I'm looking for a simple example sequence $\{X_n\}$ that converges in probability but not almost surely.

The example I have right now is Exercise 47 (1.116) from Shao:

$ X_n(w) = \begin{cases}1 &k/2^m \leq w \leq k+ 1/2^m \\ 0 &o.w. \end{cases}$

for $w \in [0,1]$ and integer $m$. In this case, since $m$ is arbitrary, you can find an infinite sequence $\{n_m\}$ where $X_{n_m} (w) = 1$.

Can you provide a simpler example? Thanks!


Define a sequence of independent rv's $X_n$ where: $$P(X_n=1)=\frac{1}{n}, \;P(X_n = 0) = 1-\frac{1}{n}$$

Let $X= 0, a.s.$

Define the event $E_n:= \{X_n=1\}$, then we get:

$$\sum_{n=1}^{\infty} P(E_n) = \sum_{n=1}^{\infty} \frac{1}{n} = \infty$$

By the "converse" Borel-Cantelli Lemma: if we have a sequence of independent events and their probabilities sum to $\infty$, then the event happens infinitely often.

So, in this case, $X_n=1$ happens infinitely often, and so $X_n$ does not converge almost surely to $X$.


$$\lim_{n\to\infty} P(|X_n-X|>\epsilon) =\lim_{n\to\infty} P(X_n>0) = 0 \;\;\forall \epsilon>0$$

So $X_n \xrightarrow{p} X$

  • $\begingroup$ Thank you Bey, this is definitely simpler! I am also interested in a "statistically meaningful" example, a case where this property comes up while making (asymptotic) inference. $\endgroup$ – nooreen Apr 26 '16 at 14:02
  • $\begingroup$ @nooreen well, the model I provided is applicable to a wide array of actual phenomena. Basically, any process that decays as $\frac{1}{n}$. Whether something is "meaningful" is quite application specific. $\endgroup$ – user75138 Apr 26 '16 at 14:28
  • $\begingroup$ @nooreen also, the definition of a "consistent" estimator only requires convergence in probability. Are there cases where you've seen an estimator require convergence almost surely? $\endgroup$ – user75138 Apr 26 '16 at 14:29
  • $\begingroup$ @nooreen see this post as well. It specifically discusses when, if ever, strong consistency is relevant to inference: stats.stackexchange.com/questions/2230/… $\endgroup$ – user75138 Apr 26 '16 at 14:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.