I was wondering if there is a way to explicitly calculate the probability that Brownian motion is at two consecutive times s,t in the upper half-plane. So I want to calculate $$\mathbb{P} (B_t>0, B_s>0)$$ if this is analytically possible.
Even an expression in terms of integrals would be helpful.
Let $p_t(x,y)$ be the Brownian transition density function $\frac 1 {\sqrt{2\pi t}}e^{-(y-x)^2/2t}$, so that its Markov property reads $$\mathbb E(f(B_{s+h})\mid B_s) = \int f(y)p_h(B_s,y) \, dy.$$
We wish to compute $\mathbb P(B_t>0,B_s>0)$. In order to use the Markov property in the last paraghraph, rewrite it as $\mathbb E(1_{\{B_s>0, B_t>0\}}) = \mathbb E\,\mathbb E(1_{\{B_s>0, B_t>0\}}\mid B_s )$, and notice that you can view $1_{\{B_s>0, B_t>0\}} = 1_{[0,\infty)\times[0,\infty)}(B_s,B_t)$ as a function of $B_{s + h}$ with $h=t-s$. Applying the property to the conditional expectation we get
$$\mathbb P(B_t>0,B_s>0) = \mathbb E\,\int1_{[0,\infty)\times[0,\infty)}(B_s,y)p_{t-s}(B_s, y) \, dy.$$
The last expectation can be computed directly using the law of $B_s$, that is, $p_s(0,x)\, dx$, so finally $$\mathbb P(B_t>0,B_s>0) = \int_0^ \infty\int_0^\infty p_{t-s}(x, y) p_s(0,x)\, dy\, dx.$$