question about a changement of probability Let $(\Omega, F, F_n, \mathbb{P})$ a filtered probability space.
Let $v_n=\mu_n+\sigma_n\varepsilon_n$ with $\mu_n ,\sigma_n$ are $F_{n-1}$ measurable and $\varepsilon_n$ ~ $N(0,1)$ are i.i.d 
Let ${Z_n} = \exp \left( { - \sum\limits_{k = 1}^n {\frac{{{\mu _k}}}{{{\sigma _k}}}{\varepsilon _k}}  - \frac{1}{2}\sum\limits_{k = 1}^n {\frac{{\mu _k^2}}{{\sigma _k^2}}} } \right)$.
I've already proved that $Z_n$ is a uniformly integrable martingale and $Z_n \xrightarrow{{ps}} Z_\infty$ with $Z_\infty = \exp \left( { - \sum\limits_{k = 1}^{+\infty} {\frac{{{\mu _k}}}{{{\sigma _k}}}{\varepsilon _k}}  - \frac{1}{2}\sum\limits_{k = 1}^{+\infty} {\frac{{\mu _k^2}}{{\sigma _k^2}}} } \right)$
Now we define $\mathbb{Q}=\mathbb{E}_{\mathbb{P}}(1_{A}Z_\infty)$ $\forall A\in F$.
I've already proved that under $\mathbb{P}$: 
$\mathbb{E}(v_n|F_{n-1})=\mu_n$ and $Var(v_n|F_{n-1})=\sigma_n^2$
I have to show that under the probability $\mathbb{Q}$:  
$\mathbb{E}(v_n|F_{n-1})=0$ and $Var(v_n|F_{n-1})=\sigma_n^2$
Some help would be appreciated
 A: Hint: There is an analogous change of measure formula for conditional expectations: If $G$ is a bounded $\mathcal F_n$-measurable random variable, then
$$
\Bbb E^{\Bbb Q}[G|\mathcal F_{n-1}]={\Bbb E[Z_nG|\mathcal F_{n-1}]\over  Z_{n-1}}.
$$
Apply this with $G=\exp(ib\nu_n)$, $b\in\Bbb R$, to see that, under $\Bbb Q$, the conditional distribution of $\nu_n$ is normal with the appropriate (conditional) mean and variance.
ADDED: Notice that $Z_n=Z_{n-1}\exp(-(\mu_n/\sigma_n)\epsilon_n-(\mu_n^2/2\sigma^2_n))$, and $\nu_n=\nu_{n-1}+\mu_n+\sigma_n\epsilon_n$. Therefore, because $\epsilon_n$ is independent of $\mathcal F_{n-1}$,
$$
{\Bbb E[Z_n\nu_n|\mathcal F_{n-1}]\over Z_n}=\mu_n+\sigma_ne^{-(\mu_n^2/2\sigma^2_n)}\Bbb E[\epsilon_n\exp(-(\mu_n/\sigma_n)\epsilon_n)|\mathcal F_{n-1}].
$$
Because  $\epsilon_n$ is standard normal and independent of $\mathcal F_{n-1}$ (while $\mu_n$, $\sigma_n$ are $\mathcal F_{n-1}$-measurable), the conditional expecation on the right is equal to (writing $\beta$ for $\mu_n/\sigma_n$)
$$
\int_{-\infty}^\infty xe^{-\beta x-x^2/2}\,{dx\over\sqrt{2\pi}}
=e^{\beta^2/2}\int_{-\infty}^\infty xe^{-(x+\beta)^2/2}\,{dx\over\sqrt{2\pi}}=-\beta e^{\beta^2/2}.
$$
Using this calculation in the previous display one gets 
$$
{\Bbb E[Z_n\nu_n|\mathcal F_{n-1}]\over Z_n}=\mu_n-\sigma_ne^{-(\mu_n^2/2\sigma^2_n)}(\mu_n/\sigma_n)e^{\mu_n^2/2\sigma^2_n}=0.
$$
