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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?

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ 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

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