# Math Help - E(Xn) -> E(X) implies E(|Xn - X|) -> 0??

1. ## E(Xn) -> E(X) implies E(|Xn - X|) -> 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!

2. Hello,

Let $X_n$ be such that $P(X_n=-1)=\frac 12=P(X_n=1)$

And let $X\equiv 0$ almost surely.

Then we have $E(X_n)=0$ and $E(X)=0$
So the condition $E(X_n)\to E(X)$ is satisfied.

But $|X_n-X|=1$ almost surely. Thus $E|X_n-X|=1$ and can't have a limit equal to 0