Related to this question, @AlexR gave a perfect answer, but left a non-trivial statement as an ending to the proof. He said that

$\forall\epsilon>0 \ ,\ \sum\limits_1^\infty P(|M_n-x^*_F|>\epsilon)<\infty\implies M_n\xrightarrow{a.s.} x^*_F$

I have tried to prove it. Have I gone wrong anywhere or assumed something wrong?

$$\forall\epsilon>0 \ ,\ \sum\limits_1^\infty P(|M_n-x^*_F|>\epsilon)<\infty\\\implies \forall\epsilon>0 \ ,\ P(\limsup_{n\to\infty}|M_n-x^*_F|>\epsilon)=0\ \ \cdots\cdots(Borel-Cantelli)\\\implies \forall\epsilon>0 \ ,\ P(|M_n-x^*_F|>\epsilon \text{ for infinitely many } n )=0\\\implies P(M_n\to x^*_F \text{ as } n\to\infty)=1\\\implies M_n\xrightarrow{a.s.} x^*_F$$


If the above is true, then I may assume

If $X_n$ is a sequence of iid random variables and $X$ is another random variable such that $\forall\epsilon>0 \ ,\ \sum\limits_1^\infty P(|X_n-x|>\epsilon)<\infty\implies X_n\xrightarrow{a.s.} X$

But Suppose $Y_n$ is a sequence of iid random variables converging in probability to $Y$ , then $$\lim_{n\to\infty}P(|Y_n-Y|>\epsilon)=0\ \ \forall \epsilon>0\\\implies P(\lim_{n\to\infty}|Y_n-Y|>\epsilon)=0\ \ \forall \epsilon>0\\\implies P(\limsup_{n\to\infty}|Y_n-Y|>\epsilon)=0\ \ \forall \epsilon>0\\\implies Y_n\xrightarrow{a.s.} Y \ \ (contradiction)$$

Where am I going wrong?


1 Answer 1


He's appealing to the first Borel-Cantelli lemma. This lemma states that for a sequence of events $\{ A_n \}_{n=1}^{\infty}$ if $\sum_{n=1}^{\infty} P(A_n) < \infty$ then $P(A_n \,\, \text{i.o.}) = 0$ (the i.o. stands for "infinitely often"). In this case, if $P(|M_n - x^\star_F| > \epsilon \,\, \text{i.o.}) = 0$ for $\epsilon > 0$ then $M_n \stackrel{\text{a.s.}}{\to} x^\star_F$ follows.

Update based on edit

The error in your argument is pulling the limit inside the probability. If this were always valid then convergence in probability and almost sure convergence would be identical, but that isn't the case. The problem here is that $\lim_{n \to \infty} |Y_n - Y|$ need not even exist, and it certainly isn't guaranteed to exist and equal zero on a set with probability one.

Take a standard counterexample where $\{ X_n \}_{n=1}^\infty$ are independent Bernoulli$(1 / n)$ random variables. The second Borel-Cantelli lemma actually states $X_n$ will almost surely bounce back and forth between zero and one infinitely often. So even though $\lim_{n \to \infty} P(X_n > \epsilon) = 0$, it makes no sense to talk about $\lim_{n \to \infty} X_n$ because the latter only exists on a null set.

  • $\begingroup$ Is my proof via $\limsup$ correct? Especially going from 2nd to 3rd line? $\endgroup$
    – Qwerty
    Commented Oct 13, 2016 at 14:58
  • $\begingroup$ It's not incorrect, but I don't think you need so many steps. Going from Borel-Cantelli to a.s. convergence is pretty much immediate. $\endgroup$
    – dsaxton
    Commented Oct 13, 2016 at 15:09
  • $\begingroup$ But I have a doubt then. I will edit my doubt in the question $\endgroup$
    – Qwerty
    Commented Oct 13, 2016 at 15:15
  • $\begingroup$ I have edited my question. Please clear my doubt( I am making a silly mistake, but still , please help me) $\endgroup$
    – Qwerty
    Commented Oct 13, 2016 at 15:26
  • $\begingroup$ I added some more to my answer. $\endgroup$
    – dsaxton
    Commented Oct 13, 2016 at 15:50

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