I want to understand the logic for why this is:

We have our model for the stock price behaviour:

$$d{S_t} = \mu {S_t}dt + \sigma {S_t}d{\tilde W_t}$$

It describes the development of a stock price over time using the risk-adjusted expected return $\mu$ and the real uncertainty in the stochastic term. We want to change the probability measure in such a way that the stochastic process remains a Brownian motion but with a drift of r instead of $\mu$. .... To repeat the manner of speaking, we want the process to change gear from an instantaneous increase of $\mu$ to r and leave the rest as before.

$$d{S_t} = r{S_t}dt + \sigma {S_t}d{\tilde W_t}$$


they change the measure to get the arbitrage free price. Check the link below, or have a look at Hull for an easy introduction.


  • $\begingroup$ can you not get the arbitrage free price while using $\mu$? $\endgroup$ – k b Apr 21 '17 at 20:58
  • $\begingroup$ @kb - the point is that a perfectly hedged option where the underlying follows the Black Scholes process would be risk free. If an asset is risk free, then it cannot earn any amount different from the risk free rate $r$. $\endgroup$ – FinanceGuyThatCantCode Apr 21 '17 at 21:23
  • $\begingroup$ risk-adjusted expected return $\mu$ is not the same as r ? $\endgroup$ – k b Apr 21 '17 at 21:27
  • $\begingroup$ mu could be any number, does not have to be the same as r. In order to get the drift to equal r, we change the measure. $\endgroup$ – mbison Apr 21 '17 at 21:50
  • $\begingroup$ @k b: who said that $\mu$ is "risk adjusted"? $\mu$ is just the expected return. $\endgroup$ – noob2 Apr 21 '17 at 22:14

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