$\{X_n\}_{n\geq 1}$ is a sequence of random variables. Want to show that \begin{align*} X_n\xrightarrow[]{P}X \text{ if and only if } \lim_{n\rightarrow\infty}E[1-\exp\{-\vert X-X_n\vert\}]=0 \end{align*} $\Rightarrow$
Convergence in probability gives that there exist a $\epsilon$ such that $\lim_{n\rightarrow\infty}P(\vert X_n-X\vert < \epsilon)=1$ or equivalent $\lim_{n\rightarrow\infty}P(\vert X_n-X\vert > \epsilon)=0$.
Hence \begin{align*} \lim_{n\rightarrow\infty}E[1-\exp\{-\vert X-X_n\vert\}]&= 1-\lim_{n\rightarrow\infty}E[\exp\{-\vert X-X_n\vert\}] \\ &= 1-\lim_{n\rightarrow\infty}\int^\infty_{-\infty} \exp\{-\vert X-X_n\vert\}P(\vert X-X_n\vert) d\vert X-X_n\vert \end{align*} Is this correct or do I have to proceed in another way? Since exp is continuous I guess somehow it should be simple. I can also see the final argument and way this is try but the relation between expected values $E$ and probability $P$ gives me problems.
$\{X_n\}_{n\geq 1}$ is a sequence of random variables. We should show that \begin{align*} X_n\xrightarrow[]{P}X \text{ if and only if } \lim_{n\rightarrow\infty}E[\exp\{-\vert X-X_n\vert\}]=1 \end{align*} $\Rightarrow$:
$X_n\xrightarrow[]{P}X$ means that there exist a $\epsilon$ such that $\lim_{n\rightarrow\infty}P(\vert X_n-X\vert > \epsilon)=0$ or $\lim_{n\rightarrow\infty}P(\vert X_n-X\vert \leq \epsilon)=1$
Hence \begin{align*} &\lim_{n\rightarrow\infty}E[\exp\{-\vert X-X_n\vert\}]\\ =& \lim_{n\rightarrow\infty}\left(E[\exp\{-\vert X-X_n\vert\}\textbf{1}\{\vert X-X_n\vert>\varepsilon\}] + E[\exp\{-\vert X-X_n\vert\}\textbf{1}\{\vert X-X_n\vert\leq \varepsilon\}] \right)\\ \geq& \exp(-\varepsilon)\lim_{n\rightarrow\infty}P(\vert X-X_n\vert\leq\varepsilon) = \exp(-\varepsilon) \end{align*} Now since $\varepsilon$ can be arbitrarily small, we got $\lim_{n\rightarrow\infty}E[\exp\{-\vert X-X_n\vert\}]\geq 1$. But $E[\exp\{-\vert X-X_n\vert\}]\leq 1$ as $\vert X-X_n\vert\geq 0$, thus $\lim_{n\rightarrow\infty}E[\exp\{-\vert X-X_n\vert\}]= 1$.
$\Leftarrow$:
For any $\varepsilon>0$ we have
\begin{align*} &E[\exp\{-\vert X-X_n\vert\}]\\ =& \left(E[\exp\{-\vert X-X_n\vert\}\textbf{1}\{\vert X-X_n\vert>\varepsilon\}] + E[\exp\{-\vert X-X_n\vert\}\textbf{1}\{\vert X-X_n\vert\leq \varepsilon\}] \right)\\ \leq& \exp(-\varepsilon)P(\vert X-X_n\vert > \varepsilon)+ P(\vert X-X_n\vert\leq\varepsilon) \\ =& \exp(-\varepsilon)[1-P(\vert X-X_n\vert \leq \varepsilon)]+ P(\vert X-X_n\vert\leq\varepsilon) \\ =& \exp(-\varepsilon)+ [1-\exp(-\varepsilon)]P(\vert X-X_n\vert\leq\varepsilon) \end{align*}
Thus $$[1-\exp(-\varepsilon)]P(\vert X-X_n\vert\leq\varepsilon)\geq E[\exp\{-\vert X-X_n\vert\}] - \exp(-\varepsilon).$$
Taking $\liminf_{n\rightarrow \infty}$ on both sides gives us
$$[1-\exp(-\varepsilon)]\liminf_{n\rightarrow \infty}P(\vert X-X_n\vert\leq\varepsilon)\geq 1 - \exp(-\varepsilon),$$
which means $\liminf_{n\rightarrow \infty}P(\vert X-X_n\vert\leq\varepsilon)\geq 1$. But on the other hand $\limsup_{n\rightarrow \infty}P(\vert X-X_n\vert\leq\varepsilon)\leq 1$, thus we get
$$\lim_{n\rightarrow \infty}P(\vert X-X_n\vert\leq\varepsilon)= 1,\ {\rm or}\ X_n\xrightarrow[]{P}X.$$