3
$\begingroup$

The Black Scholes model assumes the following dynamics for the underlying, well known as the Geometric Brownian Motion: $$dS_t=S_t(\mu dt+\sigma dW_t)$$

Then the solution is given: $$S_t=S_0\,e^{\left(\mu-\frac{\sigma^2}{2}\right)t+\sigma W_t}$$

It can be shown by Ito Lemma on function $f(t,W_t)=\ln S_t$ that this solution is correct as it leads to above dynamics.

But how do we solve the above SDE originally to find this solution?

Guessing the above solution to apply Ito seems unlikely to me.

Improve this question
$\endgroup$
5
  • $\begingroup$ Just renamed the question here because the fact that GBM are used in Black-Scholes does not make it a BS-specific question. $\endgroup$
    – SRKX
    Jul 21, 2014 at 13:15
  • $\begingroup$ @SRKX thanks for noting GBM, but since BS is based on GBM it would be BS-specific (otherwise nothing is BS-specific). $\endgroup$
    – emcor
    Jul 21, 2014 at 14:08
  • $\begingroup$ I completely disagree. What's the point in contesting and doing the re-edit? you'll get your answer anyway. We mods are trying to classify things as well as possible, not to fight with you. I'll leave it like that in the title if you want. $\endgroup$
    – SRKX
    Jul 21, 2014 at 14:21
  • $\begingroup$ @SRKX This is my question while studying Black-Scholes model. You may note that all notation is from Black-Scholes model. You gladly added that its namely a GBM, but that does not mean it had nothing to do with Black-Scholes. The BS-tag would be important for followers or subsequent questions in that category, so I will add it back if the question does not get answered. You may note that pure Math questions are also not allowed on QuantSE. $\endgroup$
    – emcor
    Jul 21, 2014 at 14:38
  • 4
    $\begingroup$ I would not die if it ended in the BS tag. The way you act (i.e. correcting what mods do on organizational purposes) without asking (like you just don't care) isn't elegant. $\endgroup$
    – SRKX
    Jul 21, 2014 at 15:18

1 Answer 1

Reset to default
7
$\begingroup$

If by 'solve' you mean how do we know that $\ln S_t$ is the right change of variable, then you can go by the following (not rigorous) line of thought:

  • Ito's fomula suggests that given an SDE $$dX_t = \mu(X_t,t)dt+\sigma(X_t,t)dW_t$$ and a function $f(x,t)$: the SDE for the process $Y_t=f(X_t,t)$ will satisfy $$dY_t = [f_t(X_t,t) + f_x(X_t,t)\mu(X_t,t) + \frac{1}{2}f_{xx}(X_t,t)\sigma^2(X_t,t)]dt+f_x(X_t,t)\sigma(X_t,t)dW_t$$
  • Now the SDE for the spot price is, as you wrote, given by $$dS_t = \mu S_tdt+\sigma S_t dW_t$$: hence a transformation $f$ applied to this SDE will have dynamics given by $$dY_t = [...]dt+f_x(S_t,t)\sigma S_t dW_t$$
  • We want the transformation to kill the dependence of volatility on the spot, therefore we want to 'solve' something like '$f_x(S_t,t)\sigma S_t = const.$' which essentially means '$f_x(x,t) = \frac{1}{x}$'. This points towards the guess $$f(x,t)=\ln x$$.
Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ Hi Kiwiakos, welcome to quant.SE! Nice answer, I would argue that being fully rigorous is not even necessary as you can show using Ito's lemma that $\ln x$ indeed is a correct solution. $\endgroup$
    – Bob Jansen ♦
    Jul 21, 2014 at 15:07
  • $\begingroup$ could you please explain, why "We want the transformation to kill the dependence of volatility on the spot"? $\endgroup$
    – emcor
    Jul 21, 2014 at 16:00
  • 1
    $\begingroup$ Sure. We want to reduce the SDE to something that has constant drift and volatility, for example if a transformation $Y_t=f(X_t,t)$ has constant mean and volatility, that is to say $dY_t = \mu_Y dt + \sigma_Y dW_t$, then we know that the solution is $Y_t = Y_0 + \mu_Y t + \sigma_Y W_t$. We can then get the solution of $X_t$ by applying the inverse function $X_t = f^{-1}(Y_t,t)$. That's what I mean by 'killing the dependence on spot'. $\endgroup$
    – Kiwiakos
    Jul 21, 2014 at 16:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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