# Ito calculus problem

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 $S_{t}^{0}$ $\quad S_{t}^{0}:=e^{rt} \quad for \quad r\geq 0$

Q1.
how to show that $\quad Y_t:=log(S_{t}^{1})$
satisfies $\quad dY_t=(r+\mu-\sigma^2/2)dt+\sigma d W_t \quad Y=0$,

Q2.
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?

• 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. – SmallChess Jan 11 '16 at 5:40
• 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??? – Michal Jan 11 '16 at 11:29
• Why $\frac{ d S_{t}^{1}}{S_{t}^{1}} + (- \frac{1}{2} \frac{(\sigma S_{t}^{1})^2)}{(S_{t}^{1})^2}) = (r+\mu)dt + \sigma dW_t -(\frac{1}{2} (- \frac{1}{2} \sigma^2 ))$? – Gordon Jan 11 '16 at 14:44
• a typo, I dropped the dt there either, I corrected the equation – Michal Jan 11 '16 at 15:25
• 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 – Bob Jansen Jan 11 '16 at 17:54

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*}