For i.i.d. $U(0,1)$ random variables $(X_i)$, $\max\limits_{1\le i \le n/2}\{(1-2i/n)X_i\}\to1$ in probability

Question

I want to show $\max_{1\le i \le \frac{n}{2}}\{(1-\frac{2i}{n})X_i\}$ converges in probability to $1$ as $n \to \infty$, where $X_i$ is an i.i.d sequence of $[0,1]$-uniformly distributed random variables.

I didn't learn any estimate to deal with maximum except Kolmogorov's inequality, but seems it does not work here. Can anyone provide any idea? Many thanks!

Answer 1

The maximum $M_n$ of interest is such that $$(1-2/\sqrt{n})K_n\leqslant M_n\leqslant1,\quad\text{where}\quad K_n=\max\{X_i\mid 1\leqslant i\leqslant\sqrt{n}\}.$$ By independence, $P(K_n\leqslant x)=P(X_1\leqslant x)^{\sqrt{n}}=x^{\sqrt{n}}\to0$ when $n\to\infty$, for every $x$ in $(0,1)$. Hence $K_n\to1$ in probability, $(1-2/\sqrt{n})K_n\to1$ in probability, and finally, $M_n\to1$ in probability.

Note the stronger result that $M_n\to1$ (and $K_n\to1$) almost surely.

Answer 2

I have some partial results; perhaps someone can continue/correct my work.

\begin{align*} &\phantom{{}={}}P\left(1-\max_{1 \le i \le n/2} \left\{\left(1-\frac{2i}{n}\right)X_i\right\}>\epsilon\right)\\ &=P\left(1-\epsilon>\max_{1 \le i \le n/2} \left\{\left(1-\frac{2i}{n}\right)X_i\right\}\right)\\ &=\prod_{i=1}^{n/2} P\left(1-\epsilon > \left(1-\frac{2i}{n}\right)X_i\right) & \text{independence}\\ &=\prod_{i=1}^{\lfloor{\epsilon n/2\rfloor}} P\left(1-\epsilon > \left(1-\frac{2i}{n}\right)X_i\right) & \text{for $i>\epsilon n /2$, the probability is $1$ since $X_i \le 1$}\\ &= \prod_{i=1}^{\lfloor{\epsilon n/2\rfloor}} \frac{1-\epsilon}{1-\frac{2i}{n}} & \text{uniform distribution}\\ &= (1-\epsilon)^{\lfloor{\epsilon n/2\rfloor}} \prod_{i=1}^{\lfloor{\epsilon n/2\rfloor}}\frac{n}{n-2i}\\ \end{align*}