Course in Probability Theory

298 1 RANDOM WALKlimit theorem, and certainly O(11n) in any event, proposition (C) above willidentify it as the right member of (23) with A = (0, oc) .Now by analogy with (27), replacing (0, cc) by (-oo, 0] and writinga(_oc,o] as 8, we have(28) ({S p} = v lim exp n 1 J'(S,, < 0)rTl n 2n=1Clearly the product of the two exponentials in (27) and (28) is just exp 0 = 1,hence if the limit in (27) were +oc, that in (28) would have to be 0 . Butsince c% (X) = 0 and (`(X2) > 0, we have T(X < 0) > 0, which implies atonce %/'(Sp < 0) > 0 and consequently cf{S f } < 0 . This contradiction provesthat the limits in (27) and (28) must both be finite, and the theorem is proved .Theorem 8 .4 .7 . Suppose that X # 0 and at least one of (,(X+) and (-,''(X- )is finite ; and let a = a ( o , oc ) , , = a (_,, o 1 .(i) If (-r(X) > 0 but may be +oo, then ASa ) =(ii) If t(X) = 0, then (F(Sa ) and o (S p) are both finite if and only if(~, '(X 2 ) < 00 .PROOF. The assertion (i) is a consequence of (18) and Wald's equation(Theorem 5 .5.3 and Exercise 8 of Sec . 5 .5) . The "if' part of assertion (ii)has been proved in the preceding theorem ; indeed we have even "evaluated"(Sa ) . To prove the "only if' part, we apply (11) to both a and ,8 in thepreceding proof and multiply the results together to obtain the remarkableequation :(29) [ 1 - c { . a e tts a}]{l - C`{rIe`rs0}] = exp -- f (t) n = 1 - r f (t) .Setting r = 1, we obtain for t ; 0 :1 - f (t) 1 - c'{eitS-} 1 - (`L{el tSf}t 2 -it +itLetting t ,(, 0, the right member above tends to (-,"ISa}c`'{-S~} by Theorem 6 .4 .2 .Hence the left member has a real and finite limit . This implies (`(X2) < 00 byExercise 1 of Sec . 6 .4 .8 .5 ContinuationOur next study concerns the r .v .'s M n and M defined in (12) and (13) ofSec . 8 .2. It is convenient to introduce a new r.v . L, which is the first time