When Björk presents the Black-Scholes model and martingale measures he starts off with a process modeling the stock price calling it $S$ with some given dynamics w.r.t some measure $P$.

Then he demonstrates that the price of any contract should satisfy the Black and Scholes PDE. Then he solves this PDE using Feynman-Kac then a new dynamic $X$ makes it's appearance.

Now for some reason he insist on relabeling this new process $X$ to $S$, what is the point of this relabeling? To me it only looks like he messes things up.

Everthing can be found around page $103$ in his third edition of Arbitrage theory in continuous time


I think everything is related to the concept of Risk Neutral measure $\mathbb{Q}$. In deriving Black- Scholes equation you use the dynamics \begin{equation} dS(t)=\mu S(t)dt + \sigma S dW(t) \end{equation} where $W$ is a brownian motion under the $\mathbb{P}$ measure, and you get the following PDE for the price $f$ of a certain derivative: \begin{equation} rS\frac{\partial f}{\partial S} +\frac{\partial f}{\partial t} + \frac{1}{2}\sigma^2S^2\frac{\partial ^2 f}{\partial S^2} =rf \end{equation} Now the Feynamn- Kac formula says that for a PDE of the form \begin{equation} \begin{cases} \frac{\partial f}{\partial t}(t,x) + \alpha(t,x)\frac{\partial f}{\partial x}(t,x) + \frac{1}{2}\beta^2(t,x)\frac{\partial^2 f}{\partial x^2}(t,x) = k(t)f(t,x) \\ f(T,x) = \Phi(x) \end{cases} \end{equation} you have a solution that can be written as expectation with respect to a particular measure $\mathbb{Q}$ \begin{equation} f(t,x) = \mathbb{E}^{\mathbb{Q}}\left[e^{-\int_t^Tk(s)ds}\Phi(X)\Big|X(t)=x\right] \end{equation} where $X$ has SDE: \begin{equation} dX(t)=\alpha(t,X)dt + \beta(t,X)d\overline{W}(t) \end{equation} where $\overline{W}$ is a $\mathbb{Q}$-brownian motion. Now, in order to match Black-Scholes equation to the general PDE that I wrote, you just put $k(t)=r$, $\alpha(t,X)=rX$,$\beta(t,X)=\sigma X$ and you rename $X$ by $S$. At this point you can forget the original specification of the dynamics of the stock and you procede by calculating the expectation to price your derivative with the new one that is: \begin{equation} dS=rSdt + \sigma Sd\overline{W}(t) \end{equation} If you want to go deeper and investigate the relationship between these two dynamics you can observe that we can change from the original SDE to the new one by changing the brownian motion in this way: \begin{equation} dW(t)=d\overline{W}(t) - \left(\frac{\mu - r}{\sigma}\right)dt \end{equation} Now the Girsanov theorem says that given this relationship between the two brownian motion the measure $\mathbb{P}$ and $\mathbb{Q}$ are equivalent.

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  • $\begingroup$ Thanks for answering. In what sense are they the same process and what does it mean to express a process in terms of a measure? $\endgroup$ – Vlad Sep 23 '19 at 17:10
  • $\begingroup$ quant.stackexchange.com/q/29614/39520 Here, we can see the Grisanov Kernel to conduct the change of measure from dS to dX. $\endgroup$ – SmurfAcco Sep 23 '19 at 19:57
  • $\begingroup$ I wrote a general case of application of Feynman Kac formula so I called the variable by $X$. When you apply this to Black-Scholes you just put $k(t)=\mu(t,X)=r$ and you name $X$ with $S$. Just think of which condition you have to impose in order to match your Black- Scholes equation with the general PDE I wrote. $\endgroup$ – ab94 Sep 23 '19 at 20:24
  • $\begingroup$ isnt the stock price dynamics already given and fixed? and we get some dynamic from Feynman-Kac, we can just change it imo. $\endgroup$ – Vlad Sep 25 '19 at 5:18
  • $\begingroup$ Yes the stock price dynamics is given under the objective (say "real world") measure and fixed. The other dynamics is not real but it is straightforwardly useful for the calculation as you can see. In fact you can calculate the price of your derivative by just taking expectation instead of solving a partial differential equation. If you want an analogous you can think of the change of coordinate system in physics: the new coordinate system does not represent your observation anymore but is really useful in modeling. $\endgroup$ – ab94 Sep 26 '19 at 12:30

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