# $\limsup \frac{X_n}{a_n}=1 \Leftrightarrow \limsup \frac{\max\{X_1,...,X_n\}}{a_n}=1$

$$\{X_n\}_{n=1}^\infty$$ is non-negative random variables, $$0\leq a_n \uparrow\infty$$. Then: $$\begin{equation} \limsup \frac{X_n}{a_n}=1 \quad a.s.\Leftrightarrow \limsup \frac{\max\{X_1,...,X_n\}}{a_n}=1 \quad a.s. \end{equation}$$

My ideas so far:

Since $$X_n$$ and $$a_n$$ are non-negative, we have: $$\begin{equation} 0\leq \frac{X_n}{a_n}\leq \frac{\max\{X_1,...,X_n\}}{a_n} \end{equation}$$ So, if $$\limsup \frac{\max\{X_1,...,X_n\}}{a_n}=1$$, then $$\limsup\frac{X_n}{a_n}\leq 1.$$

But I cannot going on to prove $$\limsup\frac{X_n}{a_n}\geq 1.$$

Thanks in advance for any help!

• Could you formulate the statement exactly? Is it "$\limsup A = 1 \ a.s. \iff \limsup B = 1\ a.s.$"?
– ajr
Mar 28, 2022 at 15:58
• Yes, and I have corrected my statement. Mar 28, 2022 at 15:59
• I have edited my answer, hopefully it's fine now.
– ajr
Mar 28, 2022 at 16:38

You have in fact shown that for any deterministic sequence $$(X_n)$$ we have $$\limsup \frac{X_n}{a_n} \leq \limsup \frac{\max (X_1,\ldots,X_n)}{a_n}.$$
Now if we show that (for $$(X_n)$$ deterministic) $$\begin{equation}\limsup \frac{\max (X_1,\ldots,X_n)}{a_n} \leq \limsup \frac{X_n}{a_n},\end{equation}$$ then it means that the limits are equal. Therefore your statement follows because the limits are equal for every fixed $$\omega$$ in the probability space.
Let $$\epsilon >0$$ and assume that $$\limsup \frac{X_n}{a_n}$$ is finite, say equal to $$1$$ (if it's infinite, then both limits are infinite by the inequality that you have shown). Then there exists $$N$$ such that for all $$n\geq N$$ we have $$\frac{X_n}{a_n} < 1+\epsilon.$$ We can also assume that $$a_{N}\geq 1$$. Since $$a_n \uparrow \infty$$, there exists $$M$$ such that $$a_{N+M}\geq \max (X_1,\ldots,X_{N-1})a_N.$$ Therefore, if $$k\geq N+M$$, the monotonicity of $$(a_n)$$ gives \begin{align*} \frac{\max(X_1,\ldots,X_k)}{a_k}&\leq \max\bigg(\frac{\max(X_1,\ldots,X_{N-1})}{a_k}, \frac{\max(X_N,\ldots,X_{k})}{a_k}\bigg)\\ &\leq \max\bigg(\frac{\max(X_1,\ldots,X_{N-1})}{\max (X_1,\ldots,X_{N-1})a_N}, \max\big(\frac{X_N}{a_N},\frac{X_{N+1}}{a_{N+1}},\ldots,\frac{X_{k}}{a_k}\big)\bigg)\\ &\leq \max\big(\frac 1{a_N}, 1+ \epsilon\big) = 1+\epsilon. \end{align*} This proves our inequality.
• Actually, the random nature of $X_n$'s is relevant. For example, $N$ is random here.