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Show that $X_n\overset{P}\to0$ iff $E(|X_n|\wedge1)\to 0$.

I have no idea of this problem. Do we need to use the equality $|X_n|\wedge1=|X_n|I_{\{|X_n|<1\}}+I_{\{|X_n|\ge 1\}}$?

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  • $\begingroup$ I did not know this before. Thank you! $\endgroup$
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    Apr 13, 2017 at 2:19

5 Answers 5

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Here are some hints~

First $P(|X_n| \wedge 1 \geq \epsilon) \rightarrow 0$ is equivalent to $P(|X_n| \geq \epsilon) \rightarrow 0$ for $0 < \epsilon <1$ (can you see why).

Second, by Markov inequality

$P(|X_n| \wedge 1 \geq \epsilon) \leq E[|X_n| \wedge 1] / \epsilon$

Thirdly,

$E[|X_n| \wedge 1] = \int_{0}^{\infty} P(|X_n| \wedge 1 \geq \epsilon) d\epsilon = \int_{0}^1 P(|X_n| \wedge 1 \geq \epsilon) d\epsilon$. Now apply Dominated Convergence Theorem you are done.

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Assume that $X_n \to 0$ in probability, and let $1 > \varepsilon > 0$ be arbitrary. We have $$ \mathbb{E}(|X_n| \wedge 1) = \mathbb{P}(|X_n| > 1) + \mathbb{E}(|X_n| I(|X_n| < \varepsilon)) + \mathbb{E}(|X_n| I ( \varepsilon \leq |X_n| \leq 1)) \ .$$ Note that $$\mathbb{E}(|X_n| I(|X_n| < \varepsilon)) \leq \varepsilon \mathbb{P}(|X_n| < \varepsilon)$$ and $$\mathbb{E}(|X_n| I ( \varepsilon \leq |X_n| \leq 1)) \leq \mathbb{P}(\varepsilon \leq |X_n|) \ .$$ Now using convergence in probability, we have $$ \lim_{n \to \infty} \mathbb{E}(|X_n| \wedge 1) \leq \lim_{n \to \infty} \left(\mathbb{P}(|X_n| > 1) + \varepsilon \mathbb{P}(|X_n| < \varepsilon) + \mathbb{P}(\varepsilon \leq |X_n|) \right) = 0 + \varepsilon \cdot 1 + 0 = \varepsilon \ .$$ Since $\varepsilon > 0$ was arbitrary, taking the limit $\varepsilon > 0$, gives us $$\lim_{n \to \infty} \mathbb{E}(|X_n| \wedge 1) = 0 \ .$$ Assume that $ \mathbb{E}(|X_n| \wedge 1) \to 0$. Let $1 \geq \varepsilon > 0$. Note that $$ \mathbb{E}(|X_n| \wedge 1) \geq \mathbb{E}(|X_n| \wedge \varepsilon) = \varepsilon \mathbb{P}(|X_n| \geq \varepsilon) + \mathbb{E}(|X_n| I(|X_n| < \varepsilon)) \geq \varepsilon \mathbb{P}(|X_n| \geq \varepsilon)\ .$$ Now taking the limit as $n \to \infty$, we have $$\lim_{n \to \infty} \mathbb{P}(|X_n| \geq \varepsilon) = 0 \ .$$

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  • $\begingroup$ Yes, this is a much simpler approach than my answer. $\endgroup$
    – Ran Wang
    Mar 2, 2017 at 15:24
  • $\begingroup$ Why is the equality in the sixth equation valid? $\endgroup$ Mar 2, 2017 at 16:43
  • $\begingroup$ Sorry about that, the equality did not hold. I, however, think I fixed it with another proof along roughly the same idea. $\endgroup$ Mar 2, 2017 at 17:43
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For the first part: Let $0<\varepsilon<1$

$$P[|X_n|\geq \varepsilon ] = P[|X_n|\wedge1\geq \varepsilon] \leq \frac{E[|X_n|\wedge 1]}{\varepsilon}$$

The last part is by Markov inequality.

For the other part: Let $0<\varepsilon<1$. $$E[|X_n|\wedge 1]=E[(|X_n|\wedge 1)1_{|X_n|\leq\varepsilon}] + E[(|X_n|\wedge 1)1_{|X_n|>\varepsilon}] \\ \leq \varepsilon + E[1_{|X_n|>\varepsilon}] \\ = \varepsilon + P[|X_n|>\varepsilon]$$

where we used $|X_n|\wedge 1 \leq 1$

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If $0<\epsilon\leq1$ then: $$\epsilon1_{\{|X_n|\geq\epsilon\}}\leq|X_n|\wedge1\leq1_{\{|X_n|\geq\epsilon\}}+\epsilon\tag1$$Taking expectations on both sides we find:$$\epsilon\Pr(|X_n|\geq\epsilon)\leq\mathbb E(|X_n|\wedge1)\leq\Pr(|X_n|\geq\epsilon)+\epsilon\tag2$$ The first inequality in $(2)$ makes it evident that: $$\lim_{n\to\infty}\mathbb E(|X_n|\wedge1)=0\implies\lim_{n\to\infty}\Pr(|X_n|\geq\epsilon)=0$$ This for every $\epsilon\in(0,1]$ so this allows us to conclude:$$\lim_{n\to\infty}\mathbb E(|X_n|\wedge1)=0\implies X_n\overset{P}\to0$$ The second inequality in $(2)$ makes it evident that the converse of this is also true.

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$\forall \epsilon\in(0,1)$, we have

$$P(|X_n|\wedge1>\epsilon)=P(|X_n|>\epsilon,1>\epsilon)=P(|X_n|>\epsilon).$$

If$X_n\overset{P}\to0$, then $P(|X_n|\wedge1>\epsilon)=P(|X_n|>\epsilon)\to0$. Notice that $|X_n|\wedge1\le1$, so by dominated convergence theorem we have $E(|X_n|\wedge1)\to0$.

If $E(|X_n|\wedge1)\to0$, then we have $P(|X_n|>\epsilon)=P(|X_n|\wedge1>\epsilon)\to0$, which means $X_n\overset{P}\to0$.

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