Show that $X_n\overset{P}\to0$ iff $E(|X_n|\wedge1)\to 0$. 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 $|X_n|\wedge1=|X_n|I_{\{|X_n|<1\}}+I_{\{|X_n|\ge 1\}}$?
 A: 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.
A: 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
$$\lim_{n \to \infty} \mathbb{P}(|X_n| \geq \varepsilon) = 0 \ .$$
A: 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 $|X_n|\wedge 1 \leq 1$
A: 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.
A: $\forall \epsilon\in(0,1)$, 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 $X_n\overset{P}\to0$.
