# Confused about an example of Brownian motion

I' am reading Introduction to Stochastic Processes by Lawler and have hit a particular example given in the book about Brownian motion that confuses me. I'll give most of example here:

Let $$t>1$$ and let us compute the probability that a standard Brownian motion crosses the x-axis sometime between 1 and $$t$$, i.e.,

$$P\{X_s = 0\mbox{ for some }\:1\:≤\:s\:≤\:t\}.$$

We first condition on what happens at time $$t=1$$. Suppose $$X_1=b>0$$. Then the probability that $$X_s=0$$ for some $$1 ≤ s ≤ t$$ is the same as the probability that $$X_s≤-b$$ for some $$0≤s≤t-1$$. This is the same (by symmetry) as the probability that $$X_s≥b$$ for some $$0≤s≤t-1$$. This probability is given by the reflection principle, so

$$P\{X_s=0\mbox{ for some }1≤s≤t\:|\:X_1=b\}=2\int_{b}^{\infty}\frac{1}{\sqrt{2\pi(t-1)}}e^{\frac{-x^2}{2(t-1)}}dx.$$

By symmetry, again the probability is the same if $$X_1=-b$$. Hence, by averaging over all possible values of b we get

$$P\{X_s=0\mbox{ for some }\:1≤s≤t\}=1-\frac{2}{\pi}\arctan\frac{1}{\sqrt{t-1}}.$$

I skipped the work that was need to demonstrate the last one. I' am confused since $$X_t$$ is a continuous random variable, so the probability it takes on an particular value is 0. So shouldn't the following be true?

$$P\{X_s=0\mbox{ for some }1≤s≤t\}=0$$

• This is the probability that $X_s$ takes the value 0 at some time from 1 to t, not at a particular time, for which the probability would be 0. Dec 20, 2015 at 14:56
• Ok thanks, sort of makes sense. That was one possibility that I considered, but wasn't really sure about. To be honest, the distinction seems very subtle and I' am still not quiet sure how that is technically different from $X_t$ taking on a particular value at a particular time. Dec 21, 2015 at 12:22
• Not P(x=0) but P crosses zero... either from above or below... So the line random walk goes up in the first instance, but then comes back down below 0 again... or goes down in the first instance, THEN goes back up above zero again. Remember, P(x=any number) = 0 for continuous distributions. P(x < any number) >= 0. Dec 23, 2015 at 1:52

This analogy suggests your intuition is correct: if you were to preselect any individual number $s$ between $1$ and $t$, and then observe many realizations of this "random walk" process, almost surely none of them would cross the x-axis precisely at $s$. We would be deeply impressed at your clairvoyance were that to happen.
Here is what some of these realizations look like. The figure plots six of them, distinguished by color. All start at $(0,0)$, but only the portions between $1$ and $t=2$ are shown.
As it happens, five of these six realizations cross the x-axis somewhere between $1$ and $2$ (at the points shown as dots).
Since these paths are almost surely continuous, the Intermediate Value Theorem (of elementary Calculus) implies that whenever one crosses the x-axis, it must actually meet it: that is, for five of these six paths there exists at least one $s$ in the interval $[1, t]$ for which the path's height actually equals $0$. We should be no more surprised that this happens than we are that at least one player hits a home run during a baseball game.
Another way to see why the chance of a crossing in $[1,t]$ is nonzero is to observe that a crossing almost certainly occurs when $X_1$ is positive and $X_t$ is negative. Thus, the chance of a crossing is at least as great as the chance that simultaneously $X_1 \gt 0$ and $X_t \le 0$. It should be obvious that these individual chances are large (each equals $1/2$, because both $X_1$ and $X_t$ have zero-mean Normal distributions). It should be almost as obvious that, despite the positive correlation between $X_1$ and $X_t$, the simultaneous chance is still pretty good--and certainly not zero.