STOCHASTIC

Since Z"=o^( w »0 = 0 and/(«,«) = 0 the last expression is equal to
« — 1 oo oo n — 1
£6(iM) I t k £ ft(„,i) X t f(n,iflk\ = £ I,
i=0 k=l fc = l i=0
k f{n,i?lh\ = H t k b{n,i)f{n,iflk\
= £ 1/*! £ t k b(n,i)f(n,iy
k=\ i=0
JAMES A. OHLSON
Consider now Y*1=Q t k b(n,i)f(n,if = S(n,k). Applying the binomial expansion
for/(n, i) k gives
S(n,k) = "£t k b(n,i) I l k )fi k - j (i° 2 ) J (n-if +J
>=o j=o\J/
= ** I (^^-•'(^ 2 y "i*(».o(»-i-) t+j .
j=0\J/ i=0
Lengthy but straightforward induction will show that 6
and since
[=0 if k+j0 if/k+y=f|>
i( k )n k - J (W) j >0 and 7 = 0,1,...,*,
.7 = 0 W/
we have that S(n, k) = 0 if and only if 2k _ «. Consequently,
and
£(*- 1)" = t" l2 t)„ + higher powers of t = 0(?" /2 ), n = 4,6,...,
£(jr-l) B = t in+l)/ \ + higher powers of? = 0(? ( " +1)/2 ),
n = 3,5,..., with r\„ > 0.
The condition JX > 0 is not crucial; for yu _ 0 the theorem still holds if we
read 0(t") as "at least of order t n " for all odd n = 3. (For n = -f