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, 2015 at 14:39
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1 Answer
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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%. $\endgroup$– DidApr 4, 2015 at 16:44
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$\begingroup$ 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. $\endgroup$– maxbaroiApr 6, 2015 at 17:32
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$\begingroup$ 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. $\endgroup$– maxbaroiMay 3, 2015 at 7:14