proving equalities in stochastic calculus

I am struggling with this question:

FIRST PART (almost done, but stuck somewhere):

Let $Z$~$N(0,1)$ be a standard normal random variable, and define a function $F$ by the formula \begin{equation} F(v,m) = \mathbb{E} \bigg[ \Big( \exp\big\{-\frac{v}{2} + \sqrt{v} Z\big\} -m \Big)^{+} \bigg]. \end{equation} Let $W$ and $Z$ be independent Brownian motions, let $r$, $\sigma_0$, $\rho$ and $S_0$ be real constants with $S_0 >0$ and $-1 \leq \rho \leq 1$ and let $a$ and $b$ be smooth functions. Suppose that \begin{equation} dS_t = S_t \Big(r dt + \sigma_t \big[ \rho dW_t + \sqrt{1- \rho^2} dZ_t \big] \Big), \quad d \sigma_t = a(\sigma_t)dt + b( \sigma_t) dW_t. \end{equation} Also, assume that the process $\sigma = (\sigma_t)_{t \geq 0}$ is bounded and adapted to the filtration $\{ \mathcal{F}_t \}$ generated by $W$, by conditioning on $\{ \mathcal{F}_t \}$, show that \begin{equation} \mathbb{E} ( e^{-rT} (S_T - K)^{+} ) = \mathbb{E} \Bigg\{ S_0 \epsilon_T F \bigg( \int_{0}^{T} (1- \rho^2 ) \sigma^2_t dt, \, Ke^{-rT}/(S_0 \epsilon_T) \, \bigg) \, \Bigg\}, \end{equation} where $\epsilon_T$ is defined by \begin{equation} \epsilon_T= \exp \bigg( -\frac{1}{2} \int_0^T \rho^2 \sigma^2_t dt + \int_0^T \rho \sigma_t dW_t \bigg). \end{equation}

$\mathbf{\text{What I obtained so far (applying Ito's formula and did some algebra)}}:$ \begin{eqnarray} &&\mathbb{E} ( e^{-rT} (S_T - K)^{+} ) \\ & = & \mathbb{E} \Bigg[ S_0 \epsilon_T \mathbb{E} \Bigg\{ \bigg( \exp \bigg[ -\frac{1}{2} \int_0^T (1-\rho^2) \sigma^2_t \,dt + \int_0^T \sqrt{1- \rho^2} \sigma_t dZ_t \bigg] - \frac{K e^{-rT}}{S_0 \epsilon_T} \bigg)^{+} \Bigg| \mathcal{F}_T \Bigg\} \Bigg]. \end{eqnarray} The only difference between my expression and the answer is the term $\int_0^T \sqrt{1- \rho^2} \sigma_t dZ_t$. Does that have something to do with the conditional expectation?

SECOND PART (don't really have a hint):

Show that \begin{equation} \mathbb{E} ( e^{-rT} (S_T - K)^{+} ) = \mathbb{E} \Bigg\{ S_0 F \bigg( \int_{0}^{T} (1- \rho^2 ) \hat{\sigma}^2_t dt, \, Ke^{-rT} \hat{\epsilon}_T/S_0 \, \bigg) \, \Bigg\}, \end{equation} where \begin{equation} d \hat{\sigma}_t= ( a(\hat{\sigma}_t) + \rho \hat{\sigma}_t b(\hat{\sigma}_t)) dt + b( \hat{\sigma}_t) d\hat{W}_t, \quad \hat{\sigma}_0 = \sigma_0 \end{equation} and \begin{equation} \hat{\epsilon}_T= \exp \bigg( -\frac{1}{2} \int_0^T \rho^2 \hat{\sigma}^2_t dt - \int_0^T \rho \hat{\sigma}_t d\hat{W}_t \bigg), \end{equation} where $\hat{W}$ is a Brownian motion.

$\mathbf{\text{I only know that the Girsanov's theorem should be used as it is in that form but }}$ $\mathbf{\text{don't know where to start. Any suggestions?}}$

• This involves changing measure? – user198044 Dec 4 '15 at 5:12