If(si ,0) f(s ,0)| + JAR.fI If(si+ ,ti) f(si,t )I
 1

 = If(sit ) f(s i+ ,t t )

AR~fl = f(si+1,t ) f(s i+ ,t +1) f(0,t ) + f(O,ti+l


 S f(si+ ',ti) f(si+1,ti+1) lf( ,ti) f(o,ti+ ) ,'


 f(0,tC ) f(0,ti+l)1 + (IAR fj > If(si+ ,1ti) f(si+ ',ti+1) .
 1

Here we diverge from the proof of (1), since the rectangles Ri, Ri

overlap (see Figure 2-12). From the first inequality we have, upon

summing over i,


j-1
SIf(s ,ti) f(si+1 ,t )
i=1


 j-1
 5 i ( R fI + If(si+1,0) f(si,0)p)


 j-1 j-1
 IE R f + E If(si+1,0) f(si'O)
 i=1 1 i=1


S Var[(,),(s,1 (f) + Var s1s]f(.,0)



 as in the proof of (1)


SI f (s,t )


(as in the proof of (1)).


Also, by a similar computation, we have