given $S^1$ satifying the SDE $\quad dS_{t}^{1}=S_{t}^{1}((r+\mu)dt + \sigma dW_t), \quad S_{0}^{1}=1 $
and the safe asset $S_{t}^{0}$ $\quad S_{t}^{0}:=e^{rt} \quad for \quad r\geq 0$

how to show that $\quad Y_t:=log(S_{t}^{1})$
satisfies $\quad dY_t=(r+\mu-\sigma^2/2)dt+\sigma d W_t \quad Y=0$,

how to find a measure Q equivalent to P (using Girsanov Theorem) such that
$dS_{t}^{1}=S_{t}^{1}(rt+\sigma d W_{t}^{*})$

I tried the fist part, is the derivation correct?

$\frac{ d S_{t}^{1}}{S_{t}^{1}}=(r+\mu)dt+\sigma dW_t$
$dY=d log(S_{t}^{1})$
by Ito
$d log(S_{t}^{1})=\ \frac{ d S_{t}^{1}}{S_{t}^{1}} + \frac{1}{2}(-\frac{1}{(S_{t}^{1})^2})(d S_{t}^{1})^2)=$ $=\ \frac{ d S_{t}^{1}}{S_{t}^{1}} + (- \frac{1}{2} \frac{(\sigma S_{t}^{1})^2)dt}{(S_{t}^{1})^2 }) = (r+\mu)dt + \sigma dW_t + (- \frac{1}{2} \sigma^2 dt) =(r+\mu-\sigma^2/2)dt+\sigma dW_t$

I am struggling with the measure change, could anybody help and explain the idea and the next steps?

  • 1
    $\begingroup$ I think this doesn't belong to the site because it's too basic for quantitative finance and any QF book would cover in details. $\endgroup$
    – SmallChess
    Commented Jan 11, 2016 at 5:40
  • $\begingroup$ topic and difficulty are two different matters, right? you were pointing some Quant Finance books as a source, could you please explain then how this question can be off-topic??? $\endgroup$
    – Michal
    Commented Jan 11, 2016 at 11:29
  • $\begingroup$ Why $\frac{ d S_{t}^{1}}{S_{t}^{1}} + (- \frac{1}{2} \frac{(\sigma S_{t}^{1})^2)}{(S_{t}^{1})^2}) = (r+\mu)dt + \sigma dW_t -(\frac{1}{2} (- \frac{1}{2} \sigma^2 ))$? $\endgroup$
    – Gordon
    Commented Jan 11, 2016 at 14:44
  • $\begingroup$ a typo, I dropped the dt there either, I corrected the equation $\endgroup$
    – Michal
    Commented Jan 11, 2016 at 15:25
  • 1
    $\begingroup$ I think it's borderline but the question has been improved. @Michal which resources are you using to learn/solve this? That would be a further and needed improvement $\endgroup$
    – Bob Jansen
    Commented Jan 11, 2016 at 17:54

2 Answers 2


For a time interval $[0,T]$, Girsanov theorem states that given a process $\lambda$ such that process $U$, defined by $$dU_t = -\lambda_tU_tdW_t, \; U_0=1,$$ is a $P$-martingale, then one can define a new measure $Q$ equivalent to $P$ by $$\frac{dQ}{dP} = U_T,$$ and a standard Brownian motion under $Q$, $W^\star$, by $$ dW^\star_t = dW_t + \lambda_tdt.$$ In your case, if we take $$ \lambda_t = \mu/\sigma \; \forall t \in [0,T],$$ then $U$ is indeed $P$-martingale (no drift) and $W^\star$ defined by $$ dW^\star_t = dW_t + \mu/\sigma dt$$ is standard Brownian motion under $Q$.

We can now re-write $S^1$ as follows (no Ito): $$ dS^1_t = (r+\mu)S^1_tdt + \sigma S_t^1 dW_t $$ $$ = rS^1_tdt + \sigma S^1_tdW^\star_t. $$

Finally, note that $Q$ is an interesting measure, a so-called EMM (equivalent martingale measure) with numeraire $S^0$, as it is equivalent to $P$ and $S^1/S^0$ (deflated $S^1$) is a $Q$-martingale. Indeed, using Ito-Leibniz, we see that $S^1/S^0$ has no drift under $Q$:

$$ d(S^1_t/S^0_t) = \sigma S^1_t/S^0_t dW^\star_t. $$


For Q2, let $\lambda = \mu/\sigma$. Moreover, we define the measure $Q$ on $(\Omega, \mathcal{F})$ such that \begin{align*} \frac{dQ}{dP}\big|_{\mathcal{F}_t} = \exp\Big(-\frac{1}{2}\lambda^2 t - \lambda W_t\Big), \mbox{ for } t \ge 0. \end{align*} Then, by Girsanov theorem, $W^*$, where \begin{align*} W_t^* = \lambda t + W_t, \end{align*} is a standard Brownian motion under the measure $Q$. Furthermore, under $Q$, \begin{align*} dS_t^1 &= S_t^1\big[(r+\mu)dt + \sigma dW_t \big]\\ &= S_t^1(rdt + \sigma dW_t^*). \end{align*}


Your Answer

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

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