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Course in Probability Theory

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292 1 RANDOM WALKBy hypothesis, the left member above is the Fourier transform of a f<strong>in</strong>ite signedmeasure v 1 with support <strong>in</strong> (0, oc), while the right member is the Fouriertransform of a f<strong>in</strong>ite signed measure with support <strong>in</strong> (-oc, 0] . It follows fromthe uniqueness theorem for such transforms (Exercise 13 of Sec . 6 .2) thatwe must have v 1 - v 2 , and so both must be identically zero s<strong>in</strong>ce they havedisjo<strong>in</strong>t supports . Thus p1 = pi and q l = q* . To proceed by <strong>in</strong>duction on n,suppose that we have proved that p j - pt and q j - q~ for 0 < j < n - 1 .Then it follows from (10) thatpn (t) + qn (t)_- pn (t) + qn (t) •Exactly the same argument as before yields that pn - pn and qn =- qn . Hencethe <strong>in</strong>duction is complete and the first assertion of the theorem is proved ; thesecond is proved <strong>in</strong> the same way .Apply<strong>in</strong>g the preced<strong>in</strong>g theorem to (8), we obta<strong>in</strong> the next theorem <strong>in</strong>the case A = (0, oo) ; the rest is proved <strong>in</strong> exactly the same way .Theorem 8 .4 .2 . If a = aA is the first entrance time <strong>in</strong>to A, where A is oneof the four sets : (0, oo), [0, oc), (-oc, 0), (-oo, 0], then we have1 - c`{ r a e `rs~} = exp - e`tsnd 7 ;r n eitsn1 = exp +n=1 n {S„EA`}elrsn d~P .From this result we shall deduce certa<strong>in</strong> analytic expressions <strong>in</strong>volv<strong>in</strong>gthe r .v . a . Before we do that, let us list a number of classical Abelian andTauberian theorems below for ready reference .(* )(A) If c, > 0 and En°- o C n r n converges for 0 < r < 1, then00lim c n rn =rT1 11=0 n=0f<strong>in</strong>ite or <strong>in</strong>f<strong>in</strong>ite .(B) If Cn are complex numbers and En°_ o c, rn converges for 0 < r < 1,then (*) is true .(C) If Cn are complex numbers such that Cn = o(n _1 ) [ or just O(n -1 )]as n --+ oc, and the limit <strong>in</strong> the left member of (*) exists and is f<strong>in</strong>ite, then(*) is true .

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