Having financial market with safe rate r and risky asset S with dynamics under physical measure P $$\frac{dS_t}{S_t}=\mu dt +\sigma dW_t$$ what is the log-stock price?

Using Ito formula it is straightforward to derive the below equation $$log(S_T)=log(S_t) + (\mu - \frac{\sigma^2}{2})(T-t) + \sigma \ (W_T - W_t) \tag{1}$$

what should be equivalent to $$log(S_T)=log(S_t) + (r - \frac{\sigma^2}{2})(T-t) + \sigma \ (W_T^* - W_t^*) \tag{2}$$

what allows to formally transition (1) into (2)?
I mean the change of dt into T-t and $\mu$ into r

  • 1
    $\begingroup$ Should (1) be $\ln S_T = \ln S_t + (\mu-\sigma^2/2)(T-t) + \sigma (W_T-W_t)$? Similarly for (2). $\endgroup$
    – Gordon
    Commented May 18, 2016 at 19:02
  • $\begingroup$ I corrected the equations (1) and (2) $\endgroup$
    – Michal
    Commented May 18, 2016 at 20:23

1 Answer 1


The dynamics \begin{align*} \frac{dS_t}{S_t} =\mu dt + \sigma dW_t. \end{align*} is under the real-world measure $\mathbb{P}$. Then, \begin{align*} d\ln S_t =\Big(\mu-\frac{1}{2}\sigma^2 \Big) dt + \sigma dW_t. \end{align*} Therefore, \begin{align*} \ln S_T = \ln S_t + \Big(\mu-\frac{1}{2}\sigma^2 \Big)(T-t) + \sigma \big(W_T-W_t\big).\tag{1} \end{align*} To obtain the dynamics under the risk-neutral probability measure $\mathbb{Q}$, we employ the Radon-Nikodym derivative \begin{align*} \frac{d\mathbb{Q}}{d\mathbb{P}}\big|_{\mathcal{F}_t} = \exp\left(-\frac{1}{2}\lambda^2 t + \lambda W_t \right), \end{align*} where $\lambda = (r-\mu)/\sigma$ is the market-risk premium. Then, from Girsanov theorem, the process $\{\widehat{W}_t, t \ge 0\}$, where $$\widehat{W}_t = W_t -\lambda t,$$ is a standard Brownian motion under measure $\mathbb{Q}$. Moreover, under measure $\mathbb{Q}$, \begin{align*} \frac{dS_t}{S_t} &=\mu dt + \sigma dW_t\\ &=rdt + \sigma d\widehat{W}_t. \end{align*} Consequently, similar to $(1)$ above, \begin{align*} \ln S_T = \ln S_t + \Big(r-\frac{1}{2}\sigma^2 \Big)(T-t) + \sigma \left(\widehat{W}_T-\widehat{W}_t\right).\tag{2} \end{align*} Note that, in $(1)$ and $(2)$, the Brownian motions $W$ and $\widehat{W}$ are different.


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