Optimal stopping
The reason is the one step ahead strategy. We work by induction onT.LetSn(T)$n
Hn(T)Xn(T)o
$n
HnXn(T)o
.AsHn(T)=maxknξkis not changing.withT,but if we increaseTthen
asHTT++ 11 HTT therefore obviouslyXn(T)Xn(T+^1 ). ThereforeSn(T+^1 ) $n
Hn(T+^1 )Xn(T+^1 )o
==n
Hn(T)Xn(T+^1 )o
n
Hn(T)Xn(T)o
=Sn(T)One must prove thatSn(T)Sn(T+^1 ).