Find $P(Z_1>0, Z_2<0)$

Question

Let $Z_t$ be standard Brownian motion.

I would do $$P(Z_1>0, Z_2<0) {= P(Z_1>0 | Z_2<0) P(Z_2<0)\\=P(Z_1-Z_0>0 | Z_2-Z_0<0) P(Z_2-Z_0<0) \\=^{(*)} P(Z_1-Z_0>0)P( Z_2-Z_0<0) P(Z_2-Z_0<0) }$$ Where the asterisk means it is because of independent increments. The figures come from the fact that $(Z_t - Z_0)$ ~ $N(0,t-0)$.

According the solutions, it doesn't have what I have written. Why can't I do what I have done here?

Answer 1

Let $Z$ be a Brownian motion.

We have that: \begin{align} \mathbb{P}\left(Z_1 > 0, Z_2 < 0\right) &= \mathbb{E}[1_{Z_1 > 0} 1_{Z_2 < 0}] \\ &= \mathbb{E}[1_{Z_1 > 0}\mathbb{E}[1_{Z_2-Z_1 < -Z_1}|Z_1] \\ &= \mathbb{E}[1_{Z_1 > 0}\mathbb{E}[1_{X < -Z_1}|Z_1]] \text{ where $X\sim \mathcal{N}(0,1)$ and independent of $Z_1$ }\\ &= \mathbb{E}[1_{Z_1 > 0}\Phi(-Z_1)] \\ &= \int_0^{+\infty}\Phi(-x)\phi(x)dx = \frac12\left[\Phi(x)^2\right]^0_{-\infty} = \frac18 \end{align} Where $\Phi$ is the c.d.f of unit normal r.v. and $\phi$ is its p.d.f

Maybe there is a simpler way to arrive at this conclusion...