$\{X_n\}_{n=1}^\infty$ 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!

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

1 Answer 1


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.

  • 1
    $\begingroup$ Actually, the random nature of $X_n$'s is relevant. For example, $N$ is random here. $\endgroup$
    – user140541
    Commented Mar 28, 2022 at 15:19
  • $\begingroup$ I guess the probabilistic aspect needs some explanation. I also missed a part of the question. $\endgroup$
    – ajr
    Commented Mar 28, 2022 at 15:56

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .