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We assume $\{X_n\}_{n\in\mathbb{N}}$ and $X$ are random variables from $\{\Omega,\mathcal{F},\mathbb{P}\}$ to $(S,d_s)$, wehre $S$ a separable metric space.

One can establish the following equivalent statement of convergence in probability.

$$\mathbb{P}(d_s(X_n,X)>\epsilon)\to 0\\ \iff d_1(X_n,X):=\mathbb{E}(d_s(X_n,X)\wedge1)\to 0 \\ \iff d_2(X_n,X):=\mathbb{E}(\frac{d_s(X_n,X)}{1+d_s(X_n,X)})\to 0, $$ as $n\to \infty$. One of the proofs is given in convergence in probability induced by a metric.

We know the following facts.

  • $d_s(X_n,X)\wedge1$ and $\frac{d_s(X_n,X)}{1+d_s(X_n,X)}$ are topologically equivalent to the original $d_s(\cdot,\cdot)$.

  • $d_1(\cdot,\cdot)$ and $d_2(\cdot,\cdot)$ metrize the $L_0(\Omega, S)$, i.e. the space of all random variables with values in $S$.

I am looking for connections between the above facts, in the perspective of topology (or metric sense).

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  • @ChristophPegel Yes, sorry for being sloppy here. Most of the setup is standard. – newbie Feb 26 '14 at 15:49
  • Sorry what does $\wedge$ mean? – Rgkpdx Feb 26 '14 at 15:55
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    @Ton, I suppose $a\wedge b=\min{a,b}$. – Christoph Feb 26 '14 at 15:58
  • It is not clear what you are looking for. Yes, those are equivalent metrics to $d$ but $d$ does not enjoy the same property. It may be that $E(d(X_n,X)) \neq 0$. This happens, for example, when doing least squares regression the error term is 0 in probability but in expectation it is a non-zero scalar multiple of the variance. – SomeEE Feb 26 '14 at 16:20

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