By Seppalainen T.

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**Example text**

2. 13. Suppose [X], [Y ], and either [X +Y ] or [X −Y ] exists. Then [X, Y ] exists, and depending on the case, [X, Y ] = 1 2 [X + Y ] − [X] − [Y ] [X, Y ] = 1 2 [X] + [Y ] − [X − Y ] . or Proof. Follows from ab = 1 2 (a + b)2 − a2 − b2 = 1 2 a2 + b2 − (a − b)2 applied to a = Xti+1 − Xti and b = Yti+1 − Yti . 14. Suppose [X], [Y ], and [X − Y ] exist. 11) 1/2 1/2 [X]t − [Y ]t ≤ [X − Y ]t + 2[X − Y ]t [Y ]t . Proof. 10) follows from Cauchy-Schwarz inequality xi yi ≤ x2i 1/2 yi2 1/2 . From [X − Y ] = [X] − 2[X, Y ] + [Y ] we get [X] − [Y ] = [X − Y ] + 2 [X, Y ] − [Y ]) = [X − Y ] + 2[X − Y, Y ] where we used the equality xi yi − yi2 = (xi − yi )yi .

D. random variables. Define the partial sums by S0 = 0, and Sn = X1 +· · ·+Xn for n ≥ 1. Then Sn is a Markov chain (the term for a Markov process in discrete time). If EXi = 0 then Sn is a martingale. With Markov processes it is natural to consider the whole family of processes obtained by varying the initial state. In the previous example, to have the random walk start at x, we simply say S0 = x and Sn = x+X1 +· · ·+Xn . The definition of such a family of processes is conveniently expressed in terms of probability distributions on a path space.

5) coincide. So we get this corollary. 9. Assume X is continuous and H is closed. Then τH is a stopping time. 10 (A look ahead). The stopping times discussed above will play a role in the development of the stochastic integral in the following way. To integrate an unbounded real-valued process X we need stopping times ζk ∞ such that Xt (ω) stays bounded for 0 < t ≤ ζk (ω). Caglad processes will be an important class of integrands. 6 shows that ζk = inf{t ≥ 0 : |Xt | > k} are stopping times, provided {Ft } is right-continuous.