In a bicycle race between two competitors, let $Y(t)$ denote the amount of time (in seconds) by which the racer that started in the inside position is ahead when $100t$ percent of the race has been completed, $0\leq t\leq 1$. (So $t$ here does not quite have the meaning of time as we usually think it. Also if $Y(t)$ is negative this means the inside racer is behind). $Y(t,0\leq t \leq 1)$ can be effectively modeled as a Brownian motion with variance parameter $\sigma^2$.
If the inside racer is leading by $\sigma$ seconds at the midpoint of the race, what is the probability that she is the winner?
My Work So Far
\begin{align*} &\mathbb{P}\left\{Y(1)\gt0\bigg|Y\left(\frac{1}{2}\right)=\sigma\right\}\\ &=\mathbb{P}\left\{Y(1)-Y\left(\frac{1}{2}\right)\gt-\sigma\bigg|Y\left(\frac{1}{2}\right)=\sigma\right\}&&\text{(1) cancel RHS}\\ &=\mathbb{P}\left\{Y(1)-Y\left(\frac{1}{2}\right)\gt-\sigma\right\}\\ &=\mathbb{P}\left\{Y\left(\frac{1}{2}\right)\gt-\sigma\right\}\\ &=\mathbb{P}\left\{\frac{Y(1/2)}{\sigma/\sqrt{2}}\gt-\frac{\sigma}{{\sigma}/\sqrt{2}}\right\}&&\text{(2) divide LHS and RHS by}\ \sigma/\sqrt{2}\\ &=\mathbb{P}\left\{Z\gt-\sqrt{2}\right\}\\ &=\Phi(\sqrt{2})\\ &\approx0.9213\\ \end{align*}
pnorm(sqrt(2))
[1] 0.9213504
I believe the reason for $\text{(2)}$ is because we'd like to make the random variables on the LHS of the inequality a standard normal random variable, and that way we can use the normal CDF to finish our calculation via the $\Phi(x)$.
Would this method be correct?
The solution in the lecture notes stated the below, but I was unsure where the lines in red were derived from. \begin{align*} &\mathbb{P}\left(Y(1)\gt0\bigg|Y\left(\frac{1}{2}\right)=\sigma\right)\\ &=\mathbb{P}\left(Y(1)-Y\left(\frac{1}{2}\right)\gt-\sigma\bigg|Y\left(\frac{1}{2}\right)=\sigma\right)\\ &=\mathbb{P}\left(Y(1)-Y\left(\frac{1}{2}\right)\gt-\sigma\right)\\ &\color{red}{=\mathbb{P}\left(N\left(0,\frac{\sigma^2}{2}\right)\gt-\sigma\right)}\\ &\color{red}{=\mathbb{P}\left(Z\gt-\frac{\sqrt{2}}{\sigma}\right)}\\ \end{align*}
The variance is proportional to the increment size [1]. Since $Y(1)$ follows a Brownian motion with variance $\sigma^2$, then $Y(1) - Y(\frac{1}{2})$ follows a Brownian motion with variance $\frac{\sigma^2}{2}$, as you have halved the increment size.