Statistics and Probability Let W(t) be the standard Brownian motion, and let X(t) = t W(1/t) fort> 0, X(O) = 0.- Show that the covariance (Cov) function of X(t) is the same as the covariance function of W(t): Cov(X(t); X(s)) = Cov(W(t); W(s)) for all s; t > 0. Assuming that the paths of X(t) are continuous with probability 1, argue that X(t) is standard Brownian motion?
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Explanation:
They used the Karhunen–Loève representation Theorem to derive Brownian path continuity. In order to determine random variables in distribution via characteristic functions, we always require the characteristic function E(eXt) to be continuous around the origin. So, axiom 4* ensures the existence of some transform?
Can we design a stochastic process W(t) that is not a Brownian motion (which is defined as a stochastic process with axiom 1,2,3,4 satisfied OR axiom 1,2,3,4* satisfied in [Karlin&Taylor]) if we merely assume axiom 1,2,3 on a stochastic process as above? Is the continuity axiom, on the other hand, redundant? (I don't believe so, but it's not clear how I'm supposed to accomplish it.)
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