Given $(U_n)_n$ a sequence of i.i.d random variables $U[0,1]$. I want to find the convergence in probability and a.s of the sequences $X_n = 1_{(0,1/n)}(U_1)$ and $Y_n = 1_{(0,1/n)}(U_n)$.

I don't understand why $X_n$ and $Y_n$ would converge to different values, given that the variables $U_i$ are i.i.d. Since $\lim_{n\to\infty}P(|X_n| > 1/n) = 0$ my guess is that $X_n \xrightarrow{P} 0$ and the same goes for $Y_n$.

What can I say about the convergence a.s. though?

  • 1
    $\begingroup$ Both converge to $0$ in probability, as you pointed out. $X_n$ converges also to $0$ pointwise a.s. $Y_n$ does too, but along SUBSEQUENCE (that is the best one can say when a sequence converges in probability) $\endgroup$
    – Mittens
    Nov 19, 2023 at 22:36
  • $\begingroup$ @Mittens You mean that there is a subsequence $X_{n_k} \xrightarrow{c.s.} 0$ but it is not true that $X_n \xrightarrow{c.s.} 0$? $\endgroup$ Nov 19, 2023 at 22:44
  • $\begingroup$ No, I am saying that $X_n\xrightarrow{n\rightarrow\infty}0$ a.s, and that there is a subsequence $Y_{n_k}$ such that $Y_{n_k}\xrightarrow{k\rightarrow\infty}0$. $\endgroup$
    – Mittens
    Nov 19, 2023 at 22:46
  • $\begingroup$ How can I show that $X_n \xrightarrow 0$ a.s though? $\endgroup$ Nov 19, 2023 at 22:50
  • $\begingroup$ Because once $U_1(\omega)$ is observed, then for $n> 1/U_1(\omega)$, $X_n(\omega)=0$. $\endgroup$
    – Mittens
    Nov 19, 2023 at 22:55

1 Answer 1


$P(Y_n>\frac 1 2)=\frac 1 n$ So Borel Cantelli Lemma (and independence of $(Y_n)$ shows that $Y_n >\frac 1 2 $ infinitely often with probability $1$. So $(Y_n)$ does not converge to $0$ a.s. Since $Y_n \to 0$ in probability it cannot converge a.s to anything other than $0$.

Of course, $X_n(\omega) \to 0$ for every $\omega$ so $X_n \to 0 $ a.s. Ignore points where $U(\omega)=0$ or $1$ because these events have probability $0$.

  • $\begingroup$ Great! Can we use Borel-Cantelli lemma as well to show that $X_n \to X$ a.s? $\endgroup$ Nov 20, 2023 at 0:33
  • $\begingroup$ Thar hardly requires a proof. If $0<x<1$ then $1_{(0,\frac 1 n)} (x)=0$ for all $n$ sufficiently large, so $1_{(0,\frac 1 n)} (x) \to 0$. @PeterSampodiras $\endgroup$ Nov 20, 2023 at 4:33

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