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发帖数: 702 | 1 对于概率空间(Omega,F,P),
若Omega为countable,
则Omega上的函数列{Xn}的almost surely收敛和in probability收敛等价。
我想由a.s.-->i.p.为显然，但是i.p.-->a.s.不知道怎么证？请达人指教。 | e**********n
发帖数: 359 | 2 Elementrary!
Assume Omega = {w_1,w_2,....}.
a.s. requires that for any epsion, k, there is an N, s.t. |X_n(w_k)} <
epsilon is satified by any n>N.
If such N does not exist, then there is a sequence j_1, j_2 .... s.t. |X_{j_
m} (w_k) |>epsilon.
P(|X_{j_m}|>epsilon ) >= P(|X_{j_m}(w_k)|>epsilon) = P(w_k), which is finite
and contridicts with the assumption of i.p. |
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