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

  1. Jun 12, 2013 #1
    As the title says, while using Stochastic Calculus, can someone explain some of the properties of differentials?

    Why does [itex]dB_t dB_t=dt[/itex]

    Also, why does [itex]dt dt=0[/itex]

    and [itex]dB_t dt=0[/itex]

    I don't really get why these work?
     
  2. jcsd
  3. Jun 12, 2013 #2

    Mute

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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.
     
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