Course in Probability Theory

6 .6 MULTIDIMENSIONAL CASE ; LAPLACE TRANSFORMS 1 201Taylor's theorem, with the remainder term in the integral form, we have(6) f (~) = (~ - ~)k-I f(j)(µ)Jj=0 J '+ (~- µ)k 1 (1 - t)kIf (k) (µ + (~ - µ)t) dt .(k-1)! 0Because of (4), the last term in (6) is positive and does not exceed~- k 1( µ)I (1 - t)k-1f(1)(µ + (Xo - µ)t)dt .(k - 1) ! oFor if k is even, then f (k) 4, and (~ - µ) k > 0, while if k is odd then f (k) Tand (~ - µ) k < 0 . Now, by (6) with ). replaced by ) . o , the last expression isequal to(7);~4 _µ - µ/k-iµ)k- f(j)f (4)(4 - A) Jj=0 J •F, (x) =[nx]n (-1)J f (j) (n) .j=0 j!~_µ(X0 -µkf (, 0 ),where the inequality is trivial, since each term in the sum on the left is positiveby (4) . Therefore, as k -a oc, the remainder term in (6) tends to zero and theTaylor series for f (A) converges .Now for each n > 1, define the discrete s .d .f. F n by the formula :This is indeed an s .d .f., since for each c > 0 and k > 1 we have from (6) :k-1 f(j)()T,j I1 = f (0+) > f (E) > n (E - n )Jj=0Letting E ~ 0 and then k T oo, we see that F,, (oc) < 1 . The Laplace transformof F„ is plainly, for X > 0 :\°O`Je - 'xdF 7 ,(x) = -~(jl~i) ( '~) f(j)(n )Lej=0`~~ - .l/») - n) J f (J) (n) = f(n (1 - e -Zl11 )),=L li (n(1 - ej=0the last equation from the Taylor series . Letting n --a oc, we obtain for thelimit of the last term f (A), since f is continuous at each A . It follows from