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

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In stochastic calculus, the property dB_t dB_t = dt arises from the behavior of Brownian motion, where the increment dB_t is of order √dt, leading to the result that the square of the increment behaves like dt. Conversely, the product dt dt = 0 reflects that the infinitesimal time intervals are negligible when squared. Additionally, the term dB_t dt = 0 indicates that the product of a stochastic increment and an infinitesimal time interval is insignificant. The discussion suggests consulting a stochastic calculus textbook for a rigorous explanation of these properties, particularly under the Ito interpretation, which may differ from the Stratonovich interpretation. Understanding these properties is crucial for applying stochastic calculus correctly in mathematical finance and other fields.
saminator910
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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?
 
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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.
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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