Suppose $\{X_n, n \ge1\}$ are iid and define $M_n=\bigvee^n_{i=1}X_i$

Question

Suppose $\{X_n,n\geq 1\}$ are iid and define $M_n=\bigvee^n_{i=1}X_i. $

1. Check that $$P[M_n >x] \leq n P[X_1 >x].$$

Proof

Note that $$[M_n >x] \subseteq \cup_{i=1}^{n}[X_i >x]$$ $$ \implies P[M_n >x] \leq P(\cup_{i=1}^{n}[X_i >x])\leq \sum_{i=1}^{n}P[X_i >x]=nP[X_1 >x]$$

2. If $E(X_1^p)< \infty$ then $\frac{M_n}{n^{\frac{1}{p}}}\rightarrow_{P} 0$

   Proof

Let $\epsilon >0$. Observe: $$P\left(\left|\frac{M_n}{n^{\frac{1}{p}}}\right| > \epsilon\right)=P(M_n >n^{\frac{1}{p}}\epsilon)\leq n P(X_1 > n^{\frac{1}{p}}\epsilon)\leq \frac{E\left(X_1^p 1_{X_1 > n^{1/p}\epsilon}\right)}{\epsilon^p} \rightarrow 0.$$ 3. If in addition to being iid, the sequence $\{X_n\}$ is non-negative, show $$\frac{M_n}{n}\rightarrow^{P} 0\> \mbox{ iff } \> nP[X_1>n] \rightarrow 0 \> \mbox{ as } \> n\rightarrow \infty.$$

I need help with part 3. But are my proofs for the previous two parts correct?

Answer 1

Your proofs for the first two parts are correct.

For the third one, the direction $\Leftarrow$ follows from 1. For the converse, define $p_n:=\mathbb P\left\{X_1>n\right\}$: we have to show that if $M_n /n\to 0$ in probability, then $np_n\to 0$. If $M_n /n\to 0$ in probability, then $1-(1-p_n)^n\to 0$. Suppose that for an increasing sequence of integers $(n_j)_j$ and some positive $\delta$, $n_j p_{n_j}\gt \delta$. Then $$1-\left(1-p_{n_j} \right)^{n_j} \geqslant 1-\left(1-\delta/ n_j\right)^{n_j}.$$ Now I think you can conclude.