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The time step typically depends on the context. Due to the self-similarity of Brownian motion the mathematics should work similarly on any time scale, although the resultant estimates might vary greatly (as you mention). Since the cited article assumes a "high-frequency market maker," the implied time step seems to be the shortest time step available or ...


In one sentence, time value has to do with the probability of crossing the strike before expiration (whether from below or above). Doesn’t matter whether the crossing results in the option being in the money or not.


Consider sequences of n flips with p=1/2 for simplicity. The sets defined based only on the first flip obviously divide the space in half, so each has positive probability (1/2) no matter what n is. But consider the outcome A (a singleton set) defined by seeing n heads. If n=2, P(A)=.25. More generally, P(A)=0.5^n, which goes to zero as n goes to infinity. ...


It can be seen that $Y_1^2+Y_2^2=-2\log{X_2}$ and that $Y_2 \over Y_1$ $=\tan(2\pi X_1)$. Therefore $X_1={1 \over{2 \pi}}{\arctan{Y_2 \over Y_1}}$ and $X_2=\exp{-(Y_1^2+Y_2^2) \over 2}$. Taking differential to get $dX_1= {1 \over{2\pi}}{{-Y_2dY_1+Y_1dY_2} \over{Y_1^2+Y_2^2}}$. Similarly, $ dX_2= {\exp {-{Y_1^2+Y_2^2} \over 2}(Y_1 dY_1 + Y_2dY_2 )}$. ...

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