S*****i
发帖数: 386 | 1 各位老大，请帮忙看看这个结论对不对？
Let {X_K} and {Y_k} be two sequences of random variables, Y is a random
variable，
if X_K - Y_K --->0 in probability and Y_k ---> Y in distribution,
then X_k ---> Y in distribution?
若是正确的话，请问哪里能找到证明？
谢谢 | B****n
发帖数: 11290 | 2 It can be proved using one of the equivalent definitions for convergence in
distribution.
Xk->Y is distribution if and only if ET(Xk)-ET(Y)->0 for any bounded
continuous function (E means expectation).
I think it is called something like Helly's Theorem.
|ET(Xk)-ET(Y)|<=|ET(Yk)-EY|+|ET(Yk)-ET(Xk)|
The first term on the right side converges to 0 due to the fact that Yk-Y in
distribution.
The second term on the right side converges to 0 because Xk-Yk converges in
probability and T is a continuous bounded function.
【在 S*****i 的大作中提到】
: 各位老大，请帮忙看看这个结论对不对？
: Let {X_K} and {Y_k} be two sequences of random variables, Y is a random
: variable，
: if X_K - Y_K --->0 in probability and Y_k ---> Y in distribution,
: then X_k ---> Y in distribution?
: 若是正确的话，请问哪里能找到证明？
: 谢谢
| S*****i
发帖数: 386 | |
|
