STOCHASTIC

182 BASIL A. KALYMON
PROOF. First we prove that for all I and 3, if r & p then
(52) /i(r|9)--/i(p|«) S (£J-ipV)(r-p)*.
Note that if r £ p S '^(g) for any 2 > 0, then
(53) /,(r I 9) - /,(p I 9) = /,(r | q, YM) - /,(p | 9, KB) = 0
so that (52) is satisfied. Also, if n*(g) > p, then
/i(r|«) -/I(P|9) -/i(r|9) - f'(P I 3. YK)
(54) g Mr I g, FK) - MP I 9, K,)
= (r - p)x + aiEit,,[fu-i(r I gi) - fi-^p | g,)].
But by Lemma 1, fi-i{r | gi) — /i_i(p | gi) £ 0 for all realizations of gi since r £ p,
so that by (54) and (53), we have that for r & p and any g, Z > 0,
(55) Mr I 9) - /I(P I 9) S (r - p)s + mE^lfUr | gi) - /(p I «i)l;
applying (55) recursively we get
fi(r\q) -flip I 9) £ (r - p)a: + aiB|,,i[(r - p)z + ofi_i£|9I.i_i[/i-s(r | g,)
- /;_2(p I g2)]]
(56) = (1 + 01) (r - p)x + aiaj-i#i,,i[/i_,(r | g2) - /,_2(p | g2)]
S (2Z'-I(3/)(J- - p)x + £,,..[/o(r I g,) - /„(p | g,)]
= HUPi\r- P)X,
where the final equality is due to the fact that/o(r \qt) = 0 = /0(p | gi).
Since (56) was established for any g, I > 0, we have
(57) /(r I 9) - /„(P | g) £ (££ fr 1 )(r ~ p)x
and hence
(58)
Mr I P, YK) - /«(r I p, Ys) = (r - p)x - (L< + etx) + aiE{„