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Channel: What does it mean that $X_T$ is a solution to a Brownian Motion driven SDE ($dX_t = ....$)? - Mathematics Stack Exchange
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Answer by user658409 for What does it mean that $X_T$ is a solution to a...

There's actually two definitions of solution to SDE. Strong and weak. Strong solution. Given a probability space $(\Omega, \mathcal F, \mathcal F_t,P)$ and a Brownian motion $B(t)$ on that space...

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What does it mean that $X_T$ is a solution to a Brownian Motion driven SDE...

Please consider the$$X_t = X_0 + \int^t_of(X_s,s)ds+\int^t_og(X_s,s)dB_s$$ where $B_t$ is Brownian Motion. This can also be expressed as:$$dX_t=f()dt+g()dB_t$$What does it mean that $X_T$ is a solution...

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