# Computing Conditional Probabilities of Brownian Motion with Strict Inequalities

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

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?

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