By Michael J. Kozdron

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Let g : [0, ∞) → R be a bounded, continuous function in L2 ([0, ∞)). If g is differentiable with g also bounded and continuous, then the integration-by-parts formula t t g(s) dBs = g(t)Bt − 0 g (s)Bs ds 0 holds. Remark. Since all three objects in the above expression are random variables, the equality is interpreted to mean that the distribution of the random variable on the left side and the distribution of the random variable on the right side are the same, namely t N g 2 (s) ds . 1) 0 is the Riemann integral of a function of Brownian motion.

Suppose that we choose t = 1 and g(s) = s. The integration-by-parts formula implies that 1 1 s dBs = B1 − 0 Bs ds. 0 If we now write 1 B1 = dBs 0 and use linearity of the stochastic integral, then we find 1 1 Bs ds = B1 − 0 1 0 1 dBs − s dBs = 0 Since 1 (1 − s) dBs 0 1 (1 − s) dBs . s dBs = 0 0 Calculating Wiener Integrals 43 is normally distributed with mean 0 and variance 1 0 1 (1 − s)2 ds = , 3 we conclude that 1 Bs ds ∼ N (0, 1/3). 0 Thus, we have a different derivation of the fact that we proved in Lecture #8.

Thus, we are forced to conclude that B1 and 1 s dBs 0 are not independent. Suppose that g and h are bounded, piecewise continuous functions in L2 ([0, ∞)) and consider the random variables t It (g) = g(s) dBs 0 and t It (h) = h(s) dBs . 0 As the previous example suggests, these two random variables are not, in general, independent. Using linearity of the Wiener integral, we can now calculate their covariance. Since t t g(s) dBs ∼ N It (g) = g 2 (s) ds , 0, 0 0 t t h(s) dBs ∼ N It (h) = h2 (s) ds , 0, 0 0 and t t [g(s) + h(s)] dBs ∼ N It (g + h) = 0 [g(s) + h(s)]2 ds , 0, 0 and since Var(It (g + h)) = Var(It (g) + It (h)) = Var(It (g)) + Var(It (h)) + 2 Cov(It (g), It (h)), we conclude that t t [g(s) + h(s)]2 ds = 0 t g 2 (s) ds + 0 h2 (s) ds + 2 Cov(It (g), It (h)).

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