Asymptotics for minimum of a sequence of random variables

Question

This is a question that I'm sure has been investigated before, but I have found no good search terms for.

Let $X_i$ be a sequence of random variables, independent and uniformly distributed in $[0,1]$. Let $Y_n=\min(X_1,\dots,X_n)$.

Is there an asymptotic which $Y_n$ almost surely follows? More precisely, is there some function $f:\mathbb N\to\mathbb R_+$ such that, almost surely, $Y_n\sim f(n)$?

If we ignore the fact that $Y_i$ are not independent, a Borel-Cantelli argument suggests $f(n)=\frac{\log n}{n}$: for any $c$, $\mathbb P(Y_n>c)=(1-c)^n\sim e^{-cn}$ for small $c$. Letting $c=d\frac{\log n}{n}$, this is asymptotic to $n^{-d}$, which has finite sum for $d>1$ and infinite for $d<1$, so we get $\limsup_{n\to\infty}\frac{Y_n}{(\log n)/n}\leq 1$ almost surely, with equality if we pretend we have independence.

Is the $\limsup$ of $\frac{Y_n}{(\log n)/n}$ actually equal to $1$ almost surely? Is the limit equal to $1$ almost surely? If not, what is the "lower" asymptotic for $Y_n$? For instance, do we at least have $Y_n=\Theta(\frac{\log n}{n})$ almost surely?

Note: I am aware that $nY_n$ converges to an exponential distribution, but I don't think that really helps answer the question, as we are interested in the entire sequence of $Y_n$ rather than their individual terms.

Answer 1

do we at least have $Y_n=\Theta(\frac{\log n}{n})$ almost surely?

The answer to this is no. It is not even true that $Y_n=\Theta(\frac{h(n)}{n})$ almost surely (a.s.) for any (deterministic or random) function $h$ such that $h(n)\to\infty$ a.s. (as $n\to\infty$). Indeed, suppose the contrary: that $\liminf_n nY_n=\infty$ a.s. Then, by the Fatou lemma, $$ \infty=E\infty\le\liminf_n EnY_n =\liminf_n n\frac1{n+1}=1, $$ a contradiction. $\quad\Box$

---

Let us now show that $Y_n\not\sim f(n)$ a.s. for any deterministic positive function $f$.

To prove this, suppose the contrary: that $Y_n\sim f(n)$ a.s. for some deterministic positive function $f$. Take any positive sequence $(a_n)$ such that $\sum_n a_n=\infty$. Then $$\sum_n P(X_n<a_n/2)=\sum_n a_n/2=\infty.$$ So, by the Borel--Cantelli lemma, events $\{X_n<a_n/2\}$ a.s. occur infinitely often (i.o.), that is, for infinitely many values of $n$. Therefore and because $\{X_n<a_n/2\}\subseteq\{Y_n<a_n/2\}$, we see that events $\{Y_n<a_n/2\}$ a.s. occur i.o. Recalling now the assumption that $Y_n\sim f(n)$ a.s., we conclude that $f(n)\le a_n$ i.o. In particular, $f(n)\le \frac1{n\ln n}=o(1/n)$ i.o.

So, $$Z_n:=nY_n\sim nf(n)=o(1) \tag{1}\label{1}$$ a.s. for $n=n_k$ and $k\to\infty$, where $(n_k)$ is a strictly increasing deterministic sequence of positive integers. Also, $EZ_n^2<2$. So, for $n$ as above, $$\begin{aligned} 1\leftarrow EZ_n&=EZ_n\,1(Z_n<4)+EZ_n\,1(Z_n\ge4) \\ &\le EZ_n\,1(Z_n<4)+EZ_n^2/4 \\ &<EZ_n\,1(Z_n<4)+1/2\to1/2 \end{aligned}$$ by \eqref{1} and dominated convergence. So, we have a contradiction. $\quad\Box$