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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: $$ B_m = ((B_m)^c)^c = \left( \cup_{n=1}^{\infty} \cap_{m=n}^\infty X_m < c_m \right)^c \quad = \left( \quad \exists n \,\, \forall m \geq n \,\, X_m < c_m \right)^c $$ $$ \text{Since } \,\, \forall m \, \text{it is equivalent to} $$ $$ \left(\exists n \in \mathbb{N} \,\, \forall m \geq n \,:\, M_n < c_n \right)^c = \left( \forall n \in \mathbb{N} \, \exists m \geq n \, : \,\, M_n > c_n \right) = \left( \cap_{n=1}^\infty \cup_{m=n}^\infty M_m > c_m \right) = (M_n > c_n \, \text{infinitely often} ) $$

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  • $\begingroup$ I've made a mistake calling the $M_m$ to be $M_n$, but they are alle $M_m$ until the last line. $\endgroup$ – nalen Sep 15 at 10:16
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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 $w\in (X_n >c_n \text{ i.o.})$.

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  • $\begingroup$ I like the subsequence part! That's the argument I needed. $\endgroup$ – nalen Sep 15 at 10:57
  • $\begingroup$ @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 $(c_n)$. $\endgroup$ – Gabriel Romon Sep 15 at 11:20

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