2
$\begingroup$

I'm trying to figure out how to find the correct equivalent martingale measure to change into. First of, since I am on mobile and find it hard to write LaTeX here, I will refer to Wikipedia's version of Girsanov's theorem.

Referring to the standard example of call options in the Black-Scholes model, for which I use standard notation for the discounted underlying $dS_t = (\mu -r)S_{t} dt + \sigma S_{t} dW_t$, the process chosen to be $X_t$ is $dX_t = \frac{\mu - r}{\sigma} dW_t $. However, this is just taking the deterministic part of the discounted GBM after putting $S_t$ of the right side in evidence and changing it to be a scaled Brownian motion instead of a predictable process. I do not understand the reasoning that leads to this, and I would like to, so that I could be able to apply similar reasoning to other models by myself. The way it is explained in the books I've read is just the usual "now, if we consider..." followed by the usual "it works!". I would like to see the logic behind this choice, since my gut would tell me to use the deterministic part in some way, but not to apply it as a Brownian motion. I suspect that this is done just to make it so that you get a non zero quadratic variation when you need to compute it, but I still don't know why they chose that particular predictable process.

If this turns out to be an ill-posed question I will try to reformat it when I am able to get back to my laptop.

$\endgroup$
  • $\begingroup$ The argument at the heart of "arbitrage-free pricing theory" is that to compute prices of derivatives under the real world measure $\Bbb{P}$ (say for an underlying with dynamics $dS_t/S_t = \mu dt + \sigma dW_t^\Bbb{P} (1)$), one can compute expectations of their future payout under an equivalent measure $\Bbb{Q}$ where discounted asset prices are martingales (more generally, the values of self-financing strategies expressed under the money market numéraire ought to be martingales). This motivates looking for a change of measure which turns $(1)$ into $dS_t/S_t = r dt + \sigma dW_t^\Bbb{Q}$. $\endgroup$ – Quantuple Jan 15 '18 at 8:37
  • $\begingroup$ ... since in that case indeed $d(e^{-rt} S_t) = \sigma dW_t^\Bbb{Q}$ which is a martingale as required under the measure $\Bbb{Q}$ (assuming no dividends). Hopefully this will help you understand the basic context behind @Lipton's answer. $\endgroup$ – Quantuple Jan 15 '18 at 8:41
1
$\begingroup$

The way I see it is follows.

  1. We want something that we can compute distribution easily and for most of people it needs to be as simple as $W_t$ or simple functions of it...So we want something like this: $$dS_t = \sigma S_t d\tilde{W}_t$$ instead of the original $dS_t=(\mu-r)S_tdt+\sigma S_tdW_t$
  2. Now we can solve for $d\tilde{W}_t$: $$\sigma S_t d\tilde{W}_t = (\mu-r)S_tdt+\sigma S_tdW_t\Rightarrow d\tilde{W}_t=\frac{(\mu-r)}{\sigma}dt +dW_t$$
  3. This is where we use Girsonov to find the Radon-Nikodym derivative of the new probability measure (in which $\tilde{W}_t$ is a Brownian motion) with respect to the old probability measure (in which $W_t$ is a Brownian motion).

    The theorem basically says that given any $d\tilde{W}_t=\theta_tdt +dW_t$, the Radon-Nikodym derivative process $Z_t$ is $dZ_t=-\theta_tZ_tdW_t$. In the example you gave, it is $-\frac{(\mu-r)}{\sigma}Z_tdW_t$ (and of course you can forget about the negative sign by definition of $W_t$).

$\endgroup$
  • $\begingroup$ Thank you, exactly what I was looking for. I forgot the supermartingale exponential was just a solution of your $dZ_t$ equation in (3)... $\endgroup$ – iamconfused Jan 14 '18 at 20:26
  • $\begingroup$ Sorry, but I was rereading your answer, and the process you write doesn't coincide with mine. There is no $S_t$ in the one I wrote. That is one of the things that was confusing me, and led me to ask this question. $\endgroup$ – iamconfused Jan 14 '18 at 20:44
  • $\begingroup$ Either way, disregarding that last part in your post, everything fits for me: since I wanted a P-local martingale $M_t$ such that $d[M,W^{P}]_{t} = \frac{\mu - r}{\sigma}$, it obviously turns out to be $M_t = \frac{\mu - r}{\sigma} dW^{P}_{t}$. $\endgroup$ – iamconfused Jan 14 '18 at 21:43
  • $\begingroup$ I made a typo in the answer, it should be $Z_t$ rather than $S_t$. So i think it should be something like $\frac{dZ_t}{Z_t} = \frac{\mu - r}{\sigma}dW_t$. Not sure what $X_t$ means in your question. $\endgroup$ – Lipton Jan 14 '18 at 22:05
  • $\begingroup$ It means the argument of the stochastic exponential, the right hand side of that SDE you just wrote. I was using Wikipedia's notation, from the link I posted. $\endgroup$ – iamconfused Jan 14 '18 at 22:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.