# How can I calculate this probability: $P(W_1<cW_2$) and $c\geq 0$?

Let $(W_t)_{t\geq 0},$ be a Brownian motion. I want to calculate the following:

$P(W_1<cW_2$) and $c\geq 0$

For $c=1$ it is easy. I just write it as an increment, but how can I do it when $c$ is not one?

• Because the joint distribution of any finite number of the $W_t$ is multivariate normal, any finite linear combination of them, such as $W_1-cW_2$, will have a Normal distribution of zero mean. What is the chance that such a random variable is less than zero? – whuber Mar 19 '15 at 14:39

Do a bit of algebraic manipulation and you'll get $$P((1-c)(W_1-W_0) < c(W_2-W_1))$$ We know the $W_t$ are Brownian motion so, $W_1-W_0$ and $W_2-W_1$ are independent normal variables. Their means are 0, and their variances are 1 (because 1-0=2-1=1. W_3-W_1 would have variance 3-1=2). So the probability is equal to $$P((1-c)Z_1 < c Z_2)$$
where the $Z_i$ are just standard normal random variables.