Risk Neutral Pricing of a Call Option with Binomial Trees with Non-Zero Interest Rates

All of our previous discussions on the binomial model have been carried out under the assumption of zero interest rates, that is, the continuously compounding rate $r$ is equal to zero. We will now relax that assumption and assume that $r$ is positive. Hence, a riskless bond will now grow as $e^{rt}$ over time $t$.

If we take $S_t$ as the price of the stock, then $S_0$ is today's price. Recalling that in the next step, the stock can take one of two values, denoted by $S_U$ and $S_D$, with $S_D < S_U$. After a time period $\Delta t$, we must have that:

\begin{eqnarray*} S_D < S_0 e^{rt} < S_U \end{eqnarray*}

If this wasn't the case, i.e. that the compounded stock price lay outside these bounds, we could construct an arbitrage opportunity.

There are three ways to find the value of a derivative paying $f(S)$ at time $\Delta t$: Risk Neutrality, Replication and Hedging. We will consider the risk neutral pricing scheme first, because it is the simplest to carry out, if slightly less intuitive than the 'constructive' methods.

Our task is to find the probability of the stock ending up in states $U$ or $D$, such that, on average, the stock grows at the compounding risk free rate $r$. Thus we need to find $p$ such that:

\begin{eqnarray*} \mathbb{E}(S_{\Delta t}) = pS_U + (1-p)S_D = S_0 e^{r\Delta t} \end{eqnarray*}

After some algebra, this leads to an expression for $p$:

\begin{eqnarray*} p = \frac{S_0 e^{r\Delta t} - S_D}{S_U - S_D} \end{eqnarray*}

Given our prior bounds on the risk free growth of the stock $S_0 e^{r\Delta t}$, we must have that $0 < p < 1$. Now we will apply our argument of risk neutrality. Previously we stated that the expectation of all possible portfolio values in the next time step had to be equal to today's value, otherwise an arbitrage opportunity would exist.

If we denote expectation under the risk neutral probability $p$, as $\mathbb{E}_{RN}$ then our risk free bond no longer satisfies this property. The bond's value at $\Delta t$ is not equal to the bond's value today:

\begin{eqnarray*} \mathbb{E}(B_{\Delta t}) \neq B_0 \end{eqnarray*}

Thus in order to make a similar argument, we need to consider instead discounted prices. This means rather than requiring that every possible portfolio to have expectation equal to today's value, we need that the expectation should be the same as the asset's valye invested at the compounding risk free rate. This is the same as saying that its discounted expectation is the same as today's value. Hence we require, for every asset A:

\begin{eqnarray*} \mathbb{E}_{RN}\left(\frac{A_{\Delta t}}{B_{\Delta t}}\right) = \frac{A_0}{B_0} \end{eqnarray*}

Clearly the bond satisfies this criterion and the stock will also satisfy this, since we have chosen the risk neutral probability $p$ to ensure this is the case. Now we wish to price the derivative, with payoff $f(S)$. We can define $C_0$ such that it satisfies the following relation:

\begin{eqnarray*} C_0 = \mathbb{E}_{RN}\left(\frac{A_{\Delta t}}{B_{\Delta t}}\right) = e^{-r\Delta t}\mathbb{E}_{RN}(f(S)) \end{eqnarray*}

In order to achieve a definitive price for this option, we must construct a portfolio which contains certain quantities of the asset, derivative and risk-free bond with initial total value of zero. Then at time $\Delta t$ the expected value of its ratio with the bond, under the risk neutral probability $p$, will also be zero.

The risk neutral expectation of the portfolio must also be zero, since the bond's value is independent of that of the asset. Further, this implies that the portfolio cannot be an arbitrage opportunity. This is because of there was a guaranteed positive outcome for this portfolio, then the risk-neutral portfolio would also be positive and thus it could not be worth zero at today's time.

Thus, we utilise the above formula but amend the expression for $\mathbb{E}_{RN}(f(S))$ such that it includes the risk-neutral probability $p$ for the payoff:

\begin{eqnarray*} C_0 = e^{-r\Delta t}(pf(S_U)+(1-p)f(S_D)) \end{eqnarray*}

The next article will be to show that we can achieve the same value using a hedging argument.

# Just Getting Started with Quantitative Trading?

## 3 Reasons to Subscribe to the QuantStart Email List:

No Thanks, I'll Pass For Now