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How do you prove that if X converges in probability and expectation that this implies convergence in mean? I think I have to use Chebyshev's Inequality, but am not sure how to incorporate the expectation convergence. Thanks

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What is the difference between convergence in expectation and convergence in mean? –  Did Nov 8 '12 at 21:26
    
Xn--> X in l1 is convergence in mean, and E[Xn]--> E[X] is convergence in expectation. Sorry I don't know of a good way to type this in using Latex. –  ben Nov 8 '12 at 23:10
    
Beware that your terminology is somewhat non standard. Anyway, the result does not hold, see my answer. –  Did Nov 9 '12 at 5:51

1 Answer 1

Here is an example which might shed some light on this question. Assume that $X_n=n$, $X_n=-n$ or $X_n=0$ with respective probabilities $a_n$, $a_n$ and $1-2a_n$. Then:

  1. $X_n\to0$ in probability if and only if $a_n\to0$.
  2. $\mathbb E(X_n)=0$ for every $n$, for every $(a_n)$.
  3. $X_n\to0$ in $L^1$ if and only if $na_n\to0$.

In particular, $X_n\to0$ in probability and $\mathbb E(X_n)=0$ for every $n$ do not imply that $X_n\to0$ in $L^1$.

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