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?

  • $\begingroup$ 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? $\endgroup$ – 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.

| cite | improve this answer | |
  • $\begingroup$ This does not answer the question, does it? But simply postpones the moment when one will apply the argument, unmentioned here, showing that the result is 50%. $\endgroup$ – Did Apr 4 '15 at 16:44
  • $\begingroup$ It's probably less than 50%, but the reduced problem is much much easier. You're no longer dealing with Wiener processes. (c-1)Z_1+cZ_2 is just a normal random variable, so calculating that is trivial. $\endgroup$ – maxbaroi Apr 6 '15 at 17:32
  • $\begingroup$ "It's probably less than 50%" Wanna bet? $\endgroup$ – Did Apr 6 '15 at 21:45
  • $\begingroup$ I'm sorry. I misspoke. I didn't mean the answer was 50%, I was referring to how much progress was made on the problem. $\endgroup$ – maxbaroi May 3 '15 at 7:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.