My question is simple, consider a European call with payoff max(S_T-K, 0), Let's suppose that the underlying stock follows a binomial tree with up and down factors I know as we take n goes to infinity that the stock is log-normally distributed at time t=T (I know how to derive it). The idea is to derive the B-S-M pricing formula as the expected value of the present value of max(S_t-K, 0) using the Lebesgue integral I write this like:
$\int_{\Omega} max(S_T-K,0) dP$
where P if I am not mistaken is the risk-neutral measure (since usually for a binomial tree we use the risk-neutral probabilities).
How can I set P to compute this? Once I have a measure P finding the distribution of max(S_t-K, 0) and then compute its density is right way to continue?. Finally, doing this should I arrive B-S-M pricing formula for a call?
2 Answers
If $Q$ is the risk-neutral measure, then take care that you need to discount your payoff appropriately, i.e. if $T$ is the time to expiry, and current time $t = 0$, then you want the integral $$ \int_{\Omega} e^{-rT}\max(S_T-K,0) dQ, $$ or $$ E^{Q}[e^{-rT}\max(S_T-K,0)]. $$ Assuming that you do not have any other assets under consideration, and a non-stochastic interest rate $r$, then you can just take $Q$ to be any probability measure that makes it so that $S_{T}$ has a lognormal distribution with the usual parameters. Rather than working with the distribution of $e^{-rT}\max(S_T-K,0)$ under $Q$, it is probably easier (but still tedious) to apply the law of the unconscious statistician together with some properties of $S_T$ that follows from the assumptions underlying the BSM model. See the first answer to a related question here.
By CLT, binomial distribution becomes normal as you increase steps $n$.
$log(S(T))=log(S0)+klog(u)+(n-k)log(d)$ is thus a shifted normal distribution, and so $S(T)$ is lognormal.