Compute $\liminf_{n\to \infty }n^{1/3}\max\{X_n,X_{n+1},X_{n+2}\}$ and $\liminf_{n\to \infty }\log(n)n^{1/3}\max\{X_n,X_{n+1},X_{n+2}\}$

Question

Let $X_n$ iid $\mathrm{Uniform} (0,1)$ r.v. How can I compute $$\liminf_{n\to \infty }n^{1/3}\max\{X_n,X_{n+1},X_{n+2}\}$$ and $$\liminf_{n\to \infty }\log(n)n^{1/3}\max\{X_n,X_{n+1},X_{n+2}\}\ \ ?$$

Since I don't have sum I think I can't use law numbers theorem (or I don't see how).

In fact, I have that if $Y_n=\max\{X_n,X_{n+1},X_{n+2}\}$, then $$\mathbb P\{Y_n\leq y\}=\mathbb P\{X_1\leq y\}^3=y^3.$$ But still, I don't see how to conclude.

Answer 1

Notice that the sequence $\left( Y_{3n}\right)_{n\geqslant 1}$ is independent and that for each positive $\varepsilon$, the series $\sum_{n\geqslant 1}\Pr\left(Y_{3n}/3n\leqslant \epsilon\right)$ diverges. Using the Borel-Cantelli lemma, we get $\Pr\left(\liminf_{n\to \infty}\left\{ 3nY_{3n} \gt \epsilon\right\} \right)=0$ and from with, we get that $\liminf_{n\to + \infty}3nY_{3n} =0 $ . A similar argument applies to the sequence $\left((3n+1)Y_{3n+1} \right)_{n\geqslant 1}$ and $\left((3n+2)Y_{3n+2} \right)_{n\geqslant 1}$.

For $\liminf_{n\to +\infty}\log(n)n^{1/3}\max\{X_n,X_{n+1},X_{n+2}\}$, the difference is that for each $R$, the series $\sum_{n\geqslant 1}\Pr\left\{\log\left(3n\right) (3n)^{1/3} Y_{3n}\leqslant R \right\}$ is convergent.