jn1 , jn2
jn1 , jn2
→ 0 in probability;
→ 0 in probability.
(2.37)
(2.38)
In order to prove (2.35) and (2.36), we need to control the differences in s and t.
Let us show (2.35). We first control |ξn (s, t) − ξn (s, t0)| for t0 − t > 1/n. By definition of ξn,
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ξn (s, t0) 6
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(as (Xi)
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−3/[
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(as (Xi)
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−
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[nt0]
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[nt]
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i=1
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i=1
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+ [ 3/02]
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(bs (Xi)
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[ 3/2]
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n
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−
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− n
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i=[nt ]+1
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nt
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X0
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nt
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