1
$\begingroup$

I was doing some reading after recently finishing a course in introductory stochastic processes where we finished by talking about Gaussian processes and Brownian motion and came across a problem I have no idea how to solve.

Let $W_t$ be standard Brownian motion. Find $\mathbb{P}\left(W_{4}<2\mid W_{5}>1\right)$ and $\mathbb{P}\left(W_{5}>1\mid W_{4}<2\right)$.

My first thought was to exploit the independence of increments by writing $W_{5}=W_{5}-W_{4}+W_{4}$ but I'm having trouble applying this idea to the first conditional probability as we are conditioning on $W_{5}$. Could someone please elaborate on how I first conditional probability could be found?

If I apply this property to the second, would it be true that I will have: $$\mathbb{P}\left(W_{5}>1\mid W_{4}<2\right)=\mathbb{P}\left(W_{5}-W_{4}>1\mid W_{4}<2\right)+\mathbb{P}\left(W_{4}>1\mid W_{4}<2\right)$$

But due to independence of increments, we have that $W_{5}-W_{4}$ is independent of $W_{4}$ and so the above would reduce to $$\mathbb{P}\left(W_{5}>1\mid W_{4}<2\right)=\mathbb{P}\left(W_{5}-W_{4}>1\right)+\mathbb{P}\left(1<W_{4}<2\right)$$

Then we could exploit the fact that increments are normally distributed to have that $\left(W_{5}-W_{4}\right)\sim\mathcal{N}\left(0,1\right)$ to get $$\mathbb{P}\left(W_{5}>1\mid W_{4}<2\right)=1-\Phi(1)+\left(\Phi\left(1\right)-\Phi\left(\frac{1}{2}\right)\right)$$

Where I have transformed $W_{4}$ to have a standard normal distribution. Is what I have done for the second conditional probability correct?

$\endgroup$
  • $\begingroup$ Aww hell yeah! I love brownies! $\endgroup$ – clathratus Oct 5 '18 at 1:13

Your Answer

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

Browse other questions tagged or ask your own question.