Inequalities with expected value on one side and probability on the other

Question

In a part of a proof I am following, the author states that

$$\displaystyle \mathbb{E}\left[\frac{|X_n - X|}{1 + |X_n - X|}\right] \leq \epsilon + \mathbb{P}(|X_n - X| > \epsilon)$$

and

$$\displaystyle \mathbb{E}\left[\frac{|X_n - X|}{1 + |X_n - X|}\right] \geq \frac{\epsilon}{2} \mathbb{P}(|X_n - X| > \epsilon)$$

$X$ and $X_n$ are random variables. I am at a loss to demonstrate these inequalities myself. I've tried a few things involving Markov's inequality along with the fact that $\displaystyle \frac{|a|}{1 + |a|} \leq |a|$, but thus far have not found success. It does not seem to me that other inequalities (e.g., Holder's or Minkowski's) would be useful here. Any help would be appreciated (even if it's pointing out that something here must be a typo.)

Answer 1

Notice that taking $c/(1+c)$, for $c > 0$, is dividing a positive number by a larger positive number. So

$$\frac{c}{1+c} < 1 \text{ and } \frac{c}{1+c} < c$$

Now

\begin{align*} \mathbb{E}\left[\frac{|X_n - X|}{1 + |X_n - X|}\right] &= \int_\Omega \frac{|X_n - X|}{1 + |X_n - X|} d\mathbb{P} \\ & = \int_{|X-X_n| \leq \varepsilon}\frac{|X_n - X|}{1 + |X_n - X|} d\mathbb{P} + \int_{|X-X_n| > \varepsilon}\frac{|X_n - X|}{1 + |X_n - X|} d\mathbb{P} \\ &\leq \int_{|X-X_n| \leq \varepsilon} |X_n - X| d\mathbb{P} + \int_{|X-X_n| > \varepsilon} 1 \, d\mathbb{P} \\ & \leq \int_{|X-X_n| \leq \varepsilon} \varepsilon \, d\mathbb{P} + \int_{|X-X_n| > \varepsilon} 1 \, d\mathbb{P} \\ &= \varepsilon \mathbb{P}(|X-X_n| \leq \varepsilon) + \mathbb{P}(|X-X_n| > \varepsilon)\\ & \leq \varepsilon + \mathbb{P}(|X-X_n| > \varepsilon) \end{align*}

This is a standard sort of thing in analysis: you break a sum or integral into a small number of large things and a large number of small things -- a "head" and a "tail."

Answer 2

Hint: divide the universe in two parts. Let $f(dx, dx_n)$ be the joint law of $(X,X_n)$. Then: $$\begin{align} \int \frac{|x-x_n|}{1+|x-x_n|} &f(dx,dx_n) \\&= \int_{|x-x_n| \le \epsilon} \frac{|x-x_n|}{1+|x-x_n|} f(dx,dx_n) + \int_{|x-x_n| > \epsilon} \frac{|x-x_n|}{1+|x-x_n|} f(dx,dx_n) \\&\le \int_{|x-x_n| \le \epsilon} {|x-x_n|} f(dx,dx_n) + \int_{|x-x_n| > \epsilon} \frac{|x-x_n|}{1+|x-x_n|} f(dx,dx_n) \\&\le \int_{|x-x_n| \le \epsilon} \epsilon f(dx,dx_n) + \int_{|x-x_n| > \epsilon} \frac{|x-x_n|}{1+|x-x_n|} f(dx,dx_n) \\&= \epsilon + E\left[ \frac{|X-X_n|}{1+|X-X_n|} \left| |x-x_n| > \epsilon \right.\right] \end{align}$$

and $$\begin{align} \int \frac{|x-x_n|}{1+|x-x_n|} &f(dx,dx_n) \\&= \int_{|x-x_n| \le \epsilon} \frac{|x-x_n|}{1+|x-x_n|} f(dx,dx_n) + \int_{|x-x_n| > \epsilon} \frac{|x-x_n|}{1+|x-x_n|} f(dx,dx_n) \\&\ge 0 + \int_{|x-x_n| > \epsilon} \frac{|x-x_n|}{2} f(dx,dx_n) \\&\ge \int_{|x-x_n| > \epsilon} \frac{\epsilon}{2} f(dx,dx_n) = \frac{\epsilon}{2} P\left[|X-X_n| > \epsilon \right] \end{align}$$