- #1
operationsres
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We all know that [itex]\int_0^t dB(s) = B(t)[/itex], where [itex]B(t)[/itex] is a standard Brownian Motion. However, is this identity true?
[itex]\int_{t_1}^{t_2} dB(s) = B(t_2) - B(t_1)[/itex]
[itex]\int_{t_1}^{t_2} dB(s) = B(t_2) - B(t_1)[/itex]