1
$\begingroup$

$\{X_n\}$ is a non-decreasing sequence of non-negative r.v.s satisfying $EX_n=an^p$ for some $a,p>0$, and $Var(X_n)\le cn^r$ with $0

Following is my attempt:
It's equivalent to prove $P(|X_n/n^p-a|>\epsilon \text{ i.o.})=0$
\begin{align}P(\cup_{n\ge m}|X_n/n^p-a|>\epsilon ) &=P(\cup_{n\ge m}|X_n-an^p|>\epsilon n^p ) \\ &= \dots \\ &= P(|X_m-am^p|>\epsilon m^p) \\ &= var(X_m)/\epsilon ^2 m^{2p} \\ &\le c/\epsilon ^2 m^{2p-r} \rightarrow 0 \end{align} But I don't know how to fill the gap using the non-decreasing and non-negative.

  • 1
    What is a non-decreasing random variable? Would you actually assume that the sequence $(X_n)$ is decreasing, meaning that $X_{n+1}\leqslant X_n$ almost surely, for every $n$? – 2017-01-12
  • 0
    I think it means $X_n \le X_{n+1}$ a.s. for every n – 2017-01-12
  • 0
    Yes, *non-decreasing*. Please modify your post. – 2017-01-12
  • 0
    @Did I didn't get your point – 2017-01-12
  • 0
    Wrong: "a sequence of non-decreasing and non-negative r.v.s" Correct: "a non-decreasing sequence of non-negative random variables". – 2017-01-12
  • 0
    @Did Oh, I see. Thanks. Any idea how to solve it? – 2017-01-12
  • 0
    Sure. What do you know about the random variables $$Y_n=\frac{X_n-an^p}{n^p}\ ?$$ – 2017-01-12

0 Answers 0