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?
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 $ , $$ \begin{align*} dS_t^1 &= S_t^1\big[(r+\mu)dt + \sigma dW_t \big]\ &= S_t^1(rdt + \sigma dW_t^). \end{align} $$