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I have problem understanding the difference when I look at the alternative definition of a.s. convergence. I know how it is defined originally, but it is the alternative definition which makes it easier to compare it to convergence in probability:

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Convergence in probability is defined that for any $\epsilon >0$.:

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We see there that the only difference is the sup. But even with the sup, I struggle to see the difference, can someone explain to me where I see it incorrect:?

For instance, if I look at convergence in probability, I would think like that. I choose an $\epsilon$. Then for any $\epsilon_2$, there is an N, such that if $n \ge N$, then $Pr(|X_n-X|\ge \epsilon)<\epsilon_2$. Now comes my problem: Since this holds for all $n\ge N$, why is it then not equal to the alternative characterisation of a.s. convergence?

UPDATE: Is it correct to say that the difference is that if $\epsilon$ and $\epsilon_2$ is given. Than for all $n \ge N$ you can in the first case use the same subset of the sample-space. But in the case of only convergence in probability, you may have to change the subset of the sample space for each $n \ge N$?

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  • $\begingroup$ en.wikipedia.org/wiki/… has a pretty good discussion of this, including examples where you have probability convergence but not a.s. convergence. $\endgroup$ Jul 12 '14 at 19:33
  • $\begingroup$ Got something from the answers below? $\endgroup$
    – Did
    Jul 15 '14 at 17:39
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The core reason is that $$\color{red}{\sup_\color{black}{n}}\color{green}{\Pr}(|X_n-X|\geqslant\varepsilon)\qquad\text{and}\qquad \color{green}{\Pr}(\color{red}{\sup_\color{black}{n}}|X_n-X|\geqslant\varepsilon)$$ have little in common. In general, the latter is (much) larger than the former. For instance, if $\Pr(X_n=1)=1-\Pr(X_n=0)=1/n$ for every $n\geqslant1$ and $(X_n)$ is independent, then $X=0$ and, for every $m\geqslant1$ and every positive $\varepsilon\leqslant1$, $$\sup_{n\geqslant m}\Pr(|X_n-X|\geqslant\varepsilon)=1/m,\qquad \Pr(\sup_{n\geqslant m}|X_n-X|\geqslant\epsilon)=1.$$

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  • $\begingroup$ Thanks, may I please ask how you calculated the last probability that is 1? $\endgroup$
    – user119615
    Jul 12 '14 at 19:58
  • $\begingroup$ Borel-Cantelli lemma shows that almost surely, $X_n=1$ for infinitely many indices $n$. $\endgroup$
    – Did
    Jul 12 '14 at 20:14
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Let $\omega=[0,1]$ and $X_n=\mathbb{1}_{[0,\frac{1}{n}]}$, you have $X_n \to X$, where $X(\omega)=0$ for $\omega \in [0,1]$ and

$$\lim_{n \to \infty} Pr(\sum_{m \geq n}|X-X_{m}|\geq \varepsilon)=1$$ for all $\varepsilon>0$

But:

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

for all $\varepsilon>0$

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