Black scholes model type I want to study the following market:
$$S_1(t)=S_1(t)(\mu_1dt + \sigma_1dW_1(t))$$
$$S_2(t)=S_2(t)(\mu_2dt+\sigma_2dW_2(t))$$
for $t\in [0,T]$, constants $\mu_i,\sigma_i$, initial values $S_i(0)>0$, $i\in\{1,2\}$ and $W_1,W_2$ two Brownian Motion with the following bracket: 
$$\langle W_1,W_2\rangle =\rho t$$ with $\rho\in\mathbb{R} $. We look at the following payoff:
$$(S_1(T)-S_2(T))^+$$
and I want to determine the price at $0$ of an option with the above payoff. It was hinted (on an old exercise from last year) to find a measure $Q$ under which the quotient $S_1/S_2$ is a martingale and calculate the price at $0$ of a European call option with strike $K=1$. Since no bank account was mentioned in the exercise text, I'm a little bit confused. 
Supposing there is a bank account $B(t)$, and assuming absence of arbitrage, we know there is a EMM $Q^1$ under which the discounted assets are martingales, i.e. $S_i(t)/B(t)$ are martingales. In this case I would define the density (of $Q$ w.r.t. to $Q^1$) on $\mathcal{F}_T$ as follows:
$$Z_T:=\frac{1}{B(T)}\frac{S_2(T)}{S_2(0)}$$
and $Z_t:=E[Z_T|\mathcal{G}_t]=\frac{1}{B(t)}\frac{S_2(t)}{S_2(0)}$. Then using Bayes formula I'm able to prove that $\frac{S_1}{S_2}$ is a $Q$-martingale. Are you think my observations are correct so far?
Now I want to compute 
$$E_Q[(\frac{S_1(t)}{S_2(t)}-1)^+|\mathcal{F}_0]$$
I wrote down:
$$E_Q[(\frac{S_1(t)}{S_2(t)}-1)^+|\mathcal{F}_0]=E_Q[(\frac{S_1(t)}{S_2(t)}-1)\mathbf1_{\{\frac{S_1(t)}{S_2(t)}>1\}}|\mathcal{F}_0]$$
Using Bayes gives
$$E_Q[(\frac{S_1(t)}{S_2(t)}-1)\mathbf1_{\{\frac{S_1(t)}{S_2(t)}>1\}}|\mathcal{F}_0]=\frac{1}{Z_0}(E_{Q^1}[\frac{S_1(t)}{S_2(t)}Z_t\mathbf1_{\{\frac{S_1(t)}{S_2(t)}>1\}}|\mathcal{F}_0]-E_{Q^1}[Z_t\mathbf1_{\{\frac{S_1(t)}{S_2(t)}>1\}}|\mathcal{F}_0])=\frac{1}{B(0)}(E_{Q^1}[\frac{S_1(t)}{B(t)}\mathbf1_{\{\frac{S_1(t)}{S_2(t)}>1\}}|\mathcal{F}_0]-\frac{1}{S_2(0)}E_{Q^1}[\frac{S_2(t)}{B(t)}\mathbf1_{\{\frac{S_1(t)}{S_2(t)}>1\}}|\mathcal{F}_0])$$
Under $Q^1$, $\frac{S_i}{B}$ is given by 
$$S_i(t)=S_i(0)\exp{(\sigma_iW^*-\frac{1}{2}\sigma_i^2t)}$$
Where $W^*=W_t+\frac{\mu_i-r}{\sigma}t$ is $Q^1$ Brownian motion. Hence the above is under $Q^1$, $\exp{(Z)}$ distributed, where $Z$ is normal distributed with mean $-\frac{1}{2}\sigma_i^2t$ and Variance $\sigma^2t$. Is it true, that this conditional expectations reduce to expectations? If not, how would you calculate this?
However I am not at all sure about this derivation so far. Since the bracket of $W_1$ and $W_2$ is explicitly given, I thought somewhere I have to use it. It would be appreciated, if I did a error in reasoning, that someone could explain how to solve the problem, i.e. calculate the price. Thanks in advance
math
 A: Let $\beta(t) := e^{rt}$, where $r$ is the domestic risk-free rate. Let $Q$ be the risk-neutral measure.
The value is:
$v_t := E^Q[\frac{\beta(t)}{\beta{T}}S_1(T) 1_{\{S_1(T) > S_2(T)\}} | \mathscr{F}_t] - E^Q[\frac{\beta(t)}{\beta{T}}S_2(T) 1_{\{S_1(T) > S_2(T)\}} | \mathscr{F}_t]$
$=: I_1 + I_2$.
To evaluate $I_1$, do a change of probability measure to $S_1(T)$ as the numeraire asset:
$ E^{P_1}[S_1(t) 1_{\{S_1(T) > S_2(T)\}} | \mathscr{F}_t] = S_1(t)P_1\{S_1(T) > S_2(T) | S_1(t),S_2(t)\}$;
where the Radon-Nikodym derivative to go from $Q$ to $P_1$ was:
$\frac{dP_1}{dQ} = \frac{S_1(T)\beta(0)}{S_1(0)\beta(T)}$
meaning that, we can read off this Doleans exponential using Girsanov's theorem to get our new Wiener process:
$dW_{P_1}(t) = dW_1^Q - \sigma_1 dt$
$W_{P_1}(t) = W_1^Q - \sigma_1 t$
Now, $S_1(t)P_1\{S_1(T) > S_2(T) | S_1(t),S_2(t)\} = S_1(t)P_1\{\text{ln}\frac{S_1(T)}{S_2(T)} > 0\}$
by the monotone increasing property of $\text{ln}$. Under measure $P_1$ and $\mathscr{F}_t$, 
$S_1(T) = S(t) e^{(r+\frac12 \sigma_1^2)(T-t) + \sigma_1 (W_1(T) - W_1(t))}$.
Under $Q$, we can decompose $S_2(T)$'s Wiener process ($W_2^Q$) into Wiener processes $W_1^Q(t)$ and $W_1^\perp(t)$, where $W_1^\perp(t)$ is a BM uncorrelated to $W_1^Q(t)$. Under $Q$:
$W_2^Q = \rho W_1^Q + \sqrt{1-\rho^2}W_1^\perp$
... You then use Levy's characterization to solve the rest of the problem. 
