Is is true that if $E|X_n - X| \to 0$ then $E[X_n] \to E[X] $?

Question

My question is motivated by the following problem:

Show that if $|X_n - X| \le Y_n$ and $E[Y_n] \to 0$ then $E[X_n] \to E[X]$.

I started off by saying that since $$|X_n - X|\ge 0 $$ then $$E[|X_n - X|]\ge 0 $$ At the same time $$E[|X_n - X|]\le E[Y_n] $$

and so

$$0 \le E[|X_n - X|] \le E[Y_n]$$

By the squeeze theorem then $E[|X_n - X|] \to 0$. I don't know how to proceed from here.

I know that if $E[|X_n - X|] = 0$ then I can set up a contradiction, like so:

Suppose that $|X_n - X| = c$, $c \ne 0$ and $E[|X_n - X|] = 0$. Then $$E[|X_n - X|] = E[c] = c \ne 0. $$

But I don't know how to show this would hold in the limit. I am planning to use this to show that $X_n \to X$ and therefore $E[X_n] \to E[X]$

score 8 · Accepted Answer · answered Sep 21 '14 at 00:01

8

$\vert E(X) - E(X_n) \vert =\vert E(X-X_n) \vert \leq E(|X-X_n|) \to 0$

answered Sep 21 '14 at 00:01

yes

by Jensen's inequality – Seyhmus Güngören Sep 21 '14 at 00:05
1

It's simpler than Jensen inequality; $-|Y| \leq Y \leq |Y|$ then by monotony and linearity of expectation $-E|Y| \leq E(Y) \leq E|Y|$ hence $|E(Y)|\leq E|Y|$. – yes Sep 21 '14 at 00:11

score 2 · Answer 2 · answered Sep 21 '14 at 00:05

2

How about trying by contrapositive: If $E[X_n] \not\to E[X]$, then by linearity, $E[X_n-X]\not\to 0$. And your result should follow.

answered Sep 21 '14 at 00:05

paw88789

Can you elaborate please? How are you applying linearity here? – badmax Sep 21 '14 at 00:10
1

I'm using $E[X_n-X]=E[X_n]-E[X]$ – paw88789 Sep 21 '14 at 00:11

2 Answers2