Let $W_t$ be a standard brownian motion, and let $0 < x < y$. We want to calculate:

$\mathbb{P}(W_y > 0 \vert W_x > 0)$.

I am pretty stuck on how to do this. The only insight I have is that we would probably want to take advantage of the fact that increments are independent in a Brownian motion to rewrite $W_x, W_y$ to evaluate this conditional probability, but I don't know how to do that or set up the integral that would probably result. I am new to this topic, so I apologize if this is a dumb question.

  • $\begingroup$ One way to get started is to note that the pair $(W_x,W_y)$ is jointly normal, with zero means, and covariance matrix $\left[\matrix{x&x\cr x&y\cr}\right]$. $\endgroup$ – John Dawkins Apr 15 at 15:42
  • $\begingroup$ @JohnDawkins hmm, that makes sense! Then I can find out what $P(W_x, W_y) > 0$ is. But I would also need to calculate what $P(W_x > 0)$ is to find the full expression, do you have a suggestion for that? Thank you very much! $\endgroup$ – 0k33 Apr 15 at 21:00
  • $\begingroup$ But $W_X$ is mean zero normal, so $P(W_x>0)=1/2$. $\endgroup$ – John Dawkins Apr 15 at 21:55
  • $\begingroup$ @JohnDawkins oh man. I was sleeping on that one. Thank you so much, I know what to do now! If you want to put your comments as an answer, I will accept it, but if not that's also ok :) $\endgroup$ – 0k33 Apr 15 at 22:27

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