Why $\mathbb E[B_t^2]=t\implies B_t\sim\sqrt t$?

Question

Let $B_t$ a standard Brownian motion. Why $$\mathbb E[B_t^2]=t\implies B_t\sim\sqrt t\ \ \ ?$$

Answer 1

Let's talk a little bit about the general situation, because in a sense this result should've been expected even if we knew very little about Brownian motion. In general, Chebyshev's inequality tells you that if you have a random variable $X$ with finite mean $m$ and finite variance $\sigma^2$, then for each $a>0$,

$$P(|X-m|>a) \leq \frac{\sigma^2}{a^2}.$$

This is often written where $a=k \sigma$, in which case you have

$$P(|X-m|>k \sigma) \leq \frac{1}{k^2}.$$

In other words, the probability that the deviation is by $k$ standard deviations is at most $\frac{1}{k^2}$, for any random variable with finite variance. In this respect the standard deviation measures the typical size of a deviation from the mean. One particular case is that the probability of being more than 10 standard deviations away is at most one percent.

Since this situation is so general, this bound is rather weak. In the case of a normal random variable, we have sharper bounds, at least for sufficiently large $k$. Suppose $X$ is normal with mean $m$ and variance $\sigma^2$, and $Z$ is normal with mean $0$ and variance $1$. Here is one estimate of these tail probabilities:

$$P(Z \geq x)=\frac{1}{\sqrt{2 \pi}} \int_x^\infty e^{-y^2/2} dy = \frac{1}{\sqrt{2 \pi}} \int_{x^2}^\infty \frac{1}{2z} e^{-z/2} dz \leq \frac{1}{\sqrt{2 \pi}} \int_{x^2}^\infty e^{-z/2} dz = \frac{2}{\sqrt{2 \pi}} e^{-x^2/2}$$

if $x^2 \geq 1/2$. It follows that $P(|X-m|>k \sigma) \leq \frac{4}{\sqrt{2 \pi}} e^{-k^2/2}$ for any $k \geq \frac{1}{\sqrt{2}}$. Play with this formula with a few values of $k$.

So a normal variable is essentially never more than a few standard deviations away from its mean. On the other hand deviations of a standard deviation or so are rather likely. (Being a bit more concrete, about 92% of the time, a normal variable deviates by more than $0.1$ standard deviations.)

For Brownian motion at each fixed time $t$ you have a normal variable with mean $0$ and variance $t$. So $B_t$ is typically on the order of $\sqrt{t}$. Because we know about the joint distribution of the increments, we can make related statements that hold for entire paths. One is that $B_t$ is almost surely Holder continuous with any exponent $\alpha < 1/2$.

Answer 2

Brownian Motion is defined to have Gaussian increments $B_u - B_v$ with variance $u - v$ and mean $0$. Since $B_0$ is $0$ that means $B_t$ is a Gaussian with variance $t$ and mean $0$. $$E[Var(B_t)] = E[B_t^2] - E[B_t]^2 = t - 0$$

I'll interpret your second statement as saying that the outcome of $B_t$ tends to be on the order of $\sqrt{t}$. I can state this more exactly by observing that $$P(-3 \sqrt{t} \leq B_t \leq 3 \sqrt{t}) \approx 99.73%$$

That comes from properties of normal random variables, it is a $\pm 3$ standard deviation wide confidence interval.