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Let $(B_t)_{t\geq0}$ be a standard Brownian motion. Show $$P(-2\leq B_t\leq 2\ \forall t\in [0,1])\geq 1-\frac{1}{\sqrt{2\pi}}$$

I know $(B_t)\overset{d}{=}N(0,t)$ so $$P(-2\leq B_t\leq 2) =\int_{-2}^2 \frac{1}{\sqrt{2\pi t}}e^{-\frac{x^2}{2t}}dx$$ Solving this integral doesn't give the needed form. And what about $t\in[0,1]$? Thanks for any help!

Alon Yariv
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Uhmm
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  • What does $(B_{t})=N(0,\sigma^{2})$ even mean? $(B_{t}){t\in [0,1]}$ is a Gaussian Process and it does NOT equal a single normal variable in distribution. In particular $N(0,\sigma^{2})$ is a probability distribution on $\Bbb{R}$ whereas $(B{t})_{t\in [0,1]}$ is a Probability distribution on $C[0,1]$ space. – Mr.Gandalf Sauron Apr 25 '23 at 16:18
  • @Mr.GandalfSauron I made an edit – Uhmm Apr 25 '23 at 16:35
  • Usually $(B_{t})$ refers to the entire process. $B_{t}$ refers to one random variable. In any case, you should really show your effort. You have just written the density of a Gaussian which by no means is some newly discovered amazing result. – Mr.Gandalf Sauron Apr 25 '23 at 17:14

2 Answers2

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$P(-2\leq B_{t}\leq 2\,\forall t\in [0,1])=P(\inf_{t\in[0,1]}B_{t}\geq -2\, , \,\sup_{t\in[0,1]}B_{t}\leq 2)$

Now just see that $P(\inf_{t\in[0,1]}B_{t}\geq -2\, , \,\sup_{t\in[0,1]}B_{t}\leq 2)\leq P(\sup_{t\in[0,1]}B_{t}\leq 2)$ .

Now $\sup_{t\in[0,1]}B_{t}\sim |B_{1}|\sim |X|$ where $X\sim N(0,1)$ .

[$\sup_{t\in[0,1]}B_{t}$ is just the Running Maximum of the Brownian motion and it has the same distribution as $|B_{1}|$]

So what you want is $P(|X|\leq 2)= 1-P(|X|\geq 2) = (1-2\cdot P(X\geq 2))$

Now just use a tail estimate for the Gaussian, namely $\int_{y}^{\infty}\frac{1}{\sqrt{2\pi}}\exp(-\frac{x^{2}}{2})\,dx \leq \frac{1}{y\sqrt{2\pi}}\exp(-\frac{y^{2}}{2})$

So $P(X\geq 2) \leq \frac{1}{2\sqrt{2\pi}}e^{-2}\leq \frac{1}{2\sqrt{2\pi}}$

Hence $-2P(X\geq 2)\geq -\frac{1}{\sqrt{2\pi}}$

Thus $P(|X|\leq 2)\geq 1-\frac{1}{\sqrt{2\pi}}$

And hence you have the required bound.

Note that using the Gaussian Tail Estimate, we could have gotten a better bound , namely,

$P(|X|\leq 2)\geq 1-\frac{1}{e^{2}\sqrt{2\pi}}$ which is obviously better than what is being asked and hence $P(-2\leq B_{t}\leq 2\,\forall t\in[0,1])\geq 1-\frac{1}{e^{2}\sqrt{2\pi}}$

Mr.Gandalf Sauron
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We will use the reflection principle to obtain a lower bound for the probability that the Brownian motion stays within the interval $[-2, 2]$ for all $t \in [0, 1]$.

Let $T = \inf\{t \in [0, 1]: B_t \notin [-2, 2]\}$ be the first time the Brownian motion exits the interval $[-2, 2]$ within the time interval $[0, 1]$. We want to find a lower bound for the probability $P(T > 1)$.

Note that if $B_t$ exits the interval $[-2, 2]$ for the first time at time $T$, then by the reflection principle, we have that $B_{2T} \overset{d}{=} -B_T$. Hence, we can write $$ P(T > 1) = P(B_1 \in [-2, 2] \mid T > 1)P(T > 1) = P(B_1 \in [-2, 2] \mid B_1 \overset{d}{=} -B_T \text{ for some } T \leq 1). $$ Now, we will show that $P(B_1 \in [-2, 2] \mid B_1 \overset{d}{=} -B_T \text{ for some } T \leq 1) \geq 1 - \frac{1}{\sqrt{2\pi}}$.

First, note that the event $B_1 \overset{d}{=} -B_T \text{ for some } T \leq 1$ implies that $|B_1| \geq 2$. Therefore, we have $$ P(B_1 \in [-2, 2] \mid B_1 \overset{d}{=} -B_T \text{ for some } T \leq 1) = P\left(\left|B_1\right| \leq 2 \mid \left|B_1\right| \geq 2\right). $$ This probability is minimized when the conditional distribution of $|B_1|$ is given by the distribution of $|Z|$ where $Z \sim N(0, 1)$, i.e., $|Z|$ has the density function $f(x) = \frac{2}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}$ for $x \geq 0$. Therefore, we obtain $$ P(B_1 \in [-2, 2] \mid B_1 \overset{d}{=} -B_T \text{ for some } T \leq 1) \geq P(|Z| \leq 2) = 1 - \frac{1}{\sqrt{2\pi}}. $$ Putting this result together, we have $$ P(-2 \leq B_t \leq 2\ \forall t\in [0,1]) = P(T > 1) \geq 1 - \frac{1}{\sqrt{2\pi}}. $$

Alon Yariv
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    You don't really need this much machinery to derive a simple union bound. You have $T\leq T_{2}$ a.s. where $T_{2}$ is the hitting time of $2$. So $P(T\geq 1)\geq P(T_{2}\geq 1) = P(|Z|\leq 2)$ as $P(T_{a}>t)=P(max_{[0,t]}B_{s}\leq a)=P(|B_{t}|\leq a)$ . In fact Gandalf's answer does almost the same thing. – Blitzkrieg Apr 25 '23 at 16:58
  • @Blitzkrieg No ! $T \le T_2$ so ${ T \ge 1 } \subset { T_2 \ge 1 } $ and $P(T \ge 1) \le P(T_2 \ge 1)$. – Kieran McShane Oct 26 '23 at 19:21