For brownian motion, why $dW_{t}=\epsilon\sqrt{dt}$?

Question

I understand ϵ is sampled from Standard Normal distribution (since standard brownian motion)

but why does this make sense?

$$ dW_{t}=\epsilon\sqrt{dt} $$

or the discrete form:

$$ ΔW_{t}=\epsilon\sqrt{Δdt} $$

is there anything to do with quadratic variation?

Also what is this formula called?

Thank you :)

Answer 1

Here is the explication why $\Delta W_t \stackrel{\mathcal{D}}{=}\epsilon \sqrt{\Delta t}$ (or $dW_t \stackrel{\mathcal{D}}{=}\epsilon \sqrt{dt}$ )

$$\Delta W_t :=W_{t+\Delta}-W_t\stackrel{\mathcal{D}}{=}\mathcal{N}(0,\Delta t)=\sqrt{\Delta t}.\mathcal{N}(0,1)=\sqrt{\Delta t}.\epsilon$$

or

$$dW_t :=W_{t+dt}-W_t\stackrel{\mathcal{D}}{=}\mathcal{N}(0,dt)=\sqrt{dt}.\mathcal{N}(0,1)=\sqrt{dt}.\epsilon$$

Here, we used the third property of Wiener process

$$W_{t+u}-W_t\stackrel{\mathcal{D}}{=}\mathcal{N}(0,u)$$