# Showing $(X_n >c_n \text{ i.o.})=(\max_{1\leq i\leq n}X_i >c_n \text{ i.o.})$

If have the following information: $$X_1, X_2, ...$$ are i.i.d., and also the distribution F of $$X_1$$ fulfills $$F(x) < 1$$ . We are given $$M_n = max\{X_1, ... ,X_n \},$$ and we also have from excercise 1 that $$P(M_n \rightarrow \infty ) = 1$$ I am suppose to show that given an increasing sequence $$c_n \rightarrow \infty$$ , then $$(M_n > c_n \,\, \text{infinitely often}) = (X_n > c_n \,\, \text{infinitely often})$$ $$\\\\$$ My answer is as follows, and I'm not sure whether it's fulfilling: $$B_m := (X_n > c_n \, \text{infinitely often} ) = \left( \cap_{n=1}^\infty \cup_{m=n}^\infty X_m > c_m \right)$$ I believe I can say: $$B_m = ((B_m)^c)^c = \left( \cup_{n=1}^{\infty} \cap_{m=n}^\infty X_m < c_m \right)^c \quad = \left( \quad \exists n \,\, \forall m \geq n \,\, X_m < c_m \right)^c$$ $$\text{Since } \,\, \forall m \, \text{it is equivalent to}$$ $$\left(\exists n \in \mathbb{N} \,\, \forall m \geq n \,:\, M_n < c_n \right)^c = \left( \forall n \in \mathbb{N} \, \exists m \geq n \, : \,\, M_n > c_n \right) = \left( \cap_{n=1}^\infty \cup_{m=n}^\infty M_m > c_m \right) = (M_n > c_n \, \text{infinitely often} )$$

• I've made a mistake calling the $M_m$ to be $M_n$, but they are alle $M_m$ until the last line. – nalen Sep 15 at 10:16

The inclusion $$(X_n >c_n \text{ i.o.}) \subset (M_n >c_n \text{ i.o.})$$ is trivial.
For the reverse inclusion, consider some $$w\in (M_n >c_n \text{ i.o.})$$ and suppose for the sake of contradiction that $$w\in (X_n >c_n \text{ i.o.})^c = \bigcup_n \bigcap_{k\geq n} (X_k \leq c_k)$$. In other words, there is some $$m\geq 1$$ such that $$\forall k\geq m, X_k(w)\leq c_k$$
Since $$w\in (M_n >c_n \text{ i.o.})$$, there is some subsequence $$(n_k)_{k\geq 1}$$ such that $$\forall k\geq 1, M_{n_k}(w)>c_{n_k}$$. Since $$(c_n)_{n\geq 1}$$ goes to $$\infty$$, $$(n_k)_k$$ can be chosen so that $$(M_{n_k}(w))_{k\geq 1}$$ is strictly increasing. By definition of $$M$$, there is a mapping $$\phi$$ such that $$M_{n_k}(w) = X_{\phi(n_k)}(w)$$. Since $$(M_{n_k}(w))_{k\geq 1}$$ is strictly increasing, $$\phi$$ is injective.
Moreover, since $$\forall k\geq m, X_k(w)\leq c_k$$, $$\phi$$ has values in $$\{1,\ldots, m-1\}$$. This contradicts the injectivity of $$\phi$$.
So $$w\in (X_n >c_n \text{ i.o.})$$.
• @nalen The only small gap left in the proof is to show that $\phi$ has values in $\{1,\ldots, m-1\}$. This stems from the monotonicity of $(c_n)$. – Gabriel Romon Sep 15 at 11:20