Multinomial Trees and Incomplete Markets

In our previous articles on pricing via hedging, risk neutrality and replication we made use of a binomial tree to value a call option. In this type of tree, each node always has at most two daughter nodes, which leads to an asset having only two values that it can take at the next step. This is clearly not a realistic situation. Here we will attempt to improve the model by adding another daughter node, only to show that the attempt will fail!

Our previous two states were $U$ and $D$, which labelled the asset as having risen or fallen from the previous time step. In our prior examples we utilised a stock priced at 100, which became 110 at U and 90 at D. We can now add a third possible state $M$, at which the stock will remain at 100 from the previous step.

We will continue with attempting to price a call option with struck at $K=100$. First of all, let's try applying the hedging argument. We will purchase $\delta$ stocks and sell a call option on our first step. At each daughter node, the portfolio will be worth $110\delta - 10$ at $U$, $100\delta$ at $M$ and $90\delta$ at $D$.

The main problem here is that we have three relations and only one unknown quantity, $\delta$, to relate them with. The system is under-specified. We can show this by setting the last two relations to be equal to each other:

\begin{eqnarray*} 100\delta = 90\delta \end{eqnarray*}

The only solution is $\delta=0$, but this leads to the portfolio at $U$ being valued at $-10$. This is not a hedged portfolio, because if $U$ occurs, we lose money. Similarly if we try the same with the portfolios for $U$ and $D$, we have:

\begin{eqnarray*} 110\delta - 10 = 90\delta \end{eqnarray*}

The solution here is $\delta = 0.5$. This values our portfolios in $U$, $M$ and $D$ at $45$, $50$ and $45$ respectively. At this point we know that the portfolio tomorrow is worth at least $45$, although uncertainty still exists. Since it can be worth more than $45$ in the next time step it must be worth more than $45$ in the first time step. This means that:

\begin{eqnarray*} 50 - C > 45 \end{eqnarray*}

Hence the call option must be valued at strictly less than $5$. The key point is that we are unable to fully replicate the portfolio in all worlds, unlike in the binomial case. However, we have only tried hedging as a price mechanism. What if we try risk neutrality? Will this change the situation? Unfortunately not.

For the risk neutral pricing argument to work we need the expected value of the stock in the next step to equal the value of the stock today. This means that states $U$ and $D$ must have equal chance of occurring. This was straightforward in a two-state binomial case, but in the case of three-states, we have $U$ and $D$ occurring with probability $p$, while $M$ occurs with probability $1-2p$ (in order to achieve a sum total of unity for our probabilities). All of these values must be non-negative. This means that $p$ can be between $0$ and $0.5$, rather than fixed at $0.5$.

What does this mean for the value of the call option? It is given by the following expression:

\begin{eqnarray*} \mathbb{E}(C) = (110-100)p + 0\cdot p + 0\cdot (1-2p) = 10p \end{eqnarray*}

Since $p \in [0,0.5]$, it implies that $\mathbb{E}(C) \in [0,5]$. However, this is the same range as provided by the hedging strategy. This leads us on to the concept of an incomplete market. This is simply a market where we are unable to construct portfolios to achieve a desired payoff function of a derivative. In such a market, a derivative is priced via the risk preferences of investors rather than by theory.

In the next article, we will add an extra step to our tree, so that it has three time states, instead of two »

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