Question
Stocks $A$ and $B$ open on trading day at the same price. Let $X(t)$ denote the dollar amount by which stock $A$'s price exceeds stock $B$'s price when $100t\%$ of the trading day has elapsed. $X(t)\ \forall\ t \in [0, 1]$ is modelled as a Brownian motion process with $\mu = 0$ and $\sigma^2 = 0.3695$. After $75\%$ of the trading day has elapsed, stock $A$'s price is $39.75$ and stock $B$'s price is $40.25$. Find the probability that $X(1) \geq 0 $.
My working
With $X(0) = 0$ and $\mu = 0$, we have $$\begin{aligned} X(t) & = X(0) + \mu t + \sigma W_t\\ & = \sigma W_t, \end{aligned}$$ where $W_t \sim \mathcal{N}(0, t)$. We can also obtain the following quantities: $$\begin{aligned} X(0.75) & = -\frac 1 2,\\ \mathbb{E}[X(0.75)] & = \mathbb{E}[X(1)]\\ & = 0,\\ Var[X(0.75)] & = 0.75\sigma^2,\\ Var[X(1)] & = \sigma^2,\\ \rho & = Corr[X(0.75), X(1)]\\ & = \frac {Cov[X(0.75), X(1)]} {\sqrt {\{Var[X(0.75)]\}\{Var[X(1)]\}}}\\ & = \frac {\sigma^2 Cov[W(0.75), W(1)]} {\sigma^2 \sqrt {0.75}}\\ & = \frac {\min\{0.75, 1\}} {\sqrt {0.75}}\\ & = \sqrt {0.75}. \end{aligned}$$ Now, let $$X(1) \mid X(0.75) = -\frac 1 2 \sim \mathcal{N}(s, t),$$ where $$\begin{aligned} s & = \mathbb{E}[X(1)] + \sqrt {\frac {Var[X(1)]} {Var[X(0.75)]}}(\rho)\{X(0.75) - \mathbb{E}[X(0.75)]\}\\ & = -\frac 1 2 \end{aligned}$$ and $$\begin{aligned} t & = (1 - \rho^2)Var[X(1)]\\ & = \frac 1 4 \sigma^2. \end{aligned}$$ With $\sigma^2 = 0.3695$, $$X(1) \mid X(0.75) = -\frac 1 2 \sim \mathcal{N}\left(-\frac 1 2, \frac {739} {8000}\right).$$ $$\begin{aligned} \therefore \mathbb{P}[X(1) \geq 0] & = 1 - \mathbb{P}[X(1) < 0]\\ & = 1 - \mathbb{P}\left(Z < \sqrt {\frac {2000} {739}}\right)\\ & = 0.04997 \end{aligned}$$
As I have just covered Brownian motion in class, I am wondering whether my answer is correct and in particular, whether my working makes sense. Any comments will be greatly appreciated :)
The problem is solved way faster if you use conditional expectations and the properties of Brownian motion. Given $X(t)=A(t)-B(t)$, we want to know $P(X(1)\geq 0|X(t))$. We know that $X(u)-X(t) \sim \mathcal{N}(0,\sigma^2(u-t))$ for $t < u \leq 1$ and this increment is independent of $\mathscr{F}_t$ (i.e. the information up to $t$). Therefore $$\begin{aligned}P(X(1)\geq 0|X(t)=x)&=P(X(1)-X(t)\geq -X(t)|X(t)=x)=\\ &=P(X(1)-X(t)\geq -x)=\\ &=P(Z\geq -x/\sqrt{\sigma^2(1-t)})=\\ &=1-\Phi(-x/\sqrt{\sigma^2(1-t)})\approx\\ &\approx 0.05 \end{aligned}$$