# New PDF release: Algebre, solutions developpees des exercices, 1ere partie,

By Mac Lane, Birkhoff (WEIL, HOCQUEMILLER)

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Extra resources for Algebre, solutions developpees des exercices, 1ere partie, ensembles, groupes, anneaux, corps [Algebra]

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0 But this is already automatically true as the ψn are orthogonal. Similarly, 1 E(A2n ) = ψn2 (s) ds = 1. 0 42 by (3) Consequently if the An are independent and N (0, 1), it is reasonable to believe that formula (4) makes sense. But then the Brownian motion W (·) should be given by ∞ t (5) ξ(s) ds = W (t) := t An 0 ψn (s) ds. 0 n=0 This seems to be true for any orthonormal basis, and we will next make this rigorous by choosing a particularly nice basis. ´ LEVY–CIESIELSKI CONSTRUCTION OF BROWNIAN MOTION DEFINITION.

Xn )g(x1 , t1 | 0)g(x2 , t2 − t1 | x1 ) −∞ . . g(xn , tn − tn−1 | xn−1 ) dxn . . dx1 . For the second equality we recalled that the random variables Yi = W (ti ) − W (ti−1 ) are independent for i = 1, . . , n, and that each Yi is N (0, ti − ti−1 ). We also changed variables using the identities yi = xi − xi−1 for i = 1, . . , n and x0 = 0. The Jacobian for this change of variables equals 1. BUILDING A ONE-DIMENSIONAL WIENER PROCESS. The main issue now is to demonstrate that a Brownian motion actually exists.

This assertion partly justiﬁes the heuristic idea, introduced in Chapter 1, that dW ≈ (dt)1/2 . 60 Proof. Set Qn := mn −1 n k=0 (W (tk+1 ) − W (tnk ))2 . Then mn −1 Qn − (b − a) = ((W (tnk+1 ) − W (tnk ))2 − (tnk+1 − tnk )). k=0 Hence mn −1 mn −1 E((Qn − (b − a)) ) = E([(W (tnk+1 ) − W (tnk ))2 − (tnk+1 − tnk )] 2 k=0 j=0 [(W (tnj+1 ) − W (tnj ))2 − (tnj+1 − tnj )]). For k = j, the term in the double sum is E((W (tnk+1 ) − W (tnk ))2 − (tnk+1 − tnk ))E(· · · ), according to the independent increments, and thus equals 0, as W (t) − W (s) is N (0, t − s) for all t ≥ s ≥ 0.