I thought of this example in hopes of helping me understand almost sure convergence a little better. So, if you could add any additional (relevant) details in your response I would greatly appreciate it.

By definition, almost sure convergence requires:

\begin{align} P(\lim_{n\to \infty}{X_n = a}) = 1 \end{align} Where $(X_n)$ is a sequence of IId random variables that converges to some real number $a$.

Let us define the $X_n$ as Bernoulli Random Variables: \begin{align} X_n = \begin{cases} 1 & \text{with probability $\frac{1}{n}$}\\ 0 & \text{with probability $1 - \frac{1}{n}$} \end{cases} \end{align}

So, as $n \to \infty \mathbb{P}(X_n = 1) = 0$ and $\mathbb{P}(X_n = 0) = 1$. Does this imply that $X_n \to 0$ almost surely?

Thank you!

  • 1
    $\begingroup$ No, $P(X_n=0)\to1$ does not imply that $X_n\to0$ almost surely, and the case when the random variables $X_n$ are independent (but certainly not i.i.d. ...) with $P(X_n=0)=1-1/n$ and $P(X_n=1)=1/n$, is a good example. Hint: By the second Borel-Cantelli lemma, the event $\limsup\{X_n=1\}$ has probability $1$. $\endgroup$ – Did Nov 16 '15 at 7:27
  • $\begingroup$ OK. This might indicate that one-on-one tutoring sessions might be more adequate to your needs than asking questions on this site. However... what is the first step in my comment that you have problems with? $\endgroup$ – Did Nov 16 '15 at 7:32
  • $\begingroup$ @Did I wish I could afford it, LOL. I go to every single office hour from my professor and TA but the rest of the class usually just wants to solve the homework. But the first thing that does not make sense is what you mean by "(but certainly not i.i.d. ...)". If they are independent, but not necessarily identically distributed, how this does form a good counter to the claim I make in my question? $\endgroup$ – David South Nov 16 '15 at 7:34
  • $\begingroup$ In the example in your question, the distribution of $X_n$ varies with $n$ hence "i." but not "i.d." Note that an i.i.d. squence never converges almost surely (except if $P(X_n=a)=1$ for some $a$, for every $n$). $\endgroup$ – Did Nov 16 '15 at 7:36
  • $\begingroup$ ahhhh..... okay. So, in the first half your comment you basically said that my claim is actually a very good example of why $P(X_n = 0) \to 1$ does not imply $X_n \to 0$. Thanks a lot @Did. $\endgroup$ – David South Nov 16 '15 at 7:37

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