# Brownian motion probability

Let $$(B_t)$$ be a standard Brownian motion. How can I compute $$P(B_2 > 0 | B_1 > 0)$$? I know that $$B_2-B_1$$ follows a $$\mathcal{N}(0,2–1)$$, but I do not know how to compute $$\int_0^{+\infty}P(B_2>0|B_1)f(B_1)dB_1$$.

• What's the distribution of $B_1$? Without that (or something it can be computed from, such as the distribution of $B_\tau$ for some $\tau \lt 1$) this question doesn't seem well-defined. Are we perhaps meant to assume that $B_0 = 0$? – Ilmari Karonen Mar 2 at 18:55
• @IlmariKaronen I think we are meant to assume that $B_0=0$ and $(B_t)_{t\geq 0}$ – Natalie_94 Mar 2 at 19:02
• Hint: No integrals necessary. The random variables $X = B_1$ and $Y=B_2$ are independent standard normal, and you are looking for $P(X+Y>0|X>0)$, which is the sum of $P(Y>0) = 1/2$ and $P(-X<Y<0|X>0)$. By symmetry the latter is $(1/2) P(|Y|<|X|)$, which is then easy to evaluate... – Lukas Geyer Mar 2 at 19:04
• @LukasGeyer I suppose you wanted to define $Y:=B_2-B_1$, right? Otherwise your comment wouldn't make much sense. – saz Mar 2 at 19:12
• @LukasGeyer Do you mean that the latter is $1/2+P(|Y|<|X|)$? – Natalie_94 Mar 2 at 19:52