I'm learning some probability theory and I've come across the following:

For an example of a sequence of random vairables that converges in the mean square sense but not almost surely: We set $$P(X_n=1) =1/n \qquad P(X_n=0)=1-1/n$$ and $X_n\to X$ in mean square but not almost sure. The not almost sure follows from the second Borel Cantelli Lemma. I got the examples from here

However, in the book Counter Examples in Probability, they give the following example for a sequence that converges almost surely but not completely $$Y_n(w) = \left\{\begin{array}{ll} 1 & 0\leq w< 1/n\\ 0 & 1/n \leq w <n \end{array}\right.$$

Don't $X_n$ and $Y_n$ define identical distributions? How does $X_n$ converse almost surely but $Y_n$ doesn't?

  • $\begingroup$ $X_n$ is only partially defined. To see why that is, consider how you might compute (say) $\Pr((X_n, X_{n+1})=(1,1)).$ What would it be? $\endgroup$
    – whuber
    May 11, 2019 at 14:03
  • $\begingroup$ What do I need to define them further? I could say they are independent. Are the $Y_n$ independent? $\endgroup$
    – Rdrr
    May 11, 2019 at 15:01
  • $\begingroup$ Two of the three replies on the page you link to are explicit about their construction. They answer your question. $\endgroup$
    – whuber
    May 16, 2019 at 20:51

1 Answer 1


There is a subtle difference between the $X$'s and the $Y$'s. In the example where the $X$'s are defined, they are constructed to be independent. This independence is essential in proving why there is no almost sure convergence. OTOH the $Y$'s are defined to ensure pointwise convergence occurs almost everywhere.

It is true that the $X$'s and $Y$'s have the same marginal distribution, i.e., each $X_n$ has the same distribution as the corresponding $Y_n$. But we've just seen that it's impossible to decide almost sure convergence by the marginal distributions alone.


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