# Pricing a Call Option with Multi-Step Binomial Trees

Pricing a Call Option with Multi-Step Binomial Trees

It is a straightforward extension from the two-step model to use multi-step trees to price call options. There are some caveats: The number of steps must be finite and each node must have exactly two immediate child nodes. Note that in a tree, child nodes can have two separate parent nodes.

We've previously covered how to use hedging or risk neutral pricing in a two-step tree in order to determine the price of a call option via a method known as backward propagation. The processes work similarly here. We start at the end step, when the option value is known for all states. If we pick two nodes in this final layer, we can always determine the arbitrage free price at the parent node via one of these methods.

## Pricing via risk neutrality

We are now going to consider a four-step binomial tree. It consists of a beginning state where a stock is worth 100, and we wish to price to a call option struck at 100 expiring after five time steps. The stock can move up or down by 2.5 at each subsequent step, so that if it goes up at each step it will reach 110 or goes down at each step will reach 90 after four steps. As in our one- and two-step tree models, we will assume interest rates are zero. This ensures that the risk neutral probabilities of an up/down move will always be $p=0.5$. See the following figure for an image of the tree with stock prices added in: Let us proceed by using a risk neutral pricing argument. The implementation of a method is far simpler than the hedging argument. We want to consider the probabilities that we land at the final nodes of the tree.

To get to the highest (110) or lowest (90) node of the final step of the tree requires, respectively, four stock up moves or four stock down moves. Each of these probabilities are:

\begin{eqnarray*} \frac{1}{2^4} = \frac{1}{16} \end{eqnarray*}

To land at either the 105 node or the 95 node there must be, respectively three stock up moves and a down move or three stock down moves and an up move. Each of these can occur in any order, so we must factor that in to the calculation. Either way, the individual probabilities are the same due to risk neutrality and the no-arbitrage argument:

\begin{eqnarray*} {4 \choose 1} \frac{1}{2^4} = \frac{1}{4} \end{eqnarray*}

To land at the final central node where the stock is 100, we need precisely two stock up moves and two stock down moves, again in any order. The probability for this is given as follows:

\begin{eqnarray*} {4 \choose 2} \frac{1}{2^4} = \frac{3}{8} \end{eqnarray*}

We have now calculated all of the possibilities. As a quick check, the values should sum to unity: $1/16 + 1/4 + 3/8 + 1/4 + 1/16 = 1$.

Now we can price the call option, because we know that its value is given by its expectation under these probabilities. We've already set the strike price of the option to be 100. The following calculation provides us with its value:

\begin{eqnarray*} (110-100)\cdot\frac{1}{16} + (105-100)\cdot\frac{1}{4} + 0\cdot\frac{3}{8} + 0\cdot\frac{1}{4} + 0\cdot\frac{1}{16} = 1.875 \end{eqnarray*}

Notice that the last three terms are equal to zero, since the payoff of the option is zero when the stock is equal to or less than the strike price.

## Generalising to N-step trees

We are now beginning to see how this method could be extended to an $N$-step binomial tree. Consider a stock with a price today (the "spot" price) given by $S$. Our tree now has $N$ steps and at each new time-step our stock can move up or down by an amount $x$. If $S_i$ is the value of the stock price after $i$ stock up moves and $N-i$ stock down moves, we have:

\begin{eqnarray} S_i = S + ix - (N-i)x = S + (2i - N)x \end{eqnarray}

Given that the stock can have any combination of $i$ stock up moves and $N-i$ stock down moves, in order to account for these separate paths, the probability of of the stock having value $S_i$ is given by:

\begin{eqnarray*} {N \choose i}\frac{1}{2^N} \end{eqnarray*}

If we now generalise our call option to a derivative which has a payoff function given by $f=f(S)$ then after $N$ steps along the tree, the value of this derivative will be equal to:

\begin{eqnarray*} \sum^{N}_{i=0} {N \choose i} \frac{1}{2^N} f(S+(2i - N)x) \end{eqnarray*}

Let us examine each component of this formula. Firstly, $f(S+(2i - N)x)$, which is the payoff of the derivative at the node when the spot price has moved up $i$ times and down $N-i$ times, for one path of the stock. This is then multiplied by $1/2^N$, which is the probability that this particular path was chosen. This is then multiplied by ${N \choose i}$, which counts the total number of ways the stock could have moved up $i$ times and down $N-i$ times. Finally, the summation term at the start counts all of the separate nodes in the final step of the tree.

The next article will discuss applying a normal stock model to the multi-step tree. »