I've looked into many books at my academic library, and very often it goes like this:

  • Brownian motion
  • Then, stochastic integration (Itô's formula etc.)
  • Application: Black-Scholes formula for price of a call option

However, I've seen a proof that doesn't require stochastic integration at all, it goes like this:

Let $Y(t) = Y(0) e^{X(t)}$ be the price of an asset, as a geometric brownian motion, i.e. $X(t) = \mu t + \sigma B(t)$ where $B$ is a standard brownian. We also assume that $\mu + \sigma^2 / 2 = 0$ so that the trend is neutral.

Now the price for a European call option (maturity $t=T$, strike price $Y(0) A$) is:

$$C = E\big( (Y(T)-Y(0) A)^+ \big) = Y(0) \cdot E\big( (e^{X(T)}-A)^+ \big).$$

But since $X(T)$ has a law $\mathcal{N}(\mu T, \sigma^2 T)$ (brownian), it's easy to see that

$$E\big( (e^{X(T)}-A)^+ \big) = \int_{\ln A}^\infty (e^x-A)\frac{1}{\sqrt{2\pi \sigma^2 T}} e^{-\frac{(x-\mu T)^2}{2 \sigma^2 T} } d x$$

and then (it's just standard integration), we get:

$$C = Y(0) (\Phi(\sigma \sqrt{T} - \alpha_T) - A \Phi(-\alpha_T))$$ with $\alpha_T := \ln(A)/(\sigma\sqrt T)+\sigma\sqrt T/2$ and $\Phi(x) = (1/\sqrt{2 \pi}) \int_{-\infty}^x e^{-t^2/2} d t$

This proves the Black-Scholes formula for a call option, without needing any stochastic integration / Itô.


Why do all textbooks use stochastic integration to prove this, when it seems we don't need it?

  • 3
    $\begingroup$ Stochastic integration is used for the dynamics of $Y(t)$ and, consequently, the distribution of $Y(T)$. Once you have the distribution, you can then compute the expectation. That is, if you have the distribution, you do not need stochastic integration. However, stochastic integration is needed to derive the distribution. $\endgroup$
    – Gordon
    Jan 19, 2016 at 15:54
  • $\begingroup$ @Gordon In the standard case of the Black-Scholes formula, don't we already have $Y(t)$ as a geometric brownian motion? $\endgroup$
    – Basj
    Jan 19, 2016 at 17:51
  • $\begingroup$ That is true. But how can we assume that $Y(t)$ is a geometric Brownian motion? It is based on the form of the SDE, where stochastic integration is used. $\endgroup$
    – Gordon
    Jan 19, 2016 at 18:40
  • $\begingroup$ @Gordon : ok, so in the Black-Scholes proof, stochastic integration is only used to prove that the price of an asset can be well modelized by a geometric Brownian motion? If you can, could you give the general idea in an answer? $\endgroup$
    – Basj
    Jan 19, 2016 at 19:19
  • 1
    $\begingroup$ What you have is not the BS formula since there has been no change of measure. Stochastic integration is required to show that a replicating portfolio exists and that a change of measure is possible. $\endgroup$
    – user9403
    Jan 20, 2016 at 12:20

6 Answers 6


you can get it as a limit of binomial trees by doing risk-evaluation on the trees and passing to the limit. See eg Baxter and Rennie or my book "Concepts and Practice of mathematical finance."

However, one should be careful when talking about "proving" a formula. You prove theorems. Theorems require assumptions. The Black--Scholes-Merton argument shows that given a GBM with any drift then no arbitrage implies that the BS formula holds. This is much stronger than assuming the drift has an unrealistic value for the starting point of the deduction.


A few years ago I asked a similar question on MO:


My take today is that you really don't need this heavy mathematical machinery for standard BS but as soon as you move on to more sophisticated (and realistic) models you surely do, so it is useful, when you develop the much needed machinery right away and use it on some toy problems instead of switching gears in the middle of your analysis.

Besides stochastic integration is also quite elegant and general and mathematicians like elegant and general solutions.


The expected value approach through integration essentially restates the expectation using the heat diffusion equation (i.e., with boundary condition $0$).

The true insight of B-S was not that the fair value of an option is the expected value of the posterior distribution, but rather that the "correct" distribution is the "risk-neutral" one. B-S's dynamic hedging argument eliminated the drift(s) -- thus the "risk-neutral" distribution. Having established the B-S PDE, B-S solve the equation by the standard convolution method for a diffusion equation. This is essentially what you have done above, without deriving a PDE. In fact, the rules of stochastic integration (Ito's Lemma, Kolmorgov, Giransov, etc...) are merely simplified restatements which were derived from measure theory.

This approach reminds me of an interesting post which restates the expected value approach graphically, and extends options analysis to product development and R&D. Read it here: http://blackswanfarming.com/product-development-payoff-asymmetry/


Stochastic integration is required to prove Girsonav's theorem; which is required to prove that you can use the "risk neutral" expectation instead of the "actual" expectation. In short, it is required so that the "assumption" that $\mu+\sigma^2/2=0$ is valid. Though typically this would be $r=\mu$ instead...


There are two holes in your proof

  1. Why value is given by expectation at all

  2. How to choose $\mu$ and $\sigma$

Girsanov tells you that $\sigma$ stays historical, 1) comes from fundamental pricing theorems on complete and arbitrage free market, $\mu$ comes from tbr definition of such market. The last two things relresent "physics" of the process, so to speak. Think Poincare's vs Eistein's contribution to special relativity.

BTW, BS didn't have any of this. They didn't care about pricing options, they aimed at constructing a market-neutral portfolio of stock and its option. That's why return was risk-free, i.e. no risk premium (on top of risk-free rate), and this was achieved by, sort of, inverse delta hedging with an option.

All beautiful theories were invited later.


In Blyth "Introduction to Quantitative Finance", the Black Scholes formula is derived without explicit use of stochastic calculus as follows: (i) show that on a binomial tree, use of risk-neutral probabilities gives arbitrage free prices (ii) show that a risk-neutral binomial tree, where a stock can go up or down by a fixed percentage at each step, converges to a risk-neutral lognormal distribution as the number of steps tends to infinity (iii) integrate as you did.

  • $\begingroup$ And just to add another text for people reading this thread: I don't know where it stands now relative to other texts but Cox and Rubintein's "Options Markets" was once a very well respected text and extremely popular. I recommend it. $\endgroup$
    – mark leeds
    May 7 at 5:57

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