Please verify the solution about Brownian motion process.

Question

Problem

• Let $Y(t)$ denote the amount of time by which the racer is ahead when $100t$ percent of the race has been completed.

• $\{Y(t), 0 \leqslant t \leqslant 1\}$ is modeled as a Brownian motion process with variance parameter $\sigma^2$. That is, for every $0 \leqslant t \leqslant 1$, $Y(t) \sim N(0, \sigma^2t)$.

  1. If the racer is leading by $\sigma$ seconds at the midpoint of the race, what is the probability that she is the winner?

  2. If the racer wins the race by a margin of $\sigma$ seconds, what is the probability that she was ahead at the midpoint?

---

Solve

1. $\Pr\left[Y(1) \gt 0 \left| Y\left(\frac{1}{2}\right) = \sigma\right]\right.$ \begin{align} &= \Pr\left.\left[Y(1)-Y\left(\frac{1}{2}\right) \gt -\sigma \right| Y\left(\frac{1}{2}\right) = \sigma\right] \\ &= \Pr\left[Y(1)-Y\left(\frac{1}{2}\right) \gt -\sigma \right] \quad \text{by the independent increments.} \\ &= \Pr\left[Y\left(\frac{1}{2}\right) \gt -\sigma \right] \quad \text{by the stationary increments.} \\ &= \Pr\left[\frac{Y\left(\frac{1}{2}\right) - 0}{\frac{\sigma}{\sqrt{2}}} \gt \frac{-\sigma - 0}{\frac{\sigma}{\sqrt{2}}}=-\sqrt{2} \right] \quad \because Y\left(\frac12\right)\sim N\left(0, \frac{\sigma^2}{2}\right)\\ &= \Phi\left(\sqrt{2}\right) \quad \text{, where } \Phi(t) = \Pr\left[N(0, 1) \le t\right] \end{align}

2. Let $\{X(t), t\geqslant0\}$ be standard Brownian motion; $X(t) \sim N(0, t)$

   Then, for $0 \leqslant t \leqslant 1$, $Y(t) = \sigma X(t)$.

   There is a theorem that, for $s \lt t$, $$X(s) |_{X(t)=c} \sim N\left(\frac{sc}{t}, \frac{s(t-s)}{t}\right)$$

   Thus, for $s \lt t$, $$\left.\sigma X\left(\frac12\right) \right|_{X(1)=1} \sim N\left(\frac{\sigma}{2}, \frac{\sigma^2}{4}\right)$$

   Therefore, $\Pr\left[\left. Y\left(\frac12\right) \gt 0 \right| Y(1) = \sigma \right]$ \begin{align} &= \Pr\left[\left. \sigma X\left(\frac12\right) \gt 0 \right| \sigma X(1) = \sigma \right] \\ &= \Pr\left[\left. \sigma X\left(\frac12\right) \gt 0 \right| X(1) = 1 \right] \\ &= \Pr\left[N\left(\frac{\sigma}{2}, \frac{\sigma^2}{4}\right) \gt 0\right] \\ &= \Phi(1) \quad \text{, where } \Phi(t) = \Pr\left[N(0, 1) \le t\right] \end{align}

---

I am sure that the above is not wrong. (but not 100% sure...)

So, I have a question in the process of solving the question #2.

Is it fine if I solve the question #2 like the following?

• From the theorem I said above, $\displaystyle \left.X\left(\frac12\right) \right|_{X(1)=1} \sim N\left(\frac{1}{2}, \frac{1}{4}\right)$.

• Therefore, \begin{align} \Pr\left[\left. Y\left(\frac12\right) \gt 0 \right| Y(1) = \sigma \right] &= \Pr\left[\left. \sigma X\left(\frac12\right) \gt 0 \right| \sigma X(1) = \sigma \right] \\ &= \Pr\left[\left. \sigma X\left(\frac12\right) \gt 0 \right| X(1) = 1 \right] \\ &= \Pr\left[\left. X\left(\frac12\right) \gt \frac{0}{\sigma} \right| X(1) = 1 \right] \\ &= \Pr\left[\left. X\left(\frac12\right) \gt 0 \right| X(1) = 1 \right] \\ &= \Pr\left[N\left(\frac{1}{2}, \frac{1}{4}\right) \gt 0\right] \\ &= \Phi(1) \quad \text{, where } \Phi(t) = \Pr\left[N(0, 1) \le t\right] \end{align}

Although the results are same as $\Phi(1)$, I am not sure the the below solve is fine.