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I'm struggling with the next problem:

"Decide reasonably whether the following statement is true or false: Let X be a random variable and $\{X_{n}\}_{n \in \mathbb{N}}$ a family of random variables.

$X_{n} \rightarrow X$ in probability if and only if $\lim_{n\to\infty} \mathbb{E}(\frac{|X_{n}-X|}{1+|X_{n}-X|})=0$ "

My intuition is that the statement is true, and my current approach is: $\lim_{n\to\infty} \mathbb{E}(\frac{|X_{n}-X|}{1+|X_{n}-X|})=\mathbb{E}(\lim_{n\to\infty}\frac{|X_{n}-X|}{1+|X_{n}-X|})$ because $X_{n}$ and $X$ are measurable by definition of random variable. Now, $\lim_{n\to\infty}|X_{n}-X| = 0$ by convergence in probability, then $\mathbb{E}(\lim_{n\to\infty}\frac{|X_{n}-X|}{1+|X_{n}-X|}) = 0$. The reciprocal is equivalent, following the same argument.

I'm not sure that this proof is complete or it has any flaw. If my intuition failed, the statement could also by false. So I would appreciate any help.

Thanks in advance!

Hallow
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    https://math.stackexchange.com/questions/533866/convergence-in-probability-induced-by-a-metric/534004 – cqfd Mar 14 '20 at 08:33
  • Thank you Thomas! I tried to look up this question before posting but I didn't find anything. – Hallow Mar 14 '20 at 08:42

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First of all you are not mentioning why you may exchange the limit and the expectation. (you may, but why?)

Then you are mixing convergence in probability and almost sure convergence. You cannot follow $$E[\lim_{n\to 0}Y_n] = 0$$ if the limit is only in probability and not almost sure.

Good news: The statement is true. To see this let denote $$Y_n := \frac{|X_{n}-X|}{1+|X_{n}-X|}$$ consider that $Y_n = |Y_n| \le 1$ and show $$ X_n \to X \text{ in probability } \iff Y_n \to 0 \text{ in probability }$$ Then for the "if" part use that for any $\tilde{\varepsilon}$ you have a suitable $\varepsilon$ s.t. $$Y_n = Y_n1_{\{Y_n \ge \varepsilon\}} + Y_n1_{\{Y_n < \varepsilon\}} \le 1_{\{Y_n \ge \varepsilon\}} + \tilde{\varepsilon}1_{\{Y_n < \varepsilon\}}$$

For the "only if" part you have $$E[Y_n] \to 0$$ hence it already follows $$Y_n \to 0 \text{ in probability }$$ and we are done.

Gono
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