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

Question

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.

Answer 1

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