Consider a standard Brownian motion/Wiener process $B_t$ with mean $0$ and variance $1$. Find $P(B_t<0$ for all $1\leq t \leq10)$.
My intuition tells me that this is a non-zero probability, but my (probably wrong) attempt shows otherwise. If someone can tell me where I messed up that would be great.
Attempt:
$$P(B_t<0 \text{ for all } 1 \leq t \leq10)=1-P(B_t>0 \text{ for some } 1 \leq t \leq10)$$ $$=1-P(B_t>0 \text{ for some } 0 \leq t \leq 9)$$
Since the probability density function for such processes is normal for some fixed time $t$, we have that the density function is $$f_t(x)=\frac{1}{\sqrt{2\pi t}}\exp\big(-\frac{x^2}{2t}\big)$$
Then by the Reflection Principle,
$$1-P(B_t>0 \text{ for some } 0 \leq t \leq 9)=1-2P(B_9>0)=1-2\int_0^\infty\frac{1}{18\pi}\exp\big(-\frac{x^2}{18})dx$$
According to Wolfram, the integral is $\frac{1}{2}$ which makes sense as the distribution of $B_t$ for any $t$ is normal with mean 0, thus the probability in question is $1-1=0$. If this isn't correct, what did I do wrong?