Probability Theory – Problem 45: Prove that defines a metric on the space of random variables

For random variables $X,~Y$, define $d(X,\ Y)=E{\frac{|X-Y|}{1+|X-Y|}}.$

(1) Prove that $d$ defines a metric on the space of random variables. That is, prove that $d$ satisfies positive definiteness, symmetry, and the triangle inequality.

(2) Prove that the random variables $X_{n}$ converge to $X$ in probability if and only if $\operatorname*{lim}_{n\to\infty}d(X_{n},\ X)=0$.

Step 1. First, prove that $d(X, Y)$ satisfies the definition of a metric. (i) Non-negativity and positive definiteness: Since $|X-Y| \ge 0$, the integrand $\frac{|X-Y|}{1+|X-Y|} \ge 0$, so $d(X, Y) \ge 0$ after taking expectations. If $d(X, Y) = 0$, then the non-negative random variable $\frac{|X-Y|}{1+|X-Y|}$ has expectation 0, which means it equals 0 almost surely, hence $|X-Y|=0$ a.s., i.e., $X=Y$ a.s. (ii) Symmetry: Clearly $|X-Y| = |Y-X|$, so $d(X, Y) = E\left[\frac{|X-Y|}{1+|X-Y|}\right] = E\left[\frac{|Y-X|}{1+|Y-X|}\right] = d(Y, X)$. (iii) Triangle inequality: Consider the function $f(t) = \frac{t}{1+t} = 1 - \frac{1}{1+t}$. For $t \ge 0$, the derivative $f'(t) = \frac{1}{(1+t)^2} > 0$, so $f(t)$ is monotonically increasing. For any real numbers $x, y, z$, by the absolute value inequality, $|x-z| \le |x-y| + |y-z|$. Let $a = |x-y|, b = |y-z|$, so $|x-z| \le a+b$. Using the monotonicity of $f(t)$ and the inequality $\frac{a+b}{1+a+b} = \frac{a}{1+a+b} + \frac{b}{1+a+b} \le \frac{a}{1+a} + \frac{b}{1+b}$, we obtain: $\frac{|X-Z|}{1+|X-Z|} \le \frac{|X-Y| + |Y-Z|}{1 + |X-Y| + |Y-Z|} \le \frac{|X-Y|}{1+|X-Y|} + \frac{|Y-Z|}{1+|Y-Z|}$. Taking expectations on both sides yields $d(X, Z) \le d(X, Y) + d(Y, Z)$. In summary, $d$ defines a metric on the space of random variables (in the sense of almost sure equality).

Step 2. Proof of sufficiency: If $\lim_{n\to\infty} d(X_n, X) = 0$, then $X_n \xrightarrow{P} X$. For any given $\epsilon > 0$, consider the event $\{|X_n - X| \ge \epsilon\}$. On this event, by the monotonicity of $f(t)=\frac{t}{1+t}$, we have $\frac{|X_n - X|}{1+|X_n - X|} \ge \frac{\epsilon}{1+\epsilon}$. By a generalization of Chebyshev's inequality or directly using properties of expectation: $d(X_n, X) = E\left[\frac{|X_n - X|}{1+|X_n - X|}\right] \ge E\left[\frac{|X_n - X|}{1+|X_n - X|} I_{\{|X_n - X| \ge \epsilon\}}\right] \ge \frac{\epsilon}{1+\epsilon} P(|X_n - X| \ge \epsilon)$. Rearranging: $P(|X_n - X| \ge \epsilon) \le \frac{1+\epsilon}{\epsilon} d(X_n, X)$. As $n \to \infty$, since $d(X_n, X) \to 0$, we get $P(|X_n - X| \ge \epsilon) \to 0$. That is, $X_n$ converges to $X$ in probability.

Step 3. Proof of necessity: If $X_n \xrightarrow{P} X$, then $\lim_{n\to\infty} d(X_n, X) = 0$. For any given $\epsilon > 0$, split the expectation into two parts: $d(X_n, X) = E\left[\frac{|X_n - X|}{1+|X_n - X|} I_{\{|X_n - X| \ge \epsilon\}}\right] + E\left[\frac{|X_n - X|}{1+|X_n - X|} I_{\{|X_n - X| < \epsilon\}}\right]$. For the first term, the integrand is always at most 1, so the first term is at most $1 \cdot P(|X_n - X| \ge \epsilon)$. For the second term, when $|X_n - X| < \epsilon$, $\frac{|X_n - X|}{1+|X_n - X|} < \frac{\epsilon}{1+\epsilon} < \epsilon$, so the second term is at most $\epsilon$. That is, $d(X_n, X) \le P(|X_n - X| \ge \epsilon) + \epsilon$. Since $X_n \xrightarrow{P} X$, as $n \to \infty$, $P(|X_n - X| \ge \epsilon) \to 0$. Therefore $\limsup_{n\to\infty} d(X_n, X) \le \epsilon$. Since $\epsilon$ is arbitrary, $\lim_{n\to\infty} d(X_n, X) = 0$.

QED.