I've been trying to come up with a intuitive, practical distinction between convergence in probability and convergence almost surely. Can someone please tell me if the following is correct?

Let $X_n$ be a statistic calculated over and over again from many random samples of size $n$ and $X$ the quantity it converges to as $n \rightarrow \infty$. Let $\epsilon$ be an arbitrary boundary around $X$. Let $A_n$ be the number of $X_n : ~ |X_n - X| > \epsilon$ (in other words, the number of $X_n$ outside a given bound) and let $B_n$ be the number of $X_n : ~ |X_n - X| \le \epsilon$ (the number of $X_n$ inside a given bound).

Then, as $n \rightarrow \infty$...

If $X_n \overset{P}{\rightarrow} X$...

  • $\frac{A_n}{B_n}\rightarrow 0$

But if $X_n \overset{a.s.}{\rightarrow} X$...

  • ${\color{red} {A_n \rightarrow k < \infty}}$

  • $\frac{A_n}{B_n}\rightarrow 0$

Is the above a valid way of interpreting the difference between convergence in probability and almost sure convergence? Thanks.

  • 3
    $\begingroup$ This cannot be true since almost sure convergence implies convergence in probability. $\endgroup$ – Did Feb 16 '12 at 18:37
  • $\begingroup$ You're right. How about: only the third statement (A_n/B_n -> 0) needs to be true for convergence in probability, while the first and third statements both need to be true for convergence almost surely? $\endgroup$ – f1r3br4nd Feb 16 '12 at 19:24
  • 4
    $\begingroup$ The way I intuitively think of convergence in probability vs convergence almost surely is that with convergence in probability you say that $X_n$ is close (within $\epsilon$) to $X$ with high probability whereas with convergence a.s. not only is $X_n$ close with high probability, but for sufficiently large $n$ all $X_m$ are close to $X$ for $m \ge n$, with high probability. In notation, $P(|X_n - X| > \epsilon) \to 0$ for convergence in probability whereas $P(\sup_{m \ge n} |X_m - X| > \epsilon) \to 0$ as $n \to \infty$ is equivalent to convergence a.s. $\endgroup$ – guy Feb 19 '12 at 5:29

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