Wiener Process $dB^2=dt$

Question

Why is $dB^2=dt$? Every online source I've come across lists this as an exercise or just states it, but why isn't this ever explicitly proved? I know that $dB=\sqrt{dt}Z$, but I don't know what squaring a Gaussian random variable means.

Answer 1

$$dB_t^2 = dt, \qquad (dt)^2 = 0, \qquad dB_t \, dt = 0 \tag{1}$$ are basically rules to simplify the calculation of the quadratic (co)variation of Itô processes - and nothing more:

Let $(B_t)_{t \geq 0}$ a one-dimensional Brownian motion and $(X_t)_{t \geq 0}$ an Itô process, i.e.

$$dX_s = \sigma(s) \, dB_s + b(s) \, ds$$

Then, by Itô's formula,

$$f(X_t)-f(X_0) = \int_0^t f'(X_s) \, dX_s + \int_0^t f''(X_s) \, \sigma^2(s) \, ds. \tag{2}$$

The point is: If we simply apply the rules in $(1)$, we obtain

$$dX_s^2 = (\sigma(s) \, dB_s + b(s) \, ds)^2 = \sigma^2(s) \underbrace{dB_s^2}_{ds} + 2b(s) \sigma(s) \underbrace{ds B_s}_{0} + b^2(s) \, \underbrace{ds^2}_{0} \\ = \sigma^2(s) \, ds.$$

Therefore, we can rewrite $(2)$ in the following way:

$$f(X_t)-f(X_0) = \int_0^t f'(X_s) \, dX_s + \int_0^t f''(X_s) \, dX_s^2$$

i.e. Itô's formula justifies the calculation rules in $(1)$.

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The (mathematical) reason why this works fine can be seen while proving Itô's formula. Actually, one can show that

$$\sum_{j=1}^n g(B_{t_{j-1}}) \cdot (B_{t_j}-B_{t_{j-1}})^2 \tag{3}$$

converges to

$$\int_0^t g(B_s) \, ds$$

as the mesh size $|\Pi|$ of the partition $\Pi=\{0=t_0<\ldots<t_n=t\}$ tends to $0$. This convergence is based on the fact that $B_t^2-t$ is a martingale. On the other hand, comparing $(3)$ with the definition of Riemann-Stieltjes integrals, it's natural to define

$$\int_0^t g(B_s) \, dB_s^2 := \lim_{|\Pi| \to 0}\sum_{j=1}^n g(B_{t_{j-1}}) \cdot (B_{t_j}-B_{t_{j-1}})^2.$$

Consequently,

$$\int_0^t g(B_s) \, dB_s^2 = \int_0^t g(B_s) \, ds.$$

A similar reasoning applies to Itô processes. Note that these integrals, i.e. integrals of the form

$$\int_0^t g(X_s) \, dX_s^2,$$

are exactly the integrals appearing in Itô's formula $(2)$ ($g \hat{=} f''$).

Answer 2

For convenience we will informally regard Brownian motion as a random walk over infinitesimal time intervals of length $\Delta t$, whose increments $$ \Delta B_{t}:=B_{t+\Delta t}-B_{t} \simeq \mathcal{N}(0, \Delta t) $$ over the time interval $[t, t+\Delta t]$ will be approximated by the Bernoulli random variable $$ \Delta B_{t}=\pm \sqrt{\Delta t} $$ with equal probabilities $(1 / 2,1 / 2)$.

The choice of the square root in $(4.2)$ is in fact not fortuitous. Indeed, any choice of $\pm(\Delta t)^{\alpha}$ with a power $\alpha>1 / 2$ would lead to explosion of the process as $d t$ tends to zero, whereas a power $\alpha \in(0,1 / 2)$ would lead to a vanishing process.

Note that we have $$ \mathbb{E}\left[\Delta B_{t}\right]=\frac{1}{2} \sqrt{\Delta t}-\frac{1}{2} \sqrt{\Delta t}=0 $$ and $$ \operatorname{Var}\left[\Delta B_{t}\right]=\mathbb{E}\left[\left(\Delta B_{t}\right)^{2}\right]=\frac{1}{2} \Delta t+\frac{1}{2} \Delta t=\Delta t $$ According to this representation, the paths of Brownian motion are not differentiable, although they are continuous by Property 2, as we have $$ \frac{d B_{t}}{d t} \simeq \frac{\pm \sqrt{d t}}{d t}=\pm \frac{1}{\sqrt{d t}} \simeq \pm \infty $$ After splitting the interval $[0, T]$ into $N$ intervals

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Notes from: https://personal.ntu.edu.sg/nprivault/MA5182/brownian-motion-stochastic-calculus.pdf

Answer 3

For independent random variables, the variance of the sum equals the sum of the variances. So $\mathbb{E}((\Delta B)^2)=\Delta t$, i.e. if you increment $t$ a little bit, then the variance of the value of $B$ before that increment plus the variance of the increment equals the variance of the value of $B$ after the increment.

Or you could say $$ \frac{\mathbb{E}((\Delta B)^2)}{\Delta t} = 1. $$ That much follows easily from the first things you hear about the Wiener process. I could then say "take limits", but that might be sarcastic, so instead I'll say that for a fully rigorous answer, I'd have to do somewhat more work.

Answer 4

Obviously $dB_t^2 \neq dt$, since $dB_t \sim \mathcal{N} (0, dt)$ is a random variable, while $dt$ is deterministic.

As Michael Hardy said, they really meant to say $\mathbb{E} \left[ dB_t^2 \right] = dt$. To convince yourself, compute $$ \mathbb{E} \left[ dB_t^n \right] = \int_{-\infty}^{+\infty} \frac{1}{\sqrt{2 \pi dt}} \exp\left(-\frac{x^2}{2 dt}\right) x^n dx \, .$$