REVIEW Example 6 v wfh goby b<l: fy t ufh goby b-l fy 2 Y, f e =-]2(=) e Thad Z h pgo byb小 fh pgo byb-l fy P(z2>LeP(znn2)ex→t+ So Buf E( Znl e t×(v em→∞+ yh, mpl, fog fh, afuf, m,.f fhyf M. h, 9g, N< pgobyb-l fy do, d. of amply m,y. dqugg, 1. h, 9g, NEREVIEW 4 Example 6 Zn =
1 with probability 1 2 0 with probability 1 2 Z1, · · · , Zn are independent. Let Zn = 1 n n i=1 Zi P     Zn − 1 2     > ε = P     1 n  Zi − 1 2     > ε ≤ E  1 n  Zi − 1 2 2 ε 2 = 1 n 2 n i=1 E  Zi − 1 2 2 = 1 n 2 × n × 1 2  1 − 1 2 → 0. Thus, Zn P→ 1 2 . Example 7 Zn =
0 with probability 1 − 1 n n 2 with probability 1 n P (|Zn| > ε) = P  Zn = n 2  = 1 n → 0. So Zn P→ 0. But E (Zn) = 0 ×  1 − 1 n + n 2 × 1 n = n → ∞. So we have an example for the statement that convergence in probability does not imply mean square convergence.