Girsanov THM and Radon-Nikodym derivative I've been having a hard time to applicate Girsanov theorem with Radon-Nikodym derivative in the demonstration of German-El Karoui-Rochet formule.
I know that $\Pi_0:=S_0\mathbb{Q}^S(S_T\geq K)-K\mathbb{Q}^T(S_T\geq K)P(0,T)$. I have to calculate these two probabilities. I start from the second assuming that the procedure is the same for the first.
Let $Z_t:=\frac{S_t}{P(t,T)}$ a process under physical measure $\mathbb{P}$ and dynamics $dZ_t:=Z_t[u_t^zdt+\sigma_t^zdW_t]$. Since the process is $(\Omega,F,\mathbb{P})$-definited while the probability to calculate is $\mathbb{Q}^T$-definited, i have to apply a change of measure with Girsanov theorem. I know that Girsanov allows me to construct (through Radon-Nikodym derivative expressed in terms of exponential martingale: $L(\omega):=\frac{d\mathbb{Q}^T}{d\mathbb{P}}:=M_t\Rightarrow E^{\mathbb{Q}^T}[X]=E^{\mathbb{P}}[M_tX]$) a martingale measure $\mathbb{Q}^T$ equivalent to the physical measure in such a way that the process $\tilde{W_t}$ under the new measure is a Brownian Motion standard. 
Let $P(T,T):=1$, i start saying that:
$\mathbb{Q}^T(S_T\geq K)=\mathbb{Q}^T(\frac{S_T}{1}\geq K)=\mathbb{Q}^T(\frac{S_T}{P(T,T)}\geq K)=\mathbb{Q}^T(Z_T\geq K)$
Now the problems.
Given that to do the change of measure it's true that $E^{\mathbb{Q}^T}[Z_T]=E^{\mathbb{P}}[M_tZ_T]$, i thought that the $\mathbb{Q}^T$-martingale resulting should have, for definition, a dynamics contains only diffusive part. Formerly: $d\mathbb{Z}_t^{\mathbb{Q}^T}=\sigma_t^zd\tilde{W}_t$ . However the text (very incomplete, actually) says the dynamics of $Z_t$ under $\mathbb{Q}^T$ measure is $d\mathbb{Z}_t^{\mathbb{Q}^T}=Z_t\sigma_t^zd\tilde{W}_t$. 
I thought to applicate the Fundamental Theorem of change of numeraire saying that, since $P(t,T)$ fulfils the conditions of theorem (always assumes values strictly positive and is a $\mathbb{Q}$-martingale for the First Fundamental Theorem of APT), for the change of numeraire $Z_t:=\frac{S_t}{P(t,T)}$ is a $\mathbb{Q}^T$-martingale. But this would contradict the hypothesis in which the process is $\mathbb{P}$-definited.
I tried to apply the Ito's formule to $Z_T^{\mathbb{Q}^T}=\left ( Z_Te^{\int_{0}^{T}\sigma_s^zdW_s-\frac{1}{2}\int_{0}^{T}(\sigma_s^z)^2ds} \right )^{\mathbb{P}}$ (in according to the change of measure) but i dont'understand why the text fixed the stochastic integral $dW_s$ under $\mathbb{Q}^T$: this contradict not only the change of measure with Radon-Nikodym measure but the definition of exponential martingale, that is a process $\mathbb{P}$-definited. 
Anyway, I can't derive the result that is $\mathbb{Q}^T(S_T\geq K)=\mathbb{Q}^T(-\tilde{Y}\leq \frac{ln(\frac{S_0}{KP(0,T)})-\frac{1}{2}\sum ^2}{\sqrt{\sum ^2}})$, where:


*

*$\sum ^2=\int_{0}^{T}(\sigma_s^z)^2ds$;

*$\tilde{Y}$ is the standardization of $Y:=e^{\int_{0}^{T}\sigma_s^zdW_s-\frac{1}{2}\int_{0}^{T}(\sigma_s^z)^2ds}$ (with $dW_s$ under $\mathbb{Q}^T$).


Any help would be really welcome. Thanks!
 A: The definition of $Z$ tell us that the process $\lbrace{Z_t\rbrace}_{t\geq0}$ is a martingale under the $T-$forward measure $\mathbb{Q}^T$. Given the dynamic of $Z$, we can find out the Radon-Nikodym derivative.
\begin{align*}
 dZ_t &= Z_t\left(u_t^zdt + \sigma_t^zdW_t \right) \\
  &= Z_t\sigma_t^z\left(dW_t + \frac{u_t^z}{\sigma_t^z}dt \right) \\
  &= Z_t\sigma_t^zd\bar{W}_t \quad \quad (1)
\end{align*}
where $\bar{W}$ is a $\mathbb{Q}^T$ Brownian motion. Therefore, we can construct the R-N measure which is 
\begin{equation*}
 M_t = \exp\left(-\frac12\int_0^t \left(\frac{u_s^z}{\sigma_s^z}\right)^2ds - \int_0^t \frac{u_s^z}{\sigma_s^z}dW_s\right)
\end{equation*}
Now we have all the ingredients to compute $\mathbb{Q}^T(S_T \geq K)$.
\begin{align*}
 \mathbb{Q}^T(S_T \geq K) &= \mathbb{Q}^T(Z_T \geq K) \\
 &=\mathbb{Q}^T\left(Z_0\exp\left(-\frac12\int_0^T(\sigma_s^z)^2ds + \int_0^T \sigma_s^zdW_s\right) \geq K\right)\\
 &=\mathbb{Q}^T\left(\int_0^T \sigma_s^zd\bar{W}_s \geq \log\left(\frac{K}{Z_0}\right) + \frac12\Sigma^2\right) \\
 &=\mathbb{Q}^T\left(\tilde{Y} \geq \frac{\log\left(\frac{K}{Z_0}\right) + \frac12\Sigma^2}{\Sigma}\right) \\
 &=\mathbb{Q}^T\left(-\tilde{Y} \leq \frac{\log\left(\frac{S_0}{KP(0,T)}\right) - \frac12\Sigma^2}{\Sigma}\right) 
\end{align*}
The fourth equality follows from the fact that $\lbrace{\int_0^t \sigma_s^zd\bar{W}_s\rbrace}_{t\geq0}$ is a Wiener process under the measure $\mathbb{Q}^T$. Hence, it follows a gaussian distribution with mean $0$ and variance $\Sigma$.
Some observations on your post :


*

*$Z$ is a lognormal process under $\mathbb{P}$ as well as under $\mathbb{Q}^T$. Hence, one cannot have $dZ_t = \sigma_t^zd\bar{W}_t$ (which is btw a normal process). That being said, you are right that under $\mathbb{Q}^T$ the process $Z$ is a real martingale and should only have the diffusive part (no finite variation process) and it is case in (1).

*Often a numeraire, say $Y$, is chosen for a given process, say $X$, such that $X$ becomes a martingale under a new measure called the martingale measure associated to the numeraire $Y$. As noted, $Y$ has to fulfill some conditions to be a numeraire. 

