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I know convergence in probability does not imply convergence in measure. I would like to see some simple example. Do you have any ideas please?

aar43
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  • Referring to the question in the title; the question in the body doesn't make sense. –  Jul 23 '15 at 17:02

1 Answers1

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Let $\{ X_n \}_{n\geq 1}$ be a sequence of independent random variables such that $X_n=1$ with probability $1/n$ and $X_n=0$ otherwise. Then $$ \mathbb P(X_n > 0)=1/n\rightarrow 0 $$ and we have convergence in probability to $0$. However, for any small $\epsilon > 0$ $$ \sum^\infty_{n=1} \mathbb P(X_n>\epsilon) = \sum^\infty_{n=1} \mathbb P(X_n=1) = \sum^\infty_{n=1} \frac{1}{n} = \infty. $$ By Borel-Cantelli this implies $\mathbb P(X_n=1 \quad\text{ i.o.})=1$. But $X_n \rightarrow 0$ a.s. if and only if $\mathbb P(X_n>\epsilon \quad \text{i.o}) = 0$, hence we don't have a.s. convergence to $0$.

  • You're assuming the $X_n$ are independent; should probably say so. Also in general saying $P(X_n>0 \text{ i.o.})=1$ does not imply that $X_n$ does not tend to $0$ as; here you say instead that $P(X_n=1 \text{ i.o.})=1$. – David C. Ullrich Jul 23 '15 at 17:00
  • @DavidC.Ullrich In general, $X_n\rightarrow 0$ a.s. if and only if $P(X_n > \epsilon \quad \text{i.o.})=0$. In this case it translates to $X_n\rightarrow 0$ a.s. if and only if $P(X_n =1 \quad \text{i.o.})=0$. So since $P(X_n =1 \quad \text{i.o.})=1$ we don't have a.s. convergence. You are right the statement doesn't hold in general but I haven't claimed it to be either. –  Jul 23 '15 at 17:15
  • Well right. I didn't say what you said was wrong; modifying the way you put it would perhaps make it more clear. (And you really do need to add independence!) – David C. Ullrich Jul 23 '15 at 17:18
  • @DavidC.Ullrich Made it pretty now –  Jul 23 '15 at 17:37
  • Lovely. I feel much better, thanks.... – David C. Ullrich Jul 23 '15 at 17:54