Pricing European Calls with Recombining Binomial Trees

# Investigating how rational price of European call option changes


# Investigating how rational price of European call option changes












Let S(0) = 100 be the initial price of the risky asset. Consider a European call option with exercise price K and expiry time T = 1 (year). Consider several binomial models and investigate how does the rational price of the option ( ie C(0) ) depends on the choice of the parameters of the model and the strike price K.

I'm trying to answer this question but I'm unsure how to start it. I'm trying to do it on a 1-step, 2-step and a 3-step binomial model and maybe see if there is a pattern but I'm unsure how to start it for the 1-step other than probably having a strike price K = 100. Also I think the 2-step will take one step in 1/2 year and the 3-step model in 1/3 year etc.

What I mean is: when I have to assign random parameters U,D and R, do I just pick any numbers I want or is there a certain interval I should stick to? Like can I say that for 1-step U=1.1 and D=0.9 and then for 2-step something like U=1.2 and D=0.8? Does that work for the question or am I not getting it?

## Answer by RRL (score 2, accepted)

https://quant.stackexchange.com/a/59844

With a (recombining) binomial tree, the terminal asset price has a binomial distribution. Given the up and down move size $u$ and $d = 1/u$, respectively, the terminal price after $n$ steps and $k$ up moves is $S_{n,k} =S_0u^kd^{n-k}= S_0u^{2k-n}$ and probability of reaching this price is $$P_{n,k} = \frac{n!}{k!(n-k)!}q^k(1-q)^{n-k},$$

where $q = \frac{r-d}{u-d}$ and $r$ is the interest rate per period associated with a single step.

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

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

This will converge to the Black-Scholes option price. Use that as a reference for assesing the accuracy. I believe that the convergence rate is $\mathcal{O}(1/n)$ as $n \to \infty$. In other words, doubling $n$ should halve the error.

You can discover this by numerical experiments using the given pricing formula (you don't have to work laboriously backwards through the tree). The time step size is $\Delta t= \frac{T}{n}$ and the size of the up move is related to volatility through $u = e^{\sigma\sqrt{\Delta t}}$. So you could hold $T$, $K$, $\sigma$, and $r$ fixed and observe the behavior of the option price as $n$ is varied.

You can also prove it analytically although that is very difficult.