fayllar.orgma'muriyatiga murojaat qiling
→ 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,