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For two random variables, define $d(X,Y)=\mathbb{E}\dfrac{|X-Y|}{1+|X+Y|}$

The text I am reading claims that $d(X, Y)$ is a metric on the a.e. equivalence class of $L_0$. I am not familiar with using metric on random variables. If I wanted to show that this were indeed a metric, how would I go about doing this?

I know already that for a general metric such as $\dfrac{|x-y|}{1 + |x+y|}$ with no randomness, it would be relatively straigforward to show it was a metric space. Is it the same idea with random variables? For example: $d(X, Y)=0 \iff \mathbb{E}(|X-Y|)=0 \iff E(X)=E(Y)$

Or is there a different process? I did see similar questions such as this one here: Show that $d_2$ defined by $d_2(x,y)=\frac{|x-y|}{1+|x-y|}$ is a metric, but none of them involve random variables.

user345
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  • What you write is not correct. You have$$d(X,Y) = 0;\Longleftrightarrow;\int\frac{|X-Y|}{1+|X+Y|},dP =0;\Longleftrightarrow;\frac{|X-Y|}{1+|X+Y|}=0\text{ a.e.};\Longleftrightarrow;X=Y\text{ a.e.}$$And what is $L_0$? – Friedrich Philipp Nov 12 '17 at 23:27
  • $\mathcal{L}_0$ means integrable, what's the problem with the expression you wrote? – Daniel Ordoñez Nov 12 '17 at 23:56

1 Answers1

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Yes, the process is the same.

1) Proof $d(X,Y)\geq 0$
2) Proof $d(X,Y)=0$ then $X=Y$ almost surely
3) Proof $d(X,Y)=d(Y,X)$
4) The triangle inequality

All of them are straightforward, perhaps the most difficult is the last one. Just remember that you are working in an equivalence class so if $E(X)=0$ then $X=0$ almost surely.

Daniel Ordoñez
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