show that $X_n \overset{\mathrm{P}}{\to} X \iff d(X_n,X) \to 0 $

Question

let $L^0$ be the space on all random variables then : $$\begin{align} d : &L^0 \times L^0 &\to &[0,1] \\ &(X,Y) &\mapsto &\mathbb{E}[\frac{|X-Y|}{1+|X - Y|}] \end{align}$$

defines a well defined metric on $L^0$

proof of first implication : considering the continuous bounded function $\phi(t) = \frac{|t|}{1+|t|}$ and using the fact that $E[\phi(X_n-X)] \to 0$ in case $X_n -X \overset{\mathrm{d}}{\to} 0$

then $$X_n \overset{\mathrm{P}}{\to} X \implies X_n - X \overset{\mathrm{P}}{\to} 0 \implies X_n - X \overset{\mathrm{d}}{\to} 0 \implies d(X_n,X) \to 0 $$

for second I wanted to use the fact that the convergence in $L^1$ is preserved under composition with continuous function coupled with the fact that $$\frac{\frac{t}{1 + t}}{1 -\frac{t}{1 + t}} = t$$

but $y \mapsto \frac{y}{1-y}$ isn't continuous at $1$.

hints ?

Answer 1

Since $t\mapsto \frac{t}{1+t}$ is increasing, we find by Chebyshev's inequality $$ \Bbb P(|X_n-X|\ge\epsilon) =\Bbb P\left(\frac{|X_n-X|}{1+|X_n-X|}\ge \frac{\epsilon}{1+\epsilon}\right)\le (1+\epsilon)/\epsilon\cdot d(X_n,X) \to 0. $$