Calculating European call option, the Bjork way

Question

We have a 3 period binomial tree with values:

                                              59.65 (C33 = 7.65)
                            56.24 (C22 = ?)
                 53.03                        53.03 (C32 = 1.03)
50                          50    (C21 = ?)
                 47.14                        47.14 (C31 = 0)
                            44.45 (C20 = ?)
                                              41.91 (C30 = 0)

W want to calculate a European call option, no arbitrage, with properties K = 52, u = 1.0606, d = 1/u = 0.943, maturity in 9 weeks, r = 0.001 per week. The value for a call option is given by $max[S_t -K, 0]$. We can calculate $C_2^{2}$ given the risk neutral formula from the literature (Bjork 3ed, 2.1.4):

$C_2^{2} = \frac{1}{1+R} (q*C_3^{3} + (1-q)*C_3^{2})$, $\frac{1}{1+R}$, given by Bjork Proposition 2.6, but as we have multiple nodes I assume we need to discount it, which gives the formula $e^{r-(T-t)}= e^{0.053348-(9/52)} = 1.009276$

$R = 1.001^{52} = 5.3348pct = 0.053348 ,$

$q = \frac{(1+R)-d}{u-d} = \frac{1.009276 - 0.943}{1.0606 - 0.943} = 0.5636$, if we then plug the values into the formula:

$C_2^{2} = \frac{1}{1+R} (q*C_3^{3} + (1-q)*C_3^{2}) = 1.009276*(0.5636*7.65 + 0.4363*1.03) = 4.8051$,

My questions are:

a) is the value of $C_2^{2}$ correct?

b) is there a faster way to calculate the option value of the tree because this takes a lot of time (yes you can write a program but I am following the theory and I believe I have to learn it by hand as well).

Answer 1

Is there a faster way to calculate the option price?

With a recombining binomial tree, the terminal asset price has a binomial distribution -- as you might have expected. For a tree with $n$ steps, the probability of reaching price $S_{n,k}$ where $k$ is the number of up moves is

$$P_{n,k} = \frac{n!}{k!(n-k)!}q^k(1-q)^{n-k}$$

The option price is the discounted risk-neutral expectation of the payoff,

$$C = \frac{1}{(1+r_s)^n}\sum_{k=0}^n\frac{n!}{k!(n-k)!}q^k(1-q)^{n-k} \max(S_{n,k}-K,0),$$

where $r_s$ is the interest rate per period associated with a single step. Using this formula avoids working backwards and computing option values at intermediate steps.

In this case we have $n= 3$ and $(1+r_s) = (1+0.001)^3$ (since each step spans 3 weeks). Hence, since $C_{31} = C_{30} = 0$,

$$C = \frac{1}{(1+r)^9} (1 \cdot q^3 C_{33} + 3 \cdot q^2(1-q) C_{32})$$

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(Note that the coefficient $1$ for the first term arises because there is one path through the tree reaching the node $(3,3)$ and the coefficient $3$ for the second term arises because there are three paths through the tree reaching the node $(3,2)$.)