Let $X_n$ $(n=1,2,\dots)$ be a sequence of discrete random variables, where the distribution of $X_n$ is the discrete uniform over $\{0, 1/n, 2/n,\dots,1 \}$. Let $U$ be a random variable whose distribution is the continuous uniform over $[0,1]$.

Does $X_n \to U$ almost surely?

My approach is to use Borel-Cantelli to check that $\sum_n\Pr[|X_n-U|>\epsilon] < \infty$, but it appears that for $n>2$ and $\epsilon_n < 1/(n+1)$, we have $\Pr[|X_n-U|>\epsilon_n] = 1-2n{\epsilon}_n/(n+1)$, whose sum does not converge. Is there another approach?

If the convergence does not happen almost surely, is it possible for any discrete random variable to converge almost surely to a continuous random variable? (It is clear in the example above that that $X_n \to^\mathrm{D} U$ i.e. convegence in distribution.)


1 Answer 1


As written, the statement is false. Suppose the sample space is $\Omega = [0,1]$, with the probability measure $P$ being the Lebesgue measure. Define the random variables $$ X_n : \Omega \to \mathbb R, \ \ \ \ X_n(\omega)=\begin{cases} \frac{\lfloor (n+1)\omega\rfloor}{n} & {\rm if \ } n {\rm \ is \ odd} \\ 1 - \frac{\lfloor (n+1)\omega\rfloor}{n} & {\rm if \ } n {\rm \ is \ even} \end{cases}$$

Notice that the distribution of $X_n$ is discrete uniform over $\{0, 1/n, 2/n, \dots, 1 \}$, and this is true regardless of whether $n$ is odd or even.

And yet, $X_n$ fails to converge pointwise a.s.

  • $\begingroup$ Nice! From your counter-example, can we conclude that there cannot exist any sequence $X_1, X_2,X_3,\dots$ with discrete uniform over $\{0,1/n,2/n,\dots,1\}$ which converges to $U([0,1])$ a.s? More generally, is it true that if $(Z_n)$ is identically distributed to $(X_n)$, and we know that $(X_n)$ does not converge to $X$ a.s., then $(Z_n)$ cannot converge to $X$ a.s? $\endgroup$
    – jII
    Jun 21, 2018 at 23:53
  • $\begingroup$ @jesterII Well, if $X_n(\omega) = \frac{\lfloor (n+1)\omega \rfloor }{n}$ for all $n$ (even or odd), and if $U(\omega) = \omega$, then $X_n$ has the correct discrete distribution and $U$ has the correct continuous distribution, and $X_n$ converges a.s. to $U$. $\endgroup$
    – Kenny Wong
    Jun 21, 2018 at 23:56
  • $\begingroup$ It seems then to determine whether a random sequence $X_n$ converges almost surely to $X$, it is not in general enough to only know the distribution $P(X_n \in A)$ of each $X_n$ in isolation, somehow we need to know more about the structure of the random variables or the sampling process. $\endgroup$
    – jII
    Jun 22, 2018 at 0:07
  • $\begingroup$ @jesterII Yes, I agree! $\endgroup$
    – Kenny Wong
    Jun 22, 2018 at 0:22

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .