$X_n \overset{a.s.}{\longrightarrow} X$ and $X_n \overset{L^1}{\longrightarrow} Y$ implies $X = Y$ a.s.?

Question

If I have a sequence of random variables $\{X_n\}_{n \geq 0}$ such that $$X_n \overset{a.s.}{\longrightarrow} X \quad\textrm{and}\quad X_n \overset{L^1}{\longrightarrow} Y$$ then is it always true that $X = Y$ a.s.? It certainly seems like it must be true. But if it is not always true, then what conditions must be imposed to make it true?

I feel like there must be an obvious proof, but I can't come up with it right off the bat. Can anyone give a pointer in the right direction?

score 5 · Accepted Answer · answered Mar 30 '13 at 22:06

5

If $X_n \to Y$ in $L^1$ then there exist a subsequence that converges almost surely to $Y$. Since a.s. limits are unique up to a set of measure zero you have that $X=Y$ a.s.

answered Mar 30 '13 at 22:06

Bunder

2,423

Bingo. That was embarrassingly obvious. Thank you! – gogurt Mar 30 '13 at 22:14

score 2 · Answer 2 · edited Apr 13 '17 at 12:20

2

Without involving subsequences:

Note that $X_n \xrightarrow{\text{a.s.}} X \ \text{ and } \ X_n \xrightarrow{L^1} Y$ each imply convergence in probability. Thus, we have $X_n \xrightarrow{\mathbb P} X \ \textrm{ and } \ X_n \xrightarrow{\mathbb P} Y$. And we know that convergence in probability is almost surely unique, i.e. $X \stackrel{\text{a.s.}}{=} Y$.

edited Apr 13 '17 at 12:20

Community

1

answered Oct 01 '14 at 22:16

ruh

513

$X_n \overset{a.s.}{\longrightarrow} X$ and $X_n \overset{L^1}{\longrightarrow} Y$ implies $X = Y$ a.s.?

2 Answers2