)) e^{-0.05 \times 3/12} = 141.93. \end{equation*}\]

8.3.2. State Prices

The risk-neutral probabilities are intimately related to the so-called Arrow-Debreu securities depicted below.

The price of each security is then the expected payoff using the risk-neutral probabilities, discounted at the risk-free rate:

\[\begin{align*} \pi^{u} & = (1 p + 0 (1 - p)) e^{-0.06 \times 6/12} = \$0.6329 \\ \pi^{d} & = (0 p + 1 (1 - p)) e^{-0.06 \times 6/12} = \$0.3375 \end{align*}\]

Example 8.9

Continuing with Example 8.8, we have that the price of an asset that pays in 3 months $1 if the stock price increases and $0 otherwise is:

\[\begin{equation*} \pi_{u} = (1 p + 0 (1 - p)) e^{-0.05 \times 3/12} = 0.5559, \end{equation*}\]

whereas the price of an asset that pays in 3 months $0 if the stock price increases and $1 otherwise is:

\[\begin{equation*} \pi_{d} = (0 p + 1 (1 - p)) e^{-0.05 \times 3/12} = 0.4317. \end{equation*}\]

Thus, the price of an asset that pays in 3 months $100 if the stock increases in price and $200 otherwise is:

\[\begin{equation*} X = 100 \pi_{u} + 200 \pi_{d} = 141.93, \end{equation*}\]

which is the same value as before.

8.4. Multiple Periods

8.4.1. The Two Period Binomial Model

We now extend the economy to two periods. We would like to price a European call option with strike price \(K,\) expiring at time \(T\) and written on a non-dividend paying stock. The figure below shows the different values that we need to compute for both the stock and the call option.

We will now build a simple two-period binomial tree. We will assume that the length of each period is the same and given by \(\Delta T = T/2.\) Note that this is not required, but makes our computations consistent with keeping the volatility constant throughout the tree.

Each period, the stock price goes up or down by by \(u\) or \(d,\) respectively. Therefore, at the end of the first period we have that

\[\begin{align*} S_{u} & = S \times u, \\ S_{d} & = S \times d. \\ \end{align*}\]

At the end of the second period the possible values are given by

\[\begin{align*} S_{uu} & = S_{u} \times u, \\ S_{ud} & = S_{u} \times d, \\ S_{dd} & = S_{d} \times d. \\ \end{align*}\]

Note that because the tree recombines, we must have that \(S_{ud} = S_{u} \times d = S_{d} \times u = S_{du}.\) The risk-neutral probability of an up-move in the first period is

\[\begin{align*} p & = \frac{S e^{r \Delta t} - S_{d}}{S_{u} - S_{d}} \\ & = \frac{S e^{r \Delta t} - S d}{S u - S d} \\ & = \frac{e^{r \Delta t} - d}{u - d}. \\ \end{align*}\]

As you can see, in our tree the risk-neutral probability of the stock going up is the same in all nodes of the tree.

The idea now is to work backwards from the end of the tree for the call option, and work our way up to the beginning. The price of the call at expiration is just the intrinsic value of the option:

\[\begin{align*} C_{uu} & = \max(S_{uu} - K, 0), \\ C_{ud} & = \max(S_{ud} - K, 0), \\ C_{dd} & = \max(S_{dd} - K, 0). \end{align*}\]

If the stock price goes up during the first period, the call price at that point must be equal to:

\[\begin{equation*} C_{u} = \left( p C_{uu} + (1 - p) C_{ud} \right) e^{-r \Delta t}. \end{equation*}\]

Similarly, if the stock price goes down during the first period we have that:

\[\begin{equation*} C_{d} = \left( p C_{du} + (1 - p) C_{dd} \right) e^{-r \Delta t}. \end{equation*}\]

Finally, the value of the option at the beginning of the tree must be equal to the expected price under the risk-neutral probabilities, discounted at the risk-free rate. Indeed, an investor that buys the call today could sell it a period later at the price observed at that point. Thus,

\[\begin{equation*} C = \left( p C_{u} + (1 - p) C_{d} \right) e^{-r \Delta t}. \end{equation*}\]

Example 8.10 (Pricing a European Call Option)

Let’s price a European call option written on a non-dividend paying stock using a two-step binomial model. The current stock price is $100, and it can go up or down by 5% each period for two periods. Each period represents 3-months, i.e. \(\Delta t = 3/12 = 0.25\). The risk-free rate is 6% per year (continuously compounded). we will price a European call option with maturity 6 months and strike $100.

The risk-neutral probability of an up-move is then:

\[\begin{equation*} p = \frac{e^{r \Delta t} - d}{u - d} = \frac{e^{0.06 \times 0.25} - 0.95}{1.05 - 0.95} = 0.6511. \end{equation*}\]

The risk-neutral probability of a down-move is just \(1 - p = 0.3489\). We can then compute:

\[\begin{align*} C_{u} & = \left( 10.25 \times p + 0 \times (1 - p) \right) e^{-0.06 \times 0.25} = \$6.57 \\ C_{d} & = \left( 0 \times p + 0 \times (1 - p) \right) e^{-0.06 \times 0.25} = \$0 \\ C & = \left( 6.57 \times p + 0 \times (1 - p) \right) e^{-0.06 \times 0.25} = \$4.22 \end{align*}\]

Example 8.11 (Pricing a European Put Option)

We can use the risk-neutral probabilities to price a European put with the same characteristics.

\[\begin{align*} P_{u} & = \left( 0 \times p + 0.25 \times (1 - p) \right) e^{-0.06 \times 0.25} = \$0.08 \\ P_{d} & = \left( 0.25 \times p + 9.75 \times (1 - p) \right) e^{-0.06 \times 0.25} = \$3.51 \\ P & = \left( 0.08 \times p + 3.51 \times (1 - p) \right) e^{-0.06 \times 0.25} = \$1.26 \end{align*}\]

Note

The value of the put in Example 8.11 can also be obtained using put-call parity. Indeed, the put is equal to a call plus a short forward:

\[\begin{align*} P & = C - S + K e^{-r T} \\ & = 4.22 + 100 - 100 e^{-0.06 \times 0.5} \\ & = \$1.26. \\ \end{align*}\]

Note that we discount the strike price using the maturity of the option which is 6 months.

8.4.2. Making the Tree Consistent with Observed Volatility

A great advantage of the binomial model is that is possible to relate the up and down movements to the risk-neutral volatility observed in the market. Over a small interval \(\Delta t\), the choice \(u = e^{\sigma \sqrt{\Delta t}}\) and \(d = 1 / u\) produces a binomial model consistent with the Black-Scholes model of a Geometric Brownian Motion (GBM).

Note that in this case \(u \times d = 1\), so the tree on average does not have a drift. The risk-neutral drift, however, is incorporated into the risk-neutral probabilities.

Proof. Let’s see how this modification works in a 1-period tree.

The binomial process for the stock implies that the gross return next period is either \(u\) or \(d\) with risk-neutral probability \(p\) or \(1 - p,\) respectively. Remember that the risk-neutral probability of the stock price going up is given by:

\[\begin{align*} p = \frac{e^{r \Delta t} - d}{u - d}. \end{align*}\]

The variance of the returns over the period can then be computed as:

\[\begin{align*} V & = p (u - (p u + (1 - p) d))^{2} + (1 - p) (d - (p u + (1 - p) d))^{2} \\ & = p (1 - p)^{2} (u - d)^{2} + (1 - p) p^{2} (u - d)^{2} \\ & = p (1 - p) (u - d)^{2} \\ & = \left( \frac{e^{r \Delta t} - d}{u - d} \right) \left( 1 - \frac{e^{r \Delta t} - d}{u - d} \right) (u - d)^{2} \\ & = (e^{r \Delta t} - d) (u - e^{r \Delta t}) \\ & = (e^{r \Delta t} - e^{-\sigma \sqrt{\Delta t}}) (e^{\sigma \sqrt{\Delta t}} - e^{r \Delta t}). \\ \end{align*}\]

Therefore,

\[\begin{align*} \lim_{\Delta t \rightarrow 0} \frac{V}{\Delta t} & = \lim_{\Delta t \rightarrow 0} \left( \frac{e^{r \Delta t} - e^{-\sigma \sqrt{\Delta t}}}{\sqrt{\Delta t}} \right) \left( \frac{e^{\sigma \sqrt{\Delta t}} - e^{r \Delta t}}{\sqrt{\Delta t}} \right) \\ & = \lim_{\Delta t \rightarrow 0} \frac{e^{r \Delta t} - e^{-\sigma \sqrt{\Delta t}}}{\sqrt{\Delta t}} \lim_{\Delta t \rightarrow 0} \frac{e^{\sigma \sqrt{\Delta t}} - e^{r \Delta t}}{\sqrt{\Delta t}} \\ & = \sigma ^{2}. \end{align*}\]

In other words, for small \(\Delta t\) we have that the variance of stock returns is \(V \approx \sigma^{2} \Delta t.\) Therefore, at each node the percentage change in the stock price is on average at \(r \Delta t\) with variance \(\sigma^{2} \Delta t.\)

Example 8.12 (Pricing a European Call Option Using Five Periods)

In this example we will price a 6-month European call option with strike price $135 written on a non-dividend paying stock that currently trades at $132 and whose volatility of stock returns is 35% per year. We will use an interest rate of 3% per year with continuous compounding.

We will use a 5-period binomial tree. Therefore, we have that \(T = 6/12 = 0.5\) and \(\Delta t = T / 5 = 0.1,\) so that \(u = e^{0.35 \sqrt{0.1}} = 1.1170\) and \(d = 1/1.1170 = 0.8952.\) The table below describes the binomial tree for the stock. Note that for convenience I wrote the tree as a lower triangle. The horizontal line describes the path where the stock price increases every period whereas the lower diagonal describes the path where the stock price goes down every period.

Time	0	0.1	0.2	0.3	0.4	0.5
Stock	132	147.45	164.71	183.98	205.52	229.57
		118.17	132.00	147.45	164.71	183.98
			105.79	118.17	132.00	147.45
				94.70	105.79	118.17
					84.78	94.70
						75.90

The risk-neutral probability of an up-move is given by:

\[\begin{align*} p = \frac{e^{0.03 \times 0.1} - 0.8952}{1.1170 - 0.8952} = 0.4859. \end{align*}\]

We can now price the call as follows.

Time	0	0.1	0.2	0.3	0.4	0.5
Call	13.16	21.23	33.18	49.79	70.92	94.57
		5.60	10.06	17.68	30.11	48.98
			1.42	2.92	6.03	12.45
				0.00	0.00	0.00
					0.00	0.00
						0.00

In the table above, the last columns contains the payoffs of the call at maturity. For example, if the stock price at maturity is 229.57, then the payoff of the call is \(229.57 - 135 = 94.57\) which is the first number on the last column. On the other hand, if the stock price at maturity 94.70, then the call is out-of-the-money and the payoff is 0.

We work backwards on the tree by computing expected payoffs using the risk-neutral probabilities and discounting at the risk-free rate for 0.1 years. For example, the first number in the before last column is computed as \(70.92 = (94.57 p + 48.98 (1 - p)) e^{- 0.03 \times 0.1}\) where \(p\) is the risk-neutral probability computed above. We keep doing this for all nodes until we reach the beginning of the tree. The price of the call is then $13.16.

A spreadsheet that shows the computations can be found here.

8.5. Appendix

In the case of European options it is possible to price a European call or put without having to work the intermediate steps. In the case of American options, however, it is necessary to check whether it is worth to exercise the option at each node of the tree. We will learn later how to handle the American option case.

Take for example a two period binomial tree for example. In this case we could use the final payoffs of the call directly to compute:

\[\begin{equation*} C = \left( \Pi_{uu} C_{uu} + \Pi_{ud} C_{ud} + \Pi_{dd} C_{dd} \right) e^{-r T}, \end{equation*}\]

where \(\Pi_{uu} = p^{2}\), \(\Pi_{ud} = 2 p (1 - p)\), and \(\Pi_{dd} = (1 - p)^{2}\). The \(\Pi\)’s represent the total risk-neutral probabilities of each final node.

We can generalize the method for an arbitrary number of periods \(n,\) which means that \(\Delta t = T / n.\) Since at each node the stock \(S\) can go up or down to \(S_{u} = S \times u\) and \(S_{d} = S \times d\), we have that the stock price at time \(T\) can take on \(n + 1\) different values:

\[\begin{equation*} S_{T} = \left( \begin{array}{c} S_{T}(0) \\ S_{T}(1) \\ S_{T}(2) \\ \vdots \\ S_{T}(n - 2) \\ S_{T}(n - 1) \\ S_{T}(n) \end{array} \right) = \left( \begin{array}{c} S u ^ {n} \\ S u ^ {n - 1} d \\ S u ^ {n - 2} d ^ {2} \\ \vdots \\ S u ^ {2} d ^ {n - 2} \\ S u d ^ {n - 1} \\ S d ^ {n} \end{array} \right) \end{equation*}\]

The risk-neutral probability of an up-move is given by:

\[\begin{equation*} p = \frac{e^{r \Delta t} - d}{u - d}. \end{equation*}\]

The total risk-neutral probability at node \(i = 0, \ldots, n\) is:

\[\begin{equation*} \Pi(i) = \binom{n}{i} p^{i} (1 - p)^{n - i}, \end{equation*}\]

where \(\binom{n}{i} = \frac{n!}{(n - i)! i!}\) is a binomial coefficient. The price of European call and put options are given by:

\[\begin{align*} C & = e^{-r T} \sum_{i = 0}^{n}{\Pi(i) \max(S_{T}(i) - K, 0)} \\ P & = e^{-r T} \sum_{i = 0}^{n}{\Pi(i) \max(K - S_{T}(i), 0)} \end{align*}\]

8.6. Practice Problems

Exercise 8.1

A stock price is currently $50. It is known that at the end of 2 months it will be either $53 or $48. The risk-free interest rate is 10% per annum with continuous compounding. What is the value of a 2-month European call option with a strike price of $49? Use the replicating portfolio argument and indicate the number of shares required to hedge the position.

Solution to Exercise 8.1

The binomial trees for the stock, the bond and the call are as follows:

Let \(N_{S}\) and \(N_{B}\) denote the number of stocks and bonds with face value $49 and expiring in two months. We replicate the call by matching the payoffs of the call when the stock price goes up and down, respectively:

\[\begin{align*} 53 N_{S} + 49 N_{B} & = 4 \\ 48 N_{S} + 49 N_{B} & = 0 \end{align*}\]

We find that:

\[\begin{align*} N_{S} & = \frac{4 - 0}{53 - 48} = 0.8000 \\ N_{B} & = -\frac{48}{49} \times 0.80 = -0.7837 \end{align*}\]

The bond price is \(B = 49 e^{-0.10 \times 2/12} = 48.19\). Therefore the call should trade at

\[\begin{equation*} C = 0.80 \times 50 - 0.7837 \times 48.19 = 2.23. \end{equation*}\]

Exercise 8.2

A stock price is currently $80. It is known that at the end of 4 months it will be either $75 or $85. The risk-free interest rate is 5% per annum with continuous compounding. What is the value of a 4-month European put option with a strike price of $80? Use the replicating portfolio argument and indicate the number of shares required to hedge the position.

Solution to Exercise 8.2

The binomial trees for the stock, the bond and the put are as follows:

Let \(N_{S}\) and \(N_{B}\) denote the number of stocks and bonds with face value $80 and expiring in four months. We replicate the put by matching the payoffs of the put when the stock price goes up and down, respectively:

\[\begin{align*} 85 N_{S} + 80 N_{B} & = 0 \\ 75 N_{S} + 80 N_{B} & = 5 \end{align*}\]

We find that:

\[\begin{align*} N_{S} & = \frac{0 - 5}{85 - 75} = -0.5000 \\ N_{B} & = - \frac{85}{80} \times (-0.50) = 0.5313 \end{align*}\]

The bond price is \(B = 80 e^{-0.05 \times 4/12} = 78.68\). Therefore the call should trade at

\[\begin{equation*} P = -0.50 \times 80 + 0.5313 \times 78.68 = 1.80. \end{equation*}\]

Exercise 8.3

A stock price is currently $40. It is known that at the end of 1 month it will be either $42 or $38. The risk-free interest rate is 8% per annum with continuous compounding. What is the value of a 1-month European call option with a strike price of $39?

Solution to Exercise 8.3

The binomial trees for the stock and the call are as follows:

The risk-neutral probability of an up move is:

\[\begin{equation*} p = \frac{40 e^{0.08 \times 1/12} - 38}{42 - 38} = 0.5669. \end{equation*}\]

We can then compute:

\[\begin{equation*} C = (3 p + 0 (1 - p)) e^{-0.08 \times 1/12} = 1.69 \end{equation*}\]

Exercise 8.4

A stock price is currently $50. It is known that at the end of 6 months it will be either $45 or $55. The risk-free interest rate is 10% per annum with continuous compounding. What is the value of a 6-month European put option with a strike price of $50?

Solution to Exercise 8.4

The binomial trees for the stock and the put are as follows:

The risk-neutral probability of an up move is:

\[\begin{equation*} p = \frac{50 e^{0.10 \times 6/12} - 45}{55 - 45} = 0.0.7564. \end{equation*}\]

We can then compute:

\[\begin{equation*} P = (0 p + 5 (1 - p)) e^{-0.10 \times 6/12} = 1.16 \end{equation*}\]

Exercise 8.5

A non-dividend stock trades for $100. Over each of the next two 6-month periods it is expected to go up or down by 10%. The risk-free interest rate is 8% per annum with continuous compounding.

1. What is the value of a 1-year European call option with a strike price of $100?

2. What is the value of a 1-year European put option with a strike price of $100?

Solution to Exercise 8.5

1. The binomial trees for the stock and the European call are as follows:

   

   
   The risk-neutral probability of an up move is:

   \[\begin{equation*} p = \frac{e^{0.08 \times 6/12} - 0.90}{1.10 - 0.90} = 0.7041. \end{equation*}\]

   We can then compute:

   \[\begin{align*} C_{u} & = (21 p + 0 (1 - p)) e^{-0.08 \times 6/12} = 14.21 \\ C_{d} & = (0 p + 0 (1 - p)) e^{-0.08 \times 6/12} = 0 \\ C & = (14.21 p + 0 (1 - p)) e^{-0.08 \times 6/12} = 9.61 \end{align*}\]

2. The binomial tree for the European put is:

   

   Therefore:

   \[\begin{align*} P_{u} & = (0 p + 1 (1 - p)) e^{-0.08 \times 6/12} = 0.28 \\ P_{d} & = (1 p + 19 (1 - p)) e^{-0.08 \times 6/12} = 6.08 \\ P & = (0.28 p + 6.08 (1 - p)) e^{-0.08 \times 6/12} = 1.92 \end{align*}\]

Exercise 8.6

Consider a non-dividend paying asset for which you have the following information: \(S_{0} = 100\), \(r = 4\%\) per year with continuous compounding, and\(\sigma = 20\%\) per year. Using a standard binomial model with four periods in which \(u = e^{\sigma \sqrt{\Delta t}}\) and \(d = 1/u\), compute the price of an at-the-money European call option expiring in 6 months.

Solution to Exercise 8.6

We start by noting that \(\Delta t = 0.5 / 4 = 0.125\), so that \(u = e^{0.20 \sqrt{0.125}} = 1.0733\) and \(d = 1 / 1.0733 = 0.9317\).

The tree for the stock is:

Period	0	1	2	3	4
					132.69
				123.63
			115.19		115.19
		107.33		107.33
	100.00		100.00		100.00
		93.17		93.17
			86.81		86.81
				80.89
					75.36

The risk-neutral probability of an up-move is

\[\begin{equation*} p = \frac{e^{0.04 \times 0.125} - 0.9317}{1.0733 - 0.9317} = 0.5177 \end{equation*}\]

Therefore, the tree for the call is:

Period	0	1	2	3	4
					32.69
				24.13
			16.19		15.19
		10.27		7.83
	6.29		4.03		0.00
		2.08		0.00
			0.00		0.00
				0.00
					0.00

The call is worth $6.29.

Exercise 8.7

Consider a non-dividend paying asset that trades for $100. Over the next six months, analysts expect that it could go up to $113 or down to $90. Compute the price of an at-the-money European call option expiring in six months. Assume that the risk-free rate is 7% per year with continuous compounding.

Solution to Exercise 8.7

The binomial trees for the stock and the call are as follows:

The risk-neutral probability of an up move is:

\[\begin{equation*} p = \frac{100 e^{0.07 \times 6/12} - 90}{113 - 90} = 0.5897. \end{equation*}\]

We can then compute:

\[\begin{equation*} C = (13 p + 0 (1 - p)) e^{-0.07 \times 6/12} = 7.40. \end{equation*}\]

Exercise 8.8

Consider a non-dividend paying asset that trades for $99. Over the next six months, analysts expect that it could go up to $114 or down to $88. Compute the price of an at-the-money European put option expiring in six months. Assume that the risk-free rate is 6% per year with continuous compounding.

Solution to Exercise 8.8

The binomial trees for the stock and the put are as follows:

The risk-neutral probability of an up move is:

\[\begin{equation*} p = \frac{99 e^{0.06 \times 6/12} - 88}{114 - 88} = 0.5390. \end{equation*}\]

We can then compute:

\[\begin{equation*} P = (0 p + 11 (1 - p)) e^{-0.06 \times 6/12} = 4.92. \end{equation*}\]

Exercise 8.9

The current price of a non-dividend paying stock is $139. Over the next year, it is expected to go up or down by 10% or 12%, respectively. The risk-free rate is 6% per year with continuous compounding. A market-maker of an important investment bank just sold 100 at-the-money European call options (i.e. one contract) expiring in one year to an important client. How many shares of the stock does she need to buy in order to hedge her exposure?

Solution to Exercise 8.9

The binomial trees for the stock and the call are as follows:

The number of shares to hedge the position is

\[\begin{equation*} N_{S} = 100 \left( \frac{13.9 -0}{152.9 - 122.32} \right) = 45.45. \end{equation*}\]

Exercise 8.10

The current price of a non-dividend paying stock is $87. Over the next year, it is expected to go up or down by 11% or 14%, respectively. The risk-free rate is 6% per year with continuous compounding. A market-maker of an important investment bank just sold 100 at-the-money European put options (i.e. one contract) expiring in one year to an important client. How many shares of the stock does she need to sell in order to hedge her exposure?

Solution to Exercise 8.10

The binomial trees for the stock and the put are as follows:

The number of shares to hedge the position is

\[\begin{equation*} N_{S} = 100 \left( \frac{0 - 12.18}{96.57 - 74.82} \right) = -56, \end{equation*}\]

meaning that the bank needs to sell 56 shares to hedge the position.

Exercise 8.11

The current price of a non-dividend paying stock is $100. Every three months, it is expected to go up or down by 4% or 6%, respectively. The risk-free rate is 8% per year with continuous compounding. Compute the price of a European call option with strike price $98 and maturity six months written on the stock.

Solution to Exercise 8.11

We have that \(\Delta t = 3/12\), \(u = 1.04\) and \(d = 0.94\). The binomial trees for the stock and the European call are as follows:

The risk-neutral probability of an up move is:

\[\begin{align*} p = \frac{e^{0.08 \times 3/12} - 0.94}{1.04 - 0.94} = 0.8020. \end{align*}\]

We can then compute:

\[\begin{align*} C_{u} & = (10.16 p + 0 (1 - p)) e^{-0.08 \times 3/12} = 7.99 \\ C_{d} & = (0 p + 0 (1 - p)) e^{-0.08 \times 3/12} = 0 \\ C & = (7.99 p + 0 (1 - p)) e^{-0.08 \times 3/12} = 6.28 \end{align*}\]

Exercise 8.12

A non-dividend paying stock costs $50. Analysts estimate that there is a 50% chance that the stock trades for $60 next year if the company succeeds in developing an important vaccine. Otherwise, the stock could fall to $40. The risk-free rate is 5% per year with continuous compounding. What should be the price of a security that pays $100 next year if the stock goes up, and $0 otherwise?

Solution to Exercise 8.12

The risk-neutral probability of the stock going up is:

\[\begin{equation*} p = \frac{50 e^{0.05} - 40}{60 - 40} \end{equation*}\]

The price of the security then is \((100 p + 0 (1 - p)) e^{-0.05} = \$59.75.\)

Exercise 8.13

In the binomial pricing model, the real probabilities of the stock going up or down do not matter to price a derivative written on the stock because:

1. We can replicate the payoffs of the derivative irrespective of the real probabilities.

2. All investors in the real economy are risk-neutral.

3. It gives a reasonable approximation to the no-arbitrage price.

4. We should always assume that the stock can go up or down with equal probability.

Solution to Exercise 8.13

The replication argument works irrespective of the real probabilities, which means that we could have done the same analysis in an economy where all investors are risk-neutral. This does not mean, however, that we are assuming that investors in the real world are risk neutral. The correct answer is alternative 1.

Exercise 8.14

In the one-period binomial model, the risk-neutral probability of the stock going up is:

1. Equal to the real probability that the stock should go up.

2. The probability that investors living in a risk-neutral world would use to assess the likelihood of the stock going up.

3. A mathematical artifact that allows us to compute the correct price of a derivative.

Solution to Exercise 8.14

Both alternatives 2 and 3 are correct.