# Basic Stochastic Calculus Question, why does dB^2 = dt?

1. Jun 12, 2013

### saminator910

As the title says, while using Stochastic Calculus, can someone explain some of the properties of differentials?

Why does $dB_t dB_t=dt$

Also, why does $dt dt=0$

and $dB_t dt=0$

I don't really get why these work?

2. Jun 12, 2013

### Mute

A suuuuuper rough heuristic is that when taking limits, $\Delta t \rightarrow 0$ and $\Delta B_t \rightarrow 0$, is that $(\Delta t)^2 \ll \Delta t^2$, and will thus not contribute (leading to $dt dt = 0$), but $\Delta B_t$ is of order $\sqrt{\Delta t}$. Hence, $(\Delta B_t)^2 \sim \Delta t$ and survives in the final stochastic equation ($dB_t dB_t = dt$), but $\Delta B_t \Delta t \sim (\Delta t)^{3/2} \ll \Delta t$, and so does not contribute ($dB_t dt = 0$).

Why exactly $\Delta B_t \sim \sqrt{\Delta t}$, I don't remember off the top of my head. You'll have to consult a stochastic calculus textbook for the rigorous explanation (or maybe someone else here can provide one). Note that I believe these differential identities assume the Ito interpretation - I'm not sure if they would be different for the Stratonovich interpretation.

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