Binomial Model

At each step two outcomes are possible, one up and one down. Two factors describe these outcomes. The relative magnitude of an up move is represented by u, and the relative magnitude of a down move is represented by d. Thus with a starting spot value of $S_o$, after one period we have ending values of $S_u$ or $S_d$ for up and down changes respectively. The underlying logic of this model is based on the ability to replicate a derivative's performance with an appropriate combination of the underlying asset and a risk-free asset. This can be structured as simultaneous equations that can be solved to find the composition of the replicating portfolio. Specifically, we wish to solve for the number of shares, N, in the replicating portfolio and the amount invested in bonds, b, in the replicating portfolio. The first of our two equations represents a starting portfolio that ends with an up move. $NS_u+B(1+r)$ illustrates the value of the replicating portfolio in this instance. The second of the simultaneous equations represents a starting portfolio that ends with a down move. $NS_d+B(1+r)$ illustrates the value of the replicating portfolio in this second instance. Each of these portfolios in turn can be equated to the value of the respective derivative outcomes $V_u \& V_d$. We combine these two sets of values into equations as follows.

(1)
\begin{equation} NS_u+B(1+r)=V_u \end{equation}
(2)
\begin{equation} NS_d+B(1+r)=V_d \end{equation}

Now solve the two simultaneous equations for the two unknowns, N and B. Solve for N first, then substitute that value into either equation to find the value of B. Recall that we can subtract one equation from the other. When we subtract the second equation from the first our first term found is $N(S_u-S_d)$. The second term involves B(1+r) - B(1+r) which cancel each other exactly leaving zero for the middle term. The last term then to the right of the = sign is simply $V_u-V_d$. The final result of the subtraction leaves us with.

(3)
\begin{equation} N(S_u-S_d) = V_u-V_d \end{equation}

Divide both sides of this equation by $(S_u-S_d)$ to isolate N on the left hand side producing the following equation.

(4)
\begin{align} N = \frac {V_u-V_d}{(S_u-S_d)} \end{align}

Now that N is known, we can rearrange the first equation to find B. First subtract $NS_u$ from both sides of the equation.

(5)
\begin{equation} B(1+r)=V_u-NS_u \end{equation}

Then divide both sides of the equation by $(1+r)$ to isolate B alone on the left.

(6)
\begin{align} B=\frac{V_u-NS_u}{(1+r)} \end{align}

Now we are prepared to find the amount to borrow in our replicating portfolio B. Note that we could equally well have solved for B by substituting our value for N into the second equation. Sometimes, depending on the values that is a preferred or simpler alternative.

At this point we know the amount we must invest in the underlying asset initially, $N\times S_o$ and the amount we must invest in the risk-free bond, B. The sum of these, the value of the replicating portfolio equals the value of the derivative at the start.

Assuming that the underlying distribution governing up and down moves is normal produces the following.

(7)
\begin{align} u=e^\sigma^\sqrt{t} \end{align}
(8)
\begin{align} d=e^{-\sigma\sqrt{t}}=\frac{1}{u} \end{align}

Outcomes at each node can be found from:

(9)
\begin{align} S_N=S_0\cdot u^{n_u-n_d} \end{align}

Call premium can be found from:

(10)
\begin{align} C=\frac{pC_u+(1-p)C_d}{1+r} \end{align}
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