I am looking into pricing options in finance and I came across the following problem: I want to simplify the probability:
$\mathbf{Q}(z_1 > - \hat{d}_1, z_2 > \hat{d}_2), $
where $z_1$ and $z_2$ are two standard normal variables such that $z_1 = \frac{1}{\sqrt{T_0}}W^{*}_{T_0}$ and $z_2 = \frac{1}{\sqrt{T}}W^{*}_{T}$, where $T>T_0$ and $W^{*}_{T}$ is the standard Brownian motion on the increment $T$. Furthermore, $\hat{d}_1$ and $\hat{d}_2$ are described as:
$\hat{d}_1 = \frac{\text{ln}\big(\frac{S(0)}{\bar{x}}\big) + (r - \frac{1}{2}\sigma^2)T_0}{\sigma\sqrt{T_0}},$
$\hat{d}_2 = \frac{\text{ln}\big(\frac{S(0)}{K}\big) + (r - \frac{1}{2}\sigma^2)T}{\sigma\sqrt{T}}.$
where $S(0)$ is the stock price at time $0$, $r$ is the risk-free rate, $\sigma^2$ is the annualized volatility and $K$ and $\bar{x}$ are constants.
I am quite sure that the distribution I am looking for is the bivariate normal, but I have no idea how to get there.
We first note that we can work directly with the underlying Brownian motion, rather than the scaled variables $z_1, \,z_2$ since
$$\mathbf Q \left( z_1 > \hat d_1, \, z_2 > \hat d_2\right) = \mathbf Q \left( W_{T_0} > \sqrt{T_0}\, \hat d_1, \, W_{T} > \sqrt{T}\, \hat d_2\right).$$
Henceforth I denote $d_1= \sqrt{T_0} \, \hat d_1, \, d_2 = \sqrt{T}\, \hat d_2$.
To compute the probability, we first use the law of total probability conditioning on the event $\{W_{T_0} \in dx\}$, and then use independence of Brownian increments. That is
\begin{align*} \mathbf Q \left( W_{T_0} > d_1, W_T > d_2\right) & = \int_{\mathbf R} \mathbf Q \left(W_{T_0} > d_1,\, W_T > d_2 \, | \, W_{T_0} \in dx\right)\, \mathbf Q\left( W_{T_0} \in dx\right)dx \\ & = \int_{d_1}^\infty \mathbf Q \left( W_T > d_2 \, | \, W_{T_0} \in dx\right)\, \mathbf Q\left( W_{T_0} \in dx\right)dx \\ & = \int_{d_1}^\infty \mathbf Q \left( W_T - W_{T_0} >d_2 -x\right)\, \mathbf Q\left( W_{T_0} \in dx\right)dx \\ \end{align*} Then we can use that $$W_{T_0} \sim N(0, T_0), \qquad W_{T} - W_{T_0} \sim N(0, T - T_0)$$ are independent and Normally distributed; let $\Phi$ denote the CDF of a standard normal distribution, then from the above we have
\begin{align*} \mathbf Q \left( W_{T_0} > d_1, W_T > d_2\right) & = \int_{d_1}^\infty \left(1 - \Phi\left( \frac{d_2 - x}{\sqrt{T - T_0} } \right) \right) \frac{1}{\sqrt{T_0}} \Phi' \left( \frac{x}{\sqrt{T_0}} \right) d x. \end{align*}
From here perhaps you can substitute in your exact formulae for $d_1,\,d_2$ and simplify the above.