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Let $X_1, X_2, ...$ be a sequence of real-valued random variables.

Prove $X_n \xrightarrow P 0$ as $n \rightarrow \infty$ iff $\lim_{n \to \infty} E(\frac{|X_n|}{|X_n|+1} )= 0$

Attempt:

Suppose $X_n \xrightarrow P 0$ as $n \rightarrow \infty$. Then since $X_1, X_2, ...$ is uniformly integrable and E(|X|)<$\infty$, then $\lim_{n \to \infty} E(X_n) = E(X)$.

Since $X_n \xrightarrow P 0$ as $n \rightarrow \infty$, then $\lim_{n \to \infty} E(\frac{|X_n|}{|X_n|+1} )$ must be equal to 0.

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    $\begingroup$ "Then since X1,X2,... is uniformly integrable" Sorry? These are not even assumed integrable. $\endgroup$ – Did May 18 '15 at 5:42
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Hint:

$X_n \xrightarrow P 0$ mean $\forall \epsilon >0, P(|X_n| > \epsilon) \to 0$.

$\frac{|X_n|}{|X_n|+1} = \frac{|X_n|}{|X_n|+1} 1_{|X_n| > \epsilon} + \frac{|X_n|}{|X_n|+1} 1_{|X_n| \leq \epsilon} \leq 1_{|X_n| > \epsilon} + \epsilon$

$\frac{|X_n|}{|X_n|+1} = \frac{|X_n|}{|X_n|+1} 1_{|X_n| > \epsilon} + \frac{|X_n|}{|X_n|+1} 1_{|X_n| \leq \epsilon} \geq 1_{|X_n| > \epsilon} \frac{\epsilon}{1+\epsilon}$

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  • $\begingroup$ Interesting. I'm a little confused as to how you were able to end up with $1_{|X_n|>\epsilon}+\epsilon$ and $1_{|X_n|>\epsilon}+\frac{\epsilon}{1+\epsilon}$ $\endgroup$ – Andrew Aug 8 '14 at 23:21
  • $\begingroup$ @Andrew remark that $\frac{y}{1+y} \leq 1$ when $y \geq 0$ and $\frac{y}{1+y} \geq \frac{\epsilon}{1+\epsilon}$ when $y \geq \epsilon$ $\endgroup$ – Petite Etincelle Aug 9 '14 at 6:35
  • $\begingroup$ Very witty, +1. $\endgroup$ – Gabriel Romon May 4 '16 at 11:04
  • $\begingroup$ @LeGrandDODOM Merci, cher camarade :) $\endgroup$ – Petite Etincelle May 4 '16 at 16:53

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