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Let $M$ be the set of all random variables from a fixed probability space to $\mathbb R$ with its borel sets.

Let's define a metric on $M$ by $d(X,Y)=E(\frac{|X-Y|}{1+|X-Y|})$

I want to prove that $d$ it's in fact a metric. The only difficult part it's to prove that $d(X,Y)=0$ if and only if $X=Y$ almost surely (a.e) and also that $X_n \to X$ in probability if and only iff $d(X_n,X)\to 0$ Thus this metric induces the probability convergence.

Shanks
  • 789

3 Answers3

16

If $d(X,Y)=0$, then $Z=\frac{|X-Y|}{1+|X-Y|}$ is a nonnegative random variable whose expectation is zero, so it must be zero a.e., so $|X-Y|=0$ a.e., so $X=Y$ a.e.

Suppose that $(X_n) \to X$ in probability. Let $\varepsilon >0$. Then for large enough $n$ we have $P(|X_n-X| > \varepsilon ) \leq \varepsilon$. For those $n$ we have

$$ \begin{array}{lcl} d(X_n,X) &=& {\bf E}(\frac{|X_n-X|}{1+|X_n-X|}{\bf 1}_{|X_n-X| \leq \varepsilon}) +{\bf E}(\frac{|X_n-X|}{1+|X_n-X|}{\bf 1}_{|X_n-X| > \varepsilon}) \\ &\leq& {\bf E}(|X_n-X|{\bf 1}_{|X_n-X| \leq \varepsilon}) +{\bf E}({\bf 1}_{|X_n-X| > \varepsilon}) \\ &\leq& {\bf E}(\varepsilon{\bf 1}_{|X_n-X| \leq \varepsilon}) +P(|X_n-X| > \varepsilon)=2\varepsilon \end{array} $$

So $d(X_n,X) \to 0$.

Conversely, suppose that $d(X_n,X) \to 0$. Let $\varepsilon >0$. Then for large enough $n$ we have $d(X_n,X) \leq \frac{\varepsilon^2}{1+\varepsilon}$. For those $n$ we have

$$ \frac{\varepsilon^2}{1+\varepsilon} \geq {\bf E}(\frac{|X_n-X|}{1+|X_n-X|}{\bf 1}_{|X_n-X| > \varepsilon}) \geq {\bf E}(\frac{\varepsilon}{1+\varepsilon}{\bf 1}_{|X_n-X| > \varepsilon})= \frac{\varepsilon}{1+\varepsilon} P(|X_n-X| > \varepsilon ) $$

So $P(|X_n-X| > \varepsilon) \leq \varepsilon$. This shows that $(X_n) \to X$ in probability.

Ewan Delanoy
  • 61,600
2

This is can be seen as part of a general result:

Theorem: Suppose $(\Omega,\mathscr{F},\mu)$ is a finite measure space and $(S,\rho)$ a separable metric space. Assume $\{f_n,\, f:n\in\mathbb{N}\}\subset S^\Omega$ is a sequence of measurable functions with respect to the outer measure $\mu^*$. Let $F:[0,\infty)\rightarrow[0,\infty)$ a bounded continuous nondecreasing function with $F(t)=0$ iff $t=0$. Then, $\rho(f_n,f)\in\mathscr{M}_{\mathbb{R}}(\mu^*)$ for all $n\in\mathbb{N}$, and $f_n$ converges in measure to $f$ if and only if $\lim_n\int F(\rho(f_n,f))\,d\mu=0$.

Here is a short proof:

Let $\varepsilon>0$ arbitrary and $\|F\|_\infty:=M$. Notice that $$ F(\varepsilon)\mathbb{1}_{\{\rho(f_n,f)>\varepsilon\}} \leq F(\rho(f_n,f))\leq F(\varepsilon) + M\mathbb{1}_{\{\rho(f_n,f)>\varepsilon\}} $$ and denote by $D(f_n,f)=\int F(\rho(f_n,f))\,d\mu$. Then

$$ \begin{align} F(\varepsilon)\mu(\{\rho(f_n,f)>\varepsilon\})\leq D(f_n,f) \leq F(\varepsilon)\mu(\Omega)+ M \mu(\{\rho(f_n,f)>\varepsilon\})\tag{1}\label{mean-metric} \end{align} $$ Necessity follows by letting $n\nearrow\infty$ and then $\varepsilon\searrow0$. Sufficiency follows by letting $n\nearrow\infty$.

Corollary: Suppose $(\Omega,\mathscr{F},\mu)$ is a finite measure space. Assume $F:[0,\infty)\rightarrow[0,\infty)$ is a bounded nondecreasing subadditive function such that $F(x)=0$ iff $x=0$, then $$ D_F(f,g)=\int_\Omega F(|f-g|)\,d\mu $$ is a complete metric on $L_0$ which results in the same topology as in $(L_0,d_0)$.

Here is a short proof:

It is easy to check that $D_F$ is a metric on $L_0$. By Theorem above it is enough to show that $(L_0,D_F)$ is complete. If $\{f_n:n\in\mathbb{N}\}$ is a Cauchy sequence w.r.t. $D_F$, then
$$ \lim_{M\rightarrow\infty} \sup_{n,m\geq M} \mu(|f_n-f_m|>\varepsilon)\leq \tfrac{1}{F(\varepsilon)}\lim_{M\rightarrow\infty}\sup_{n,m\geq M}D_F(f_n,f_m)=0 $$ by $\eqref{mean-metric}$. Hence there are integers $n_k< n_{k+1}$ such that $\sup_{n,m\geq n_k}\mu(|f_n-f_m|>2^{-k})< 2^{-k}$ and so, $\sum_k\mu(|f_{n_{k+1}}-f_{n_k}|>2^{-k})<\infty$. By the Borel--Cantelli lemma, the set $$ A=\{|f_{n_{k+1}}-f_{n_k}|>2^{-k},\,i.o\} $$ has $\mu$--measure zero. It follows that $\{f_{n_k}\}$ is $\mu$--a.s. a Cauchy sequence in $\mathbb{R}$; thus, $f_{n_k}$ converges $\mu$--a.s to some $f\in\mathscr{M}_{\mathbb{R}}(\mu^*)$. By dominated convergence, $\lim_kD_F(f_{n_k},f)\rightarrow0$. Therefore $\lim_nD_F(f_n,f)=0$.

Examples

  1. In $\mathbb{R}$, take $\rho(x,y)=|x-y|$ and $F(t)=\min(t,1)$
  2. In $\mathbb{R}$, take $\rho(x,y)=|x-y|$, and $F(t)=\frac{t}{1+t}$
Mittens
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  • How are $\mu$ and $\mathscr{M}_{\mathbb{R}}(\mu^)$ defined? – Learner Nov 19 '21 at 09:55
  • @Learner: $\mu^$ is the outer measure induced by $\mu$ on the power set of $\Omega$. This induced a mean ($|f|_{\mu^}=\inf{\int h}$, where $0\leq h\leq |f|$, and $h$ can be expressed as the supremum of a sequence of elementary functions ) on the space of all numerical functions $\overline{R}^\Omega$. The space $\mathcal{F}$ numerical functions that have finite mean is a linear lattice and closed under the operation $\min(f,1)$ (Stone lattice), and $(\mathcal{F},|;|_{\mu^*})$ is complete. $L_1(\mu)$ is the closure of the space of elementary functions on $\mathcal{F}$ (to be continue) – Mittens Nov 19 '21 at 16:04
  • (continuation) Measurability with respect to the mean $|;|{\mu^*}$ is defined in terms of uniformities. Let $\mathcal{E}$ be space of integrable simple functions. An $(S,d)$-valued function is $\mu$-measurable if for any integrable set $A$ and $\varepsilon>0$, there an integrable set $A_0\subset A$ with $\mu(A\setminus A_0)<\varepsilon$ such that for for any $\eta>0$, there is $\delta>0$ and a finite collection $\phi_1,\ldots,\phi_n$ such that for any $x,y\in A$, $\sup{1\leq j\leq n}|\phi_j(x)-\phi_j(y)|<\delta$ implies $\rho(f(x),f(y))<eta$. (to be continued) – Mittens Nov 19 '21 at 19:59
  • (continuation) When $(S,d)$ is separable, it can be show that of $f$ and $g$ are $S$-valued measurable (in the sense described above) then $x\mapsto d(f(x),g(x))$ is $\mathbb{R}$-valued measurable (in the sense define above). Real--valued functions that are measurable in the sense above and have finite mean, are integrable in the sense of Lebesgue. There is an extension of Egoroff's theorem for measurable functions (in the sense above). For $(S,d)=(\mathbb{R},|;|)$ things are much easier. – Mittens Nov 19 '21 at 20:04
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Here is another proof that $d(X_n,X)\to 0$ implies $X_n\to X$ in probability. By subtracting $X$, we can assume $X_n \ge 0$, $d(X_n,0)\to 0$ and prove that $X_n\to 0$ in probability. Using the "layer cake" representation of the integral, the fact that $\frac{X_n}{1+X_n} \le 1$ a.s., and the change of variable $x=\frac{\lambda}{1-\lambda}$ our assumption $d(X_n,0) \to 0$ can be rewritten as \begin{align*} 0 = \lim_{n\to\infty}E\bigg[\frac{X_n}{1+X_n}\bigg] &= \int_0^1 P\bigg(\frac{X_n}{1+X_n}>\lambda\bigg)\,d\lambda\\ &= \int_0^\infty P(X_n>x)(1+x)^{-2}\,dx. \end{align*} (By the dominated convergence theorem, it is now easy to see that $X_n\to 0$ in probability implies that $d(X_n,0)\to 0$.)

Suppose to the contrary that $x_0 > 0$ and $\limsup_{n\to\infty}P(X_n > x_0) = c > 0$. By the definition of the limsup of a numerical sequence, there is a subsequence $\{n_k\}_{k=1}^\infty$ so that $P(X_{n_k} > x_0) \to c$. Because the cdf $F_k(x) \equiv P(X_{n_k} > x)$ is nonincreasing and nonnegative, we have $$F(x) \equiv \liminf_{k\to\infty}F_k(x) \ge \liminf_{k\to\infty}F_k(x_0) \ge c\,\mathbf 1_{[0,x_0]}(x).^{\text{[1]}}$$

By Fatou's lemma, and our assumption that $d(X_n,0)\to 0$, \begin{align*} 0 = \liminf_{k\to\infty}\int_0^\infty F_k(x)(1+x)^{-2}\,dx &\ge \int_0^\infty F(x)(1+x)^{-2}\,dx \\ &\ge \int_0^{x_0}c\,(1+x)^{-2}\,dx\\ &= c\,\frac{x_0}{1+x_0} > 0, \end{align*} a contradiction.

[1]: For a Borel set $A\subseteq\mathbf R$, $\mathbf 1_A(x) = 1$ if $x\in A$ and $\mathbf 1_A(x) = 0$ if $x\in \mathbf R\smallsetminus A$. $\mathbf 1_A$ is called the indicator function of $A$.

Alex Ortiz
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