# How to price and find a replicating portfolio for a call spreads using a two-period binomial model?

Consider a two-period binomial model for a risky asset with each period equal to a year and take $S_0 = 1$, $u = 1.03$ and $l = 0.98$.

a.) If the interest rate for both periods is $R = .01$, find the price of the option with the payoff shown in Figure 2.4 with $K_1 = 1.00$ and $K_2 = 1.05$ at all nodes of the tree.

b.) Find the replicating portfolio at each nodes of the tree.

Solution for a.) First we need to calculate the first period probabilities:
$$\hat{\pi}_u = \frac{1 + R - k}{u - l} = .6 \ \ \text{and} \ \ \hat{\pi}_l = \frac{1 + R - k}{u - l} = .4$$ The second-period probabilities are the same since $R = .01$ is constant in both periods. Lastly, the probability at the end nodes are .36 for the $S_0 u^2$, .24 for $S_0 u l$, and .16 at $S_0 l^2$. The payoff we have in the graph is a call spread, so we are long a call $C_1$ at strike $K_1$ and short a call $C_2$ at strike $K_2$. The price of the instrument is thus $V = C_1 - C_2$. Now we can calculate the end nodes: $$V(S_0 u^2) = (S_T - K_1)^{+} - (S_T - K_2)^{+} = .05 \ \ \ V(S_0 u l) = .0094 \ \ \ V(S_0 l^2) = 0$$ Next, we work backwards one period and compute $$V(S_0 u) = \frac{1}{1 + .01}(.6(.05) + .4(.0094)) = .033426 \ \ \ V(S_0 l) = \frac{1}{1 + .01}(.6(.0094) + 0) = .005584$$ Lastly, we can calculate $V_0$ which is $$V_0 = \frac{1}{1+.01}(.6^2 (.05) + 2(.24)(.0094)) = .022289$$ Finally we can calculate the expectations of the payoff at each node to get the prices: $$E(V(S_0 u)) = \frac{1}{1+.01}(.6(.05) + .4(.0094)) = .033426$$ $$E(V(S_0 l)) = 0$$ $$E(V(S_0 u^2)) = \frac{1}{(1.01)^2}(.36(.05)) = .017645$$ $$(V(S_0 u l)) = \frac{1}{(1.01)^2}(.24(.0094)) = .002212$$ $$E(V(S_0 l^2)) = 0$$

• The answer below for the payoff decomposition is fine, That is, at maturity $T=2$, the payoff is given by \begin{align*} & \ (K_2-K_1)1_{S_T > K_2} + (S_T-K_1)1_{K_1 \le S_T < K_2} \\ =& \ (K_2-K_1)1_{S_T > K_2} +(S_T-K_1)1_{S_T \ge K_1}-(S_T-K_1)1_{S_T \ge K_2} \\ =&\ (S_T-K_1)1_{S_T \ge K_1} - (S_T-K_2)1_{S_T \ge K_2} \\ =&\ (S_T-K_1)^+-(S_T-K_2)^+. \end{align*} Mar 16 '16 at 12:56
• @Gordon so far I have calculated the risk neutral probabilities and the stock price nodes for all periods. Now, I drew out the option price tree from the answer provided I need to calculate: $V(S_0 u^2)$, $V(S_0 u d)$, $V(S_0 d^2)$. Then I work my way backwards, to calculate $V(S_0 u)$, and $V(S_0 d)$. Then finally I can calculate $V_0$. Is this correct? If so I am not sure how to calculate $V(S_0 u^2)$, $V(S_0 u d)$, $V(S_0 d^2)$ the pay off graph. Mar 16 '16 at 17:16
• That is correct. To compute $V(S_0u^2)$, you only need to replace $S_T$ in your payoff, as is shown in my comments above, by $S_0u^2$. Mar 16 '16 at 17:19
• @Gordon what do you mean by replace $S_T$, do you mean this: $V(S_0 u^2) = (S_T - K_1)^{+} - (S_T - K_2)^{+} = (1.0609 - 1) - (1.0609 - 1.05)$? Mar 16 '16 at 17:26
• You can edit my answer to make things clearer if you need Gordon. But indeed, what I meant when writing "use the graph" is exactly what Gordon mentions: the graph describes a function, you simply need to evaluate that function at given abscissae. Mar 16 '16 at 17:27

The payoff you mention is that of a call spread, i.e. long a call $C_1$ struck at $K_1$ and short a call $C_2$ struck at $K_2$, with $K_2>K_1$. The price of the instrument is therefore: $V = C_1 - C_2$.

[First way] If you are stuck because this payout seems 'unsual' to you, an easy way to reach your goal (assuming you know how to use binomial trees to price standard call options, if you don't please see below), would be to:

• Use a 2-period binomial tree to price the call $C_1$;
• Use a 2-period binomial tree to price the call $C_2$;
• Create a final option tree whose leaves will receive the difference between the leaves' values of the previous ones, since we've established that at all times $V = C_1-C_2$ by absence of arbitrage opportunity.

[Second way] It is also possible to do it directly with a single option tree of course (as mentioned in my comments and explained below).

How to use a binomial tree to price (European) options

• Grow a recombining stock price tree. Over each period, the stock price can either evolve upwards or downwards (hence the term binomial). Assuming you start a period with a stock worth $S$, you'll observe either $S_u = u S$ or $S_d = d S$ at the end of that period.

Here you are asked to work with 2 periods, so starting from $S_0$, you'll end up with 6 nodes in total: the first node $S_0$, then the first-period nodes $S_0 u$ and $S_0 d$, and the final period nodes $S_0 u^2$, $S_0 u d$, $S_0 d^2$.

• Grow yet another tree with the same structure. Call this the option price tree. As the name indicates, its nodes will figure the option value for each state of the stock described by the stock price tree. One objective is obviously to determine the value of the option $V_0$ when the stock is worth $S_0$ at inception. Yet, at this stage, you only know the value of the option at expiry. Indeed, at expiry, by absence of arbitrage opportunity, the option should be worth its payoff. This can be used to place values on the terminal nodes of the option price tree.

In your case, the payoff function is represented by a graph. You can use this graph to find what values to attribute to each terminal node of the option price tree. For instance, in the situation where the stock is worth $S_0 u^2 = (1.03)^2$ at expiry, find the corresponding payoff ($f(S_0u^2)$ in the payout graph) and plug it as the option value at the relevant terminal tree node. Repeat for each terminal node.

• Finally, work backwards from the terminal tree nodes, by taking risk-neutral expectations. The idea is the opposite of when you grew the stock price tree: instead of moving from 1 stock price node (start of period) to 2 stock price nodes (end of period), you now work in the option price tree and proceed backwards from 2 option price nodes (end of period) to 1 option price node (start of period).

For instance, assume you have identified - from the payoff function - the values $V_{uu} = f(S_0u^2)$ and $V_{ud} = f(S_0ud)$ of the option at expiry. You are now looking for the value $V_u$ at the end of the first period, knowing that the stock finished in the upward state. This values is not given by $f(S_0 u)$ because you are not at the option expiry any more: no arbitrage argument cannot be used here. However, one can show that

$$V_u = \frac{1}{1+R} (q V_{uu} + (1-q) V_{ud})$$

where $q$ figures a risk-neutral probability of going in the upwards state over each period. Mathematically, $q$ computes as:

$$q = \frac{(1+R) -d}{u-d}$$

You can see that these risk-neutral probabilities are constant provided the interest rates are constant. Now you are done with $V_u$. You can repeat the process to compute $V_d$, the value of the option at the end of the first period knowing that the stock finished in the downward state, from the values of $V_{ud}$ and $V_{dd}$. This writes:

$$V_d = \frac{1}{1+R} (q V_{ud} + (1-q) V_{dd})$$

Using the same rationale, from the quantities $V_u$ and $V_d$ you just computed, you can further deduce $V$, the option value at inception where the stock price is $S_0$, by once again taking a discounted risk-neutral expectation.

$$V = \frac{1}{1+R} (q V_{u}+ (1-q) V_{d})$$

This is how you work your way to the root of the option tree using a 'backwards induction' process.

To see where this concept of risk-neutral probability comes from, assume you were to build a portfolio of primary assets at the start of a given period, whose objective will be to perfectly mimic the value of the option position in all possible states of the economy. At the start of a period, your portfolio of primary assets is worth

$$\Pi = \alpha S + \beta$$

At the end of a period, in the upward state, because we want it to be replicating we need to have:

$$\alpha S_u + \beta (1 + R) = V_u$$

Similarly, in the downard state:

$$\alpha S_d + \beta (1 + R) = V_d$$

Solve these 2 equations for the 2 uknowns $\alpha$ and $\beta$. You end up with:

\begin{align*} \alpha &= \frac{V_u - V_d}{S_u-S_d} \\ \beta &= \frac{1}{1+R} \frac{u V_d - d V_u}{(u-d)} \end{align*}

Now, by construction the portfolio is replicating. Hence by no arbitrage arguments its value at the start of the period should be exactly the same as that of the option:

$$V = \alpha S + \beta$$

With the values you have found for $\alpha$ and $\beta$ (functions of $R$, $V_u$ and $V_d$), you can re-write this equation in the form $V = \frac{1}{1+R} (q V_u + (1-q) V_d)$ introduced earlier, hence the concept of risk-neutral probabilities.

If you have correctly understood everything until now, then this also answers the point (b) of your question: on each period, the replicating portfolio corresponds to holding $\alpha$ shares and $\beta$ bonds.

• I understand the theory behind the binomial model. I can calculate the upper pay off node and lower pay off node but I do not understand how to relate that to the figure Mar 15 '16 at 20:48
• I don't think you do. Because if you did, no offence but, this would be trivial. You mention 2 periods. So start by growing a proper stock price tree. As far as the option price tree is concerned, start by the terminal leaves using the payoff function described in the graph (the option value equals it's payoff at maturity by absence of arbitrage opportunity)... then work your way backwards up to the root of the option price tree taking risk-neutral expectations at each step. If anything of this does not make sense, please re - read through theory and ask questions ! Mar 15 '16 at 21:16
• I need more help Mar 16 '16 at 0:28
• Yes you do. But the key with binomial tree (and martingale pricing in general) is that you do not need the 'real world' probabilities for the stock to go up or down (this is why none are given), but rather, 'risk-neutral' probabilities. I will edit my answer to give details on that. Mar 16 '16 at 8:40
• I posted my solution for a.) I am now trying to do part b. I am not exactly sure how I use what you wrote for the replication part to find a replicating portfolio for each node. Mar 16 '16 at 18:21