$X_n \to 0$ in probability iff $E\left( \frac{|X_n|}{1+|X_n|}\right) \to 0$ Good day,
In class we had the exercise as stated in the title: 
Show that $X_n \to 0$ in probability for $n\to \infty$ iff $E\left( \frac{|X_n|}{1+|X_n|}\right) \to 0$ for $n \to \infty$.
And the solution is the following:
$"\Leftarrow"$ Assume that $E\left( \frac{|X_n|}{1+|X_n|}\right)\to 0$ and let $\epsilon>0$. Then, according to Markovs inequality, we have 
$$P(|X_n|>\epsilon)\color{red}{=}P\left(\frac{|X_n|}{1+|X_n|}>\frac{\epsilon}{1+\epsilon}\right) \leq \frac{\epsilon}{1+\epsilon} E\left( \frac{|X_n|}{1+|X_n|}\right) \to 0 $$
$"\Rightarrow"$ Assume $X_n \to 0$ in probability, then
$$E\left( \frac{|X_n|}{1+|X_n|}\right)\color{orange}{\leq}\frac{\epsilon}{1+\epsilon}\cdot P(|X_n|\leq \epsilon)+1\cdot P(|X_n|>\epsilon) \to \frac{\epsilon}{1+\epsilon} \to 0$$
To be honest I have some problems with the solution. They are as following:


*

*$"\color{red}{=}"$ How? My try: $P(|X_n|>\epsilon)=P\left(\frac{|X_n|}{1+|X_n|}>\frac{\epsilon}{1+|X_n|}\right)$ but how do I get this to above formulation?

*$"\color{orange}{\leq}"$ It seems a bit like $E(X)=\sum x_i p_i$ but I don't get how this inequality is proved.


As a hint to the exercise there was: $f(u)=\frac{u}{1+u}$ is strictly increasing on $[0,\infty)$. So this should come into play somewhere in the answer to my questions I guess.
Can someone please hep me with these two questions?
Thanks a lot,
Marvin
 A: *

*The hint is the key your first question. If $f(u)=\frac{u}{1+u}$ is a strictly increasing function, then $|X_n|>\varepsilon$ implies $f(|X_n|)>f(\varepsilon)$, i.e. $$\frac{|X_n|}{1+|X_n|}>\frac{\varepsilon}{1+\varepsilon}.$$

*This is shown using indicator functions. I will abbreviate $Y_n=\frac{|X_n|}{1+|X_n|}$. Since $1_{\{|X_n|\leq\varepsilon\}}+1_{\{|X_n| >\varepsilon\}}=1$, and by linearity of the expctation function, we have
$$
\mathbb{E}[Y_n]=\mathbb{E}[Y_n\cdot 1_{\{|X_n|\leq\varepsilon\}}]+\mathbb{E}[Y_n\cdot 1_{\{|X_n|>\varepsilon\}}].
$$


*

*Now, $\mathbb{E}[Y_n\cdot 1_{\{|X_n|\leq\varepsilon\}}]\leq \frac{\varepsilon}{1+\varepsilon}\cdot \mathbb{E}[1_{\{|X_n|\leq\varepsilon\}}]=\frac{\varepsilon}{1+\varepsilon}\cdot \mathbb{P}[|X_n|\leq\varepsilon]$ (where we have used again the monotonicity of $f$, and the fact that $\mathbb{E}[1_A]=\mathbb{P}[A]$).

*$\mathbb{E}[Y_n\cdot 1_{\{|X_n|>\varepsilon\}}]\leq 1\cdot \mathbb{P}[|X_n|>\varepsilon]$ where we have used the fact that $f(u)\leq 1$ for any $u$. 



Hope this clarifies things.
A: You can use the hint to prove the equality you have issues with: another way of saying that $f$ is strictly increasing on $[0, \infty)$ is that
$$ x > u \Longleftrightarrow \frac{x}{1+x} > \frac{u}{1+u}$$
for all $x \geq 0, u \geq 0$. Now take $x = |X_n|$ and $u = \epsilon$ and you obtain the aforementioned result.
The inequality is obtained by finding an upper bound $A\; (= \frac{\epsilon}{1+\epsilon})$ for $\frac{|X_n|}{1 + |X_n|}$ when $|X_n| < \epsilon$, and an upper bound $B\; (= 1)$ for $\frac{|X_n|}{1 + |X_n|}$ when $|X_n| \geq \epsilon$.
Then the claim is made that 
\begin{equation}\label{eq:1}
E\left(\frac{|X_n|}{1 + |X_n|} \right) \leq A \cdot \mathbb{P}(|X_n| < \epsilon) + B \cdot \mathbb{P}(|X_n| \geq \epsilon).
\end{equation}
So, you need to answer a couple of questions:


*

*Are the specific values of the upper bounds clear? (Hint: for $A$, look back at the equivalence I've written above. For $B$ you need to show that $\frac{x}{1+x} \leq 1$.)

*Do you understand the split of the expected value into a part with $|X_n| < \epsilon$ and one with $|X_n| \geq \epsilon$?

