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A sequence ${X_n}$ of random variables converges in probability towards the random variable $X$ if for all $\epsilon > 0$

$$\lim_{n\to\infty}\Pr\big(|X_n-X| > \epsilon\big) = 0$$

But why use $\epsilon > 0$ and not just take $\epsilon = 0$?

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simonzack
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1 Answers1

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Let $X,X_1,X_2,\dots$ be constant random variables: $X(\omega)=x$ and $X_n(\omega)=x_n$ for each $\omega\in\Omega$.

If $x_n$ converges to $x$ then for each $\epsilon>0$: $$\lim_{n\rightarrow\infty}P(|X_n-X|>\epsilon)=0$$showing that $X_n$ converges in probability to $X$.

However for each $n$ with $x_n\neq x$ we have $P(|X_n-X|>0)=1$.

drhab
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  • So I guess the actual question he meant to ask was are the two equivalent? And not about the "usefulness" of convergence in probability? – Ilham Apr 25 '15 at 19:49
  • @Ilham No, I'm asking about the usefulness, drhab's answer gives a reasonable example which doesn't converge in my definition. – simonzack Apr 25 '15 at 19:53
  • @simonzack So yes, basically you were asking whether your definition was equivalent to the definition of convergence in probability: whether both imply each other or whether one doesn't imply the other, which it doesn't in this case... – Ilham Apr 25 '15 at 19:56
  • @Ilham Not quite, I was after some way to demonstrate why the standard definition is nicer in a way. An argument or an example are both fine. But thanks for taking the time to look at my question (and spot a typo!). – simonzack Apr 25 '15 at 19:57
  • @simonzack I'm not really sure what you mean by nicer. We have weaker modes of convergence simply because we want to see whether some properties still hold with weaker conditions. For example dominated convergence still holds with weaker conditions in $\sigma$-finite measure spaces:

    http://math.stackexchange.com/questions/206851/generalisation-of-dominated-convergence-theorem

    That way we can generalize some properties instead of demanding stricter assumptions.

     – Ilham Apr 25 '15 at 20:01
  • @Ilham That's also a great argument for a weaker definition, do you know of any? – simonzack Apr 25 '15 at 20:04
  • @simonzack I don't understand what you mean by "do I know of any?" Any what? Mode of convergence? Argument? Nor do I know what you mean by "weaker definition". I was talking about weaker modes of convergence, not "weaker definition". When a mode of convergence is weaker, it means that it does not imply the stronger mode but the stronger mode implies the weaker. I'm sure you know this but a weaker mode of convergence than convergence in probability is convergence in distribution.

    Also, if you have more questions, ask a new one since this discussion is getting overlong. And make it precise!

     – Ilham Apr 25 '15 at 20:10