P is convex let a e 0 1 recall holders inequality

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<P is convex. Let a e [0 , 1]. Recall Holder's inequality from Subsection 6.5.2: If p > 0, q > 0, p- 1 + q- 1 = 1, then Set p = 1/a, and q = 1/(1- a) and we have <jJ(au 1 + (1-a)uz) = logE(eau1Y1e(l-a)u2Y1) E(e"JYJ)) a ( E(e"2Y1)) 1-a = a</J(ut) + (1 - a)</J(uz) . 2. The set {u : </J(u) < oo} = : [¢ < oo) is an interval containing 0. (This interval might be [0, OJ = {0}, as would be the case if Yt were Cauchy distributed). If u1 < uz and </J(u;) < oo, i = 1, 2, then for 0 1, So if u; e [¢ < oo), i = 1, 2, then [ut. uz} C < oo ]. Note ¢(0) = logE (e 0 Y1) = log 1 = 0. 3. If the interior of [ <P < oo] is non-empty, <P is analytic there, hence infinitely differentiable there, and so ¢' (0) = E (Yt). One may also check that ¢"(0) = Var(Yt). 4. On[¢< oo), <Pis strictly convex and <P' is strictly increasing.
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400 10. Martingales 10.14.1 The Basic Martingales Here is a basic connection between martingales and the random walk. Proposition 10.14.1 For any u e [</> < oo 1 define euXn Mn(u) = exp{uXn- n<J>(u)} = ( )n. E(e"Yl) Then {(Mn(u), Bn). n e N} is a positive martingale with E(Mn(u)) = 1. Also, Mn(u)--. 0 a.s . as n--. oo and hence {Mn(u)} is a non-regular martingale. Proof. The fact that {Mn(u)} is a martingale was discussed earlier. See item 4 of Section 10.5. Now we check that Mn(u) --. 0 almost surely as n --. oo . We have that u e [</> < oo] and 0 e [</> < oo] implies e [</> < oo], and by strict con- vexity (10.50) Also is a positive martingale and L1-bounded, so there exists a random variable Z such that and = exp{uXn- Z 2 < oo. Therefore Mn(u) = exp{uXn- n<J>(u)} u u = exp{uXn - 2n<J>(2) + n[2</>( 2)- </>(u)]} 2 u 1 = (X + o(1)) exp{n[2(</>( 2 ) - z<J>(u) )]} --.o !<J>(u) < 0 from (10.50). 0 MORE MARTINGALES. From Proposition 10.14.1 we can get many other mar- tingales. Since </> is analytic on [ </> < oo ], it is also true that u exp{ux- n<J>(u)} is analytic. We may expand in a power series to get oo uk exp{ux- n<J>(u)} = L- fk(n, x). k=O k! (10.51)
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10.14 Wald's Identity and Random Walks 401 If we expand the left side of (10.51) as a power series in u, the coefficient of uk is fk (n, x) I k!. As an example of this procedure, we note the first few terms: and fo(n,x) = exp{ux- n<f>(u)Jiu=O = 1 a ft(n,x) =au exp{ux -n<f>(u)Jiu=O = exp{ux- n<f>(u)}(x- n<f>'(u))iu=O = (x -nEYt) az f2(n,x) = auz exp{ux -n<f>(u)Jiu=O = aau {eux-mp(u)(x- n<f>'(u))Jiu=O = eux-mp(u)(-n<f>"(u)) + eux-mp(u)(x- n<f>'(u))21u=0 = (x- n£(Yt)) 2 - nVar(Yt). Each of these coefficients can be used to generate a martingale. Proposition 10.14.2 For each k 2: 1,{(/k(n, Xn), Bn), n :::: 0} is a martingale. In particular k = 1, k=2, are martingales. {(ft (n, Xn) = Xn- nE(Yt) = Xn- E(Xn), Bn), n eN} {((Xn- E(Xn)) 2 - Var(Xn). Bn), n eN} (lfVar(Yt) = a 2 and E(Yt) = 0, then {X;- na 2 } is a martingale.) For the most important cases where k = 1 or k = 2, one can verify directly the martingale property (see Section 10.5). Thus we do not give a formal proof of the general result but only give the following heuristic derivation. From the martingale property of {Mn(u)} we have form < n that is, E (exp{uXn- n<f>(u)}IBm) = e"Xm-m4J(u) so that
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402 10. Martingales Now differentiate inside E( IBm) k times and set u = 0. This needs justification which can be obtained from the Dominated Convergence Theorem. The differen- tiation yields 0 10.14.2 Regular Stopping Times We will call the martingale {(Mn(u), Bn), n eN}, where Mn(u) = exp{uXn- ucf>(u)} = e"Xn j(Ee"Y' )n the exponential martingale. Recall from Proposition 10.14.1 that if u ::/; 0, and u e [4> < oo), then Mn(u) -..
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