# Given $(U_n)_n$ a sequence of i.i.d random variables $U[0,1]$, find convergence of $1_{(0,1/n)}(U_1)$ and $1_{(0,1/n)}(U_n)$

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?

• 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) Nov 19, 2023 at 22:36
• @Mittens You mean that there is a subsequence $X_{n_k} \xrightarrow{c.s.} 0$ but it is not true that $X_n \xrightarrow{c.s.} 0$? Nov 19, 2023 at 22:44
• 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 $Y_{n_k}\xrightarrow{k\rightarrow\infty}0$. Nov 19, 2023 at 22:46
• How can I show that $X_n \xrightarrow 0$ a.s though? Nov 19, 2023 at 22:50
• Because once $U_1(\omega)$ is observed, then for $n> 1/U_1(\omega)$, $X_n(\omega)=0$. Nov 19, 2023 at 22:55

$$P(Y_n>\frac 1 2)=\frac 1 n$$ 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 $$0$$.
• Great! Can we use Borel-Cantelli lemma as well to show that $X_n \to X$ a.s? Nov 20, 2023 at 0:33
• 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 Nov 20, 2023 at 4:33