Hello,
Let be such that
And let almost surely.
Then we have and
So the condition is satisfied.
But almost surely. Thus and can't have a limit equal to 0
Let X be a random variable. Let Xn be a sequence of random variables.
Then E(|Xn - X|) -> 0 as n->∞ implies E(Xn) -> E(X) as n->∞.
[I can prove this by |E(W)|≤E(|W|) and squeeze theorem.]
Is it also true that E(Xn) -> E(X) implies E(|Xn - X|) -> 0??
How can we prove it?
Any help is much appreciated!