Let $\textbf{B(t)} = (B_{1}(t), B_{2}(t))$ be a standard two dimensional Brownian motion starting from $(0, 0)$.
(a) For each $x = (x_{1}, x_{2}) \in \mathbb{R}^{2} \setminus \{0\}$, show that there exists a number $\alpha > 0$ such that $X(t) = α(x_{1}B_{1}(t)+ x_{2}B_{2}(t))$ is a standard one dimensional Brownian motion.
(b) Let $\tau_{1}$ be the first time that $\textbf{B(t)}$ hits the line $m_{1}x + y = c$ and $\tau_{2}$ be the first time that $\textbf{B(t)}$ hits the line $m_{2}x - y = c$. Find $\mathbb{P}[\tau_{1} \land \tau_{2} < 1]$. Here, $m_{1}m_{2} = -1$.
(c) Let $\tau$ be the first time that $\textbf{B(t)}$ hits either of the lines $ax + y = b$ or $ax + y = −b$. Show that
$$\mathbb{E}(\tau) = b^2/(1+a^2) $$
(d) Let $\tilde{\tau}$ be the first time that $\textbf{B(t)}$ hits the line $x + y = \sqrt{2}$. Find $$\mathbb{E}(e^{\tilde{\tau}/2})$$
$\textbf{My Attempt}$
a) we know that $\textbf{B(t)}$ has a bivariate normal distribution so any linear combination would be a univariate normal distribution.
$$\mathbb{E}(X(t)) = 0$$.
For $s < t$, $$Cov[X(s), X(t)] = Cov[\alpha x_{1}B_{1}(t), \alpha x_{1}B_{1}(s)] + Cov[\alpha x_{1}B_{1}(t), \alpha x_{2}B_{2}(s)] + Cov[\alpha x_{2}B_{2}(t), \alpha x_{1}B_{1}(s)] + Cov[\alpha x_{1}B_{1}(s), \alpha x_{2}B_{2}(s)] = \alpha^2 x_{1}^2 s + \alpha^2 x_{2}^2 s = \alpha^2(x_{1}^2+x_{2}^2)s.$$ If we let $\alpha^2 = 1/(x_{1}^2+x_{2}^2)$, then $Cov[X(s), X(t)] = s.$ Thus, we have shown that X(t) is a Gaussian process with mean 0, and $Cov[X(s), X(t)] = min(s,t)$. Hence, it is a one dimensional BM.
b) I know that the lines are perpendicular. Define the new processes,
$$X_{1}(t) = B_{2}(t) + m_{1}B_{1}(t)$$ $$X_{2}(t) = -B_{2}(t) + m_{2}B_{1}(t)$$
$X_{1}(t)$ and $X_{2}$ are BM, so they are Gaussian processes, and we only need to show that their covariance is 0.
$$Cov(X_{1}(t) , X_{2}(t)) = Cov(B_{2}(t) + m_{1}B_{1}(t), - B_{2}(t) + m_{2}B_{1}(t) = - Cov(B_{2}(t), B_{2}(t)) + Cov(B_{2}(t), m_{1}B_{1}(t)) + Cov(m_{1}B_{t}, B_{2}(t)) + Cov(m_{1}B_{1}(t),m_{2}B_{1}(t)) = t-t=0 $$
Hence, $X_{1}(t)$ , $X_{2}(t)$ are independent of each other. Next, we calculate the distance between where X_{1}(t) starts and the line $m_{2}x - y = c$. The distance equals $c/\sqrt{m_{1}^2 + 1}$. Thus,
$$\mathbb{P}(\tau_{1} < 1) = 2\mathbb{P}\Bigg(X_{1}(t) > \frac{c}{\sqrt{m_{1}^2 + 1}}\Bigg) = 2 - 2\Phi\Bigg(\frac{c}{\sqrt{m_{1}^2 + 1}}\Bigg),$$
where $\Phi(.)$ is the CDF of a standard normal random variable.
Similarly,
$$\mathbb{P}(\tau_{2} < 1) = 2\mathbb{P}\Bigg(X_{2}(t) > \frac{c}{\sqrt{m_{2}^2 + 1}}\Bigg) = 2 - 2\Phi\Bigg(\frac{c}{\sqrt{m_{2}^2 + 1}}\Bigg).$$
Since, $X_{1}(t)$ and $X_{2}(t)$ are independent, so $\tau_{1}$ and $\tau_{2}$ are also independent.
$$\mathbb{P}[\tau_{1} \land \tau_{2} < 1] = 1 - \mathbb{P}[\tau_{1} > 1]\mathbb{P}[\tau_{2} > 1] = 1 - \Bigg[2 - 2\Phi\Bigg(\frac{c}{\sqrt{m_{1}^2 + 1}}\Bigg)\Bigg]\Bigg[2 - 2\Phi\Bigg(\frac{c}{\sqrt{m_{2}^2 + 1}}\Bigg)\Bigg]$$
c) The distance between the two given (parallel) lines is $\frac{2b}{\sqrt{a^2+1}}.$ We make use of the optional stopping theorem (OST) and part a). We choose two appropriate martingales and the two points as $\frac{-b}{\sqrt{a^2+1}}$ and $\frac{b}{\sqrt{a^2+1}}$. The distance between those points is $\frac{2b}{\sqrt{a^2+1}}$. Using OST for the first martingale gives us, the probability that the BM hits $\frac{-b}{\sqrt{a^2+1}}$ first as $1/2$. Thus, by symmetry the probability that the BM hits $\frac{b}{\sqrt{a^2+1}}$ first is also $1/2$.
$$\mathbb{E}[\tilde{\tau}] = \frac{1}{2}\Bigg(\frac{b^2}{a^2+1}\Bigg) + \frac{1}{2}\Bigg(\frac{b^2}{a^2+1}\Bigg) = \Bigg(\frac{b^2}{a^2+1}\Bigg).$$
d) If I can get the density of $\tilde{\tau}$, then I just need the moment generating function evaluated at 1/2.
Alternatively, I know in 1-D that we can also define a new martingale and then calculate this quantity directly.
Any help will be appreciated.