qtbabe9876a9

Answered

2022-05-29

Suppose we have a sequence of positive random variables ${X}_{1},{X}_{2},...,X$. I am trying to prove a characterization of almost sure convergence.

It states that ${X}_{n}\to X$ almost surely iff for every $\u03f5>0$, $\underset{n\to \mathrm{\infty}}{lim}P[\underset{k\ge n}{sup}\frac{{X}_{k}}{X}>1+\u03f5]=0$ and $\underset{n\to \mathrm{\infty}}{lim}P[\underset{k\ge n}{sup}\frac{X}{{X}_{k}}>1+\u03f5]=0$.

If I assume almost sure convergence, then the implication is easy but I am not being able to prove the other way round.

It states that ${X}_{n}\to X$ almost surely iff for every $\u03f5>0$, $\underset{n\to \mathrm{\infty}}{lim}P[\underset{k\ge n}{sup}\frac{{X}_{k}}{X}>1+\u03f5]=0$ and $\underset{n\to \mathrm{\infty}}{lim}P[\underset{k\ge n}{sup}\frac{X}{{X}_{k}}>1+\u03f5]=0$.

If I assume almost sure convergence, then the implication is easy but I am not being able to prove the other way round.

Answer & Explanation

concludirgt

Expert

2022-05-30Added 11 answers

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MathJax(?): Can't find handler for document
Since all the random variables are positive,

$$

Thus,

$$

But

$$

That is, $\mathsf{P}(lim\u2006sup{X}_{n}/X>1)=0$ is equivalent to

$$

and, similarly, $\mathsf{P}(lim\u2006inf{X}_{n}/X<1)=0$ is equivalent to

$$

The last condition is equivalent to $\underset{n\to \mathrm{\infty}}{lim}\mathsf{P}\phantom{\rule{negativethinmathspace}{0ex}}(\underset{k\ge n}{sup}X/{X}_{k}\ge 1+\u03f5)=0\phantom{\rule{1em}{0ex}}\mathrm{\forall}\u03f5>0$.

$$

Thus,

$$

But

$$

That is, $\mathsf{P}(lim\u2006sup{X}_{n}/X>1)=0$ is equivalent to

$$

and, similarly, $\mathsf{P}(lim\u2006inf{X}_{n}/X<1)=0$ is equivalent to

$$

The last condition is equivalent to $\underset{n\to \mathrm{\infty}}{lim}\mathsf{P}\phantom{\rule{negativethinmathspace}{0ex}}(\underset{k\ge n}{sup}X/{X}_{k}\ge 1+\u03f5)=0\phantom{\rule{1em}{0ex}}\mathrm{\forall}\u03f5>0$.

Trevor Wood

Expert

2022-05-31Added 5 answers

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MathJax(?): Can't find handler for document
Let

$$

Then ${A}_{n+1}^{\u03f5}\subseteq {A}_{n}^{\u03f5}$ and ${B}_{n+1}^{\u03f5}\subseteq {B}_{n}^{\u03f5}$, thus

$$

and

$$

Since the above hold for every $\u03f5>0$ we conclude that

$$

These two are equivalent to the following two

$$

Now observe that $\underset{n\to \mathrm{\infty}}{lim\u2006sup}\frac{{X}_{n}}{X}=\frac{1}{X}\cdot \underset{n\to \mathrm{\infty}}{lim\u2006sup}{X}_{n}$ and $\underset{n\to \mathrm{\infty}}{lim\u2006sup}\frac{X}{{X}_{n}}=X\cdot \underset{n\to \mathrm{\infty}}{lim\u2006sup}\frac{1}{{X}_{n}}=X\cdot \frac{1}{\underset{n\to \mathrm{\infty}}{lim\u2006inf}{X}_{n}}.$. Therefore,

$$

taking intersections we conclude that ${X}_{n}\stackrel{\text{a.s}}{\u27f6}X$

$$

Then ${A}_{n+1}^{\u03f5}\subseteq {A}_{n}^{\u03f5}$ and ${B}_{n+1}^{\u03f5}\subseteq {B}_{n}^{\u03f5}$, thus

$$

and

$$

Since the above hold for every $\u03f5>0$ we conclude that

$$

These two are equivalent to the following two

$$

Now observe that $\underset{n\to \mathrm{\infty}}{lim\u2006sup}\frac{{X}_{n}}{X}=\frac{1}{X}\cdot \underset{n\to \mathrm{\infty}}{lim\u2006sup}{X}_{n}$ and $\underset{n\to \mathrm{\infty}}{lim\u2006sup}\frac{X}{{X}_{n}}=X\cdot \underset{n\to \mathrm{\infty}}{lim\u2006sup}\frac{1}{{X}_{n}}=X\cdot \frac{1}{\underset{n\to \mathrm{\infty}}{lim\u2006inf}{X}_{n}}.$. Therefore,

$$

taking intersections we conclude that ${X}_{n}\stackrel{\text{a.s}}{\u27f6}X$

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