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?

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

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?

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

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.,

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

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

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

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

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

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?

$$P\{X_s=0\:for\:some\:1≤s≤t\}=0$$

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?

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