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).

  • $\begingroup$ I think you need to fix your discounting in a few places. In general, multiplying by $1+r(T-t)$ is similar to multiplying by $e^{r(T-t)}$. You are mixing those up here. $\endgroup$ – kurtosis Sep 12 at 22:50
  • $\begingroup$ When I follow the method given by @RRL, it seems that the discount factor is $\frac{1}{(1+r)^9} = \frac{1}{(1+0.001)^9} = 0.9910$, can you point out where the mistakes are? $\endgroup$ – wecandothis Sep 13 at 7:40
  • $\begingroup$ It depends on if you do exponential or periodically-compounded discounting. I've seen trees built with both, but Bjork is using one so for replication you should use that. $\endgroup$ – kurtosis Sep 13 at 7:52
  • $\begingroup$ Are you referring to proposition 2.11? $\endgroup$ – wecandothis Sep 13 at 7:58

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 $C_{31} = C_{30} = 0$,

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

(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 $(3,2)$.)

| improve this answer | |
  • $\begingroup$ Many many thanks: if we apply your formula $C = \frac{1}{(1+r)^9} (1 \cdot q^3 C_{33} + 3 \cdot q^2(1-q) C_{32})$ = $\frac{1}{(1+0.001)^9} (1 \cdot 0.5636^3 (7.65) + 3 \cdot 0.5636^2(0.4363)(1.03))$ = $0.9910 \cdot ((0.1790 \cdot 7.65) + (3 \cdot (0.3176 \cdot 0.4363)1.03)) = 1.4918$ as the option price. Is it possible that you check the value of the price? Also because @kurtosis pointed out that my discounting is not correct $\endgroup$ – wecandothis Sep 13 at 7:34
  • $\begingroup$ You’re welcome. First, to get the risk neutral probability you have the correct formula $q = (1+R-d)/(u-d)$ but the $R$ here must be the interest rate for the single step period. It seems you are not given a continuously compounded rate but rather a weekly compounded rate $r = 0.001$. Since a single step is 3 weeks we have $R = (1+r)^3 -1$. $\endgroup$ – RRL Sep 13 at 14:08
  • $\begingroup$ Look at the Wikipedia entry for compound interest and in particular the example when a monthly rate is specified. There is a distinction between nominal and effective annual rate. If I am told that the interest rate per week is $0.001$ then I would assume this is a rate that is compounded at a weekly frequency. The nominal annual rate here is $52\times 0.001$. $\endgroup$ – RRL Sep 13 at 14:18
  • $\begingroup$ The problem in the notes says time to maturity is 9 weeks, rfi = 0.1% per week thus a 10000 means 10 usd every week. If I follow your line of thinking, this means that we have indeed $r = 0.001$, you say $R$ must be the interest rate for the single step period. We have 9 weeks left, and 3 steps is indeed 3 periods; $R = (1+r)^3 -1= (1.001^3 -1) = 0.003$, plug into q, gives us (1.003 - 0.943)/(1.0606−0.943) = 0.5102. If we plug this in your formula we get 0.9910⋅((0.1328⋅7.65)+(3⋅(0.2603⋅0.4898)1.03))= 3.46 which seems a bit high.. $\endgroup$ – wecandothis Sep 13 at 14:55
  • 1
    $\begingroup$ @wecandothis: I get $q= 0.5108$ and $q^3C_{33} + 3q^2(1-q)C_{32} = 1.4140$. Dividing by $(1+r)^9 = 1.009036$ I get $C = 1.4013$. This is the same as the result obtained by working backwards through the tree. $\endgroup$ – RRL Sep 15 at 0:40

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