What is $P(B_1 > 0, B_2 > 0)$ , where $B_t$ is a Brownian Motion at time $t$?

Question

The following question is found from this MSE post. For completeness, I restate the problem below.

Question: What is $P(B_1 > 0, B_2 > 0)$ , where $B_t$ is a Brownian Motion at time $t$?

From that post, the OP calculates the probability as follows:

$P(B_1 > 0, B_2 > 0) = P(B_1 > 0, B_2 - B_1 > -B_1) = P(Z_1 > 0, Z_2 > -Z_1) = \frac{3}{8}$ by applying a symmetry argument to the $(Z_1, Z_2) \sim N(0, I_2)$ distribution.

I can understand all equalities except the last one which leads to the answer $\frac{3}{8}.$

In other words, I do not understand how symmetry argument to bivariate normal distribution is at play here.

Answer 1

After reading this Cross Validated post, I think I know how to obtain $$P(Z_1 > 0, Z_2 > -Z_1) = \frac{3}{8}.$$ Following the same idea as in the Cross Validated post, we draw an $Z_2$ against $Z_1$ plane. We also draw two lines $Z_2 = Z_1$ and $Z_2 = -Z_1$ to split the plane into $8$ quadrants. Then the required probability $$P(Z_1 > 0, Z_2 > -Z_1)$$ is the intersection between two regions $Z_1 > 0$ and $Z_2 > -Z_1,$ which consist of $3$ quadrants out of $8$. (I think the bold statement is true if the two random variables $Z_1$ and $Z_2$ are independent?) So, the probability is $\frac{3}{8}.$

Answer 2

First of all $Z_1$ has to be positive which happens with $\frac{1}{2}$ probability. Then we have to have $Z_2>-Z_1$ which conditioned on the previous event means that $Z_2$ is either positive (which has again $\frac{1}{2}$ probability) or it has to be negative (again $\frac{1}{2}$) and greater than $Z_1$ in absolute value - by symmetry it's once again $\frac{1}{2}$ yielding $\frac{3}{4}$ probability for that case. So the probability of the entire event is $\frac{1}{2} \cdot \frac{3}{4} = \frac{3}{8}$.