# Black Scholes model without using Girsanov's theorem? It might happen?

We can calculate the stock price by the equation: $$\frac{dS_t}{dt} = \mu dt + \sigma dB_t$$,where $$B_t$$ is a Brownian motion.

First i create a portfolio that consists of $$\Phi$$ units of stock share and $$\phi$$ units of cash. Denote the amount of share and cash at time t as $$\Phi_t$$,$$\phi_t$$.Then, the value of the portfolio at time t $$(V_t)$$ will be the sum of the value of stock share $$(φ_t*S_t)$$ and the amount of real interest that can be earned by possessing the cash for dt amount of time $$(rP dt)$$ so that $$V_t = \Phi_t S_t + \phi_t r P dt$$. I do the calculations without using Girsanov's theorem and i get the Black-Scholes equation:$$\frac{\partial V}{\partial t} + \frac{1}{2}\sigma^2 S_t^2\frac{\partial^2 V}{\partial x^2} + r S_t \frac{\partial V}{\partial x} - r V = 0$$. For $$\mu \ne 0$$, the process $$S_t$$ is not a martingale, right? In many bibliographies authors uses the Girsanov's and Novikov's theorem something that i didn't use. I can 't understand the difference between my solution and thw other way. Can somebody help me; I hope i didn't confuse you.

## 1 Answer

A very interesting topic ! Black-Scholes originally did not make use of the Girsanov theorem and arrived at the equation the way you described.

Later theoretical work on arbitrage pricing uncovered the concepts of the risk-neutral measure and derivatives pricing as an expectation under that measure. That work relies on stochastic calculus far more and one could argue is a more “satisfying” approach. This is where the Girsanov theorem comes into play. See Harrison, Kreps, Pliska. Here is an overview: https://www.fields.utoronto.ca/programs/scientific/09-10/finance/courses/pliska2.pdf

• Thank you very much! So i "work" on propability space $(\Omega,F,P)$ ,where $\Omega$ is the sample space , F is σ-algebra and P propability measure and there is no problem to not change the propability measure if i don't want to use Girsanov's theorem to my project? But $S_t$ will not be a martingale,right? – Maria Oikon May 18 '20 at 20:24
• Well the discounted price process of $S$ is only a martingale under the risk-neutral measure in any case. If you don’t go via the change of measure route (Girsanov) then you don’t need to use the martingale property. But it depends on what exactly you need to do. If you need to derive the BS formula then it isn’t necessary and you can follow what they did in the paper (although personally I find the measure route much neater). There are other topics that can’t be worked on without the arbitrage pricing framework (martingales and measures) though. – Ivan May 19 '20 at 16:29
• Thank you very much. You helped me a lot! – Maria Oikon May 20 '20 at 11:43