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, and suppose that ${(Y(t),0 \leq t \leq 1)}$ can be effectively modelled as a Brownian motion process with variance parameter $\sigma^2$

I want to find

(a)If the inside racer is leading by $\sigma$ seconds at 25% of the race what is the probability that she is the winner?

(b)If the inside racer wins the race by a margin of $\sigma$ seconds, what is the probability that she was ahead at 25% point of the race?

Solutions: (a)$P{(Y(1)>0|Y(1/4)=\sigma)}$

=$P{(Y(1)-Y(1/4)> -\sigma|Y(1/4 =\sigma)}$

= $P{(Y(1) - Y(1/4)>-\sigma)}$ by independent increments

=$P{(Y(3/4) > -\sigma)}$ by stationary increments





1 Answer 1


(b) Because we must compute $P(Y(1/4)>0|Y(1)=\sigma)$, let us first determine the conditional distribution of Y(s) given that Y(t)=C when s < t.Now, since${[X(t),t\geq 0]}$ is standard Brownian motion when $X(t)=Y(t)/\sigma$, we obtain that the conditional distribution of X(s), given that $X(t)=C/\sigma$, is normal with mean $\frac{sC}{t\sigma}$ and variance s(t-s)/t.Hence the conditional distribution of $Y(s)=\sigma Xs$ given that Y(t)=C is normal with mean sC/t and variance $\sigma^2 \frac{s(t-s)}{t}$. Hence

$P[Y(1/4)>0|Y(1)=\sigma ]$=$[P(N(\frac{\sigma}{4},\frac{3\sigma^2}{16}>0)]$




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