1
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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 | |
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.