I want to apply the Girsanov theorem for change of measure for geometric Brownian motion under real world measure $\mathbb{P}$ to risk-neutral probability measure $\mathbb{Q}$ where the drift is given by stochastic interest rate modelled by CIR process, i.e. to show
$dS(t) = S(t) \cdot (\mu dt + \sigma_S dW_S^{\mathbb{P}}(t)) \iff dS(t) = S(t) \cdot (r(t) dt + \sigma_S dW_S^{\mathbb{Q}}(t))$
where
$dr(t) = \kappa_r \cdot (\theta_r - r(t))dt + \sigma_r \cdot \sqrt[]{r(t)}dW_r^{\mathbb{P}} (t) \:$ and $\: W_S^{\mathbb{P}}(t) = \rho_{S,\,r} W_r^{\mathbb{P}}(t) + \sqrt{1-\rho_{S,\,r}^2} W_{Z_1}^{\mathbb{P}}(t)$.
$W_r^{\mathbb{P}}(t)$ and $W_{Z_1}^{\mathbb{P}}(t)$ are independent Brownian motions. Correlation factor fulfills $|\rho_{S,\,r}|\leq 1$.
For this I need to show that the Novikov condition is satisfied for
$\gamma_S(t)=\dfrac{\mu_S-r(t)}{\sigma_S} $
With help of Can I apply the Girsanov theorem to an Ornstein-Uhlenbeck process? I already showed that
$\mathbb{E} \left[ \exp \left (\dfrac{1}{2}\int_s^{s+\varepsilon} \gamma_S^2(t) dt \right) \right] \leq \dfrac{1}{\varepsilon} \int_s^{s+\varepsilon} \mathbb{E} \left[ \exp \left (\dfrac{\varepsilon}{2} \gamma_S^2(t) \right) \right] dt $
Unfortuneately, I have no idea how to find the upper bound for the expectation under the integral in case of CIR process, especially non-central chi squared distribution. Is it possible at all? Any help appreciated!
I found the solution by applying Theorem 1 of Cheridito, Filipovic and Kimmel (2007) from the paper "Market price of risk specications for affine models: Theory and evidence".
With notation as above we consider the process
\begin{align*} M_T = \exp (-\int_0^T \gamma_S(s)dW_S^{\mathbb{P}}(s) - \dfrac{1}{2} \int_0^T \gamma_S^2(s)ds) \, , \, t \in [0,\,T] \end{align*}
which is a well-defined positive local martingale with respect to $\mathbb{P}$, and thus also a $\mathbb{P}$-supermartingale. If we show that it is a $\mathbb{P}$-martingale then it defines the Radon-Nikodym derivative of $\mathbb{Q}$ with respect to $\mathbb{P}$ and we can apply the Girsanov theorem. For this we need
\begin{equation*} \mathbb{E}[M_T]=1 \: . \end{equation*}
One of difficulties we face is that during the measure change also the dynamics of interest rate process change (details are discussed in the paper). Therefore we consider a continuous, well-defined process $\tilde{r}(t)$ on the probability space $(\Omega, \mathbb{P}, \mathbb{F})$ whose dynamics and distribution are the same as those of $r(t)$ under $\mathbb{Q}$. Its existence and uniqueness follow from above mentioned theorem. The following part of the proof is based on its (c) part. We write \begin{equation*} \tilde{\gamma}_S(t) = \dfrac{\mu_S - \tilde{r}(t)}{\sigma_S} \, , \, t\geq0 \: . \end{equation*} Now, for each $n \geq 1$ we define two stopping times
\begin{equation*} \tilde{\tau}_n = \inf \{t > 0 \, | \, \tilde{\gamma}_S(t) \geq n\} \wedge T \end{equation*}
and
\begin{equation*} \tau_n = \inf \{t > 0 \, | \, \gamma_S(t) \geq n\} \wedge T \: . \end{equation*}
Because both measure shifts are continuous processes that do not explode with time, the infimum in the definition of stopping times is taken since some $n_M$ over the empty set. We have $\inf \{\emptyset\} \wedge T = \infty \wedge T = T$ and it holds (Limit 1.) \begin{equation}~\label{eq:stoppingConvergence} \underset{n \rightarrow \infty}{\lim} \mathbb{P} (\tilde{\tau}_n = T) = \underset{n \rightarrow \infty}{\lim} \mathbb{P} (\tau_n = T) = 1 \: . \end{equation} Further we define the sequence of truncated measure shifts
\begin{equation*} \gamma_S^n(t) = \gamma_S(t)1_{\{t\leq \tau_n \} } \, , \, t \in [0,\,T] ,\, n \geq 1 \end{equation*}
By the definition of $\tau_n$ the integral $\int_0^t (\gamma_S^n(s))^2 ds$ is bounded by $n^2t$, hence our truncated measure shift satisfies the Novikov condition
\begin{align*} \mathbb{E}[\exp ( \dfrac{1}{2}\int_0^T (\gamma_S^n(t))^2 dt )] \leq \exp (\dfrac{n^2T}{2}) < \infty \: \end{align*} and for each $n\geq 1$ the process \begin{equation*} M_T^n = \exp ( \int_0^T \gamma_S^n(s) dW_S^{\mathbb{P}} (s) - \dfrac{1}{2} \int_0^T ( \gamma_S^n(s))^2ds) \, , \, t \in [0,\,T] \end{equation*} is a martingale with respect to $\mathbb{P}$. Furthermore using Limit 1., we get
\begin{equation*} M_T^n 1_{\{t\leq \tau_n \} } = M_T 1_{\{t\leq \tau_n \} } \underset{n \rightarrow \infty}{\longrightarrow}M_T \: \text{a.s.} \end{equation*}
Now we construct a probability measure $\mathbb{Q}^n = M^n_T\cdot\mathbb{P}$ equivalent to $\mathbb{P}$. According to Girsanov theorem it holds
\begin{equation*} W_S^n(t) = \int_0^t \gamma_S^n(s) ds + W_S^{\mathbb{P}}(t) , \, t \in [0,\,T] \end{equation*}
and $W_S^n(t)$ is a Brownian motion under $\mathbb{Q}^n$. Using again above mentioned theorem, we can deduce that under $\mathbb{Q}^n$ the stopped process $(r(t \wedge \tau_n))_{t\geq 0}$ has the same distribution as the stopped process $(\tilde{r}(t \wedge \tilde{\tau}_n))_{t\geq 0}$ under $\mathbb{P}$. This allows us to arrive at the main result
\begin{align*} \mathbb{E}[M_T]= \underset{n \rightarrow \infty}{\lim}\mathbb{E}[ M_T^n 1_{\{\tau_n = T \} }] = \underset{n \rightarrow \infty}{\lim} \mathbb{Q}^n( \tau_n = T ) = \underset{n \rightarrow \infty}{\lim} \mathbb{P}( \tilde{\tau}_n = T ) = 1 \: . \end{align*}
The first equation follows from Monotone convergence theorem. The second step uses the definition of probability measure $\mathbb{Q}^n$. The next one is possible because the distributions of $(r(t \wedge \tau_n))_{t\geq 0}$ and $(\tilde{r}(t \wedge \tilde{\tau}_n))_{t\geq 0}$ are the same under $\mathbb{Q}^n$ and $\mathbb{P}$, respectively. The last equation follows from the Limit 1.