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I am generally able to prove that a sequence of random variables $X_n$ converge almost surely to a random variable $X$ by using the following strategy:

Take any typical sample point $\omega\in\Omega-A$ where $A$ is the union of all possible null sets. Then for this selected $\omega$, $X_n(\omega)$ and $X(\omega)$ are real numbers, so we just have to show that $X_n(\omega)\to X(\omega)$ by usual laws of real analysis i.e. given $\varepsilon>0$ there exists $N\in\mathbb N$ such that $\forall n\geq N$, $|X_n(\omega)-X(\omega)|<\varepsilon$.

However, I am confused as to how one can show that a sequence of random variables DOES NOT converge to a random variable almost surely. In particular, consider the following question:

Let $\{X_n\}$ be a sequence of independent random variables such that $P(X_n=n)=P(X_n=-n)=\dfrac{1}{2\sqrt{n}}$ and $P(X_n=0)=1-\dfrac{1}{\sqrt{n}}$. Show that $\{\dfrac{X_n}{n}\}$ does not converge almost surely to $0$.

I am aware that if $Y_n$ are independent random variables then $$\sum_{n=1}^\infty P(|Y_n|\geq \epsilon)<\infty\space\space\forall\epsilon>0\iff P(Y_n\to 0)=1$$

So in this question, for any $\epsilon>0$ $$P(\dfrac{|X_n|}{n}\geq\epsilon)=P(X_n\neq0)=\dfrac{1}{\sqrt{n}}$$

Since $$\sum_{n=1}^\infty \dfrac{1}{\sqrt{n}}=\infty$$ we have that $X_n/n$ cannot converge almost surely to $0$.

However, I spent a lot of time trying to show that $X_n/n$ does not converge a.s. to $0$ using the definition of a.s. convergence. That is, I wanted to show that there exists $A=\{\omega: X_n(\omega)/n$ does not converge to $0\}$ with $P(A)>0$. Somehow I find this quite hard to do. What is the general method to show that a sequence does not converge almost surely from the definitions only?

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    $\begingroup$ You're essentially there as, for any $\epsilon>0$, you have basically shown that $|X_n/n|\ge\epsilon$ will a.s. recur (since it is expected to recur an infinite number of times and these are independent). $\endgroup$ Commented May 24, 2015 at 7:17

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Define $$A_n := \left\{ \left| \frac{X_n}{n} \right| \geq 1 \right\}.$$ The set $$A := \limsup_{n \to \infty} A_n$$ satisfies, by the Borel-Cantelli lemma, $\mathbb{P}(A)>0$ and $\frac{X_n(\omega)}{n}$ does not converge to $0$ for each $\omega \in A$.

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  • $\begingroup$ So you used independence of $A_n$ and the fact that $\sum_{n=1}^\infty P(A_n)=\infty$ which immediately implies that $P(\limsup A_n)=1$, right? $\endgroup$ Commented May 24, 2015 at 7:23
  • $\begingroup$ @LandonCarter Yes, that's actually what you have already shown. My point is simply that the Borel-Cantelli lemma does not only yield that $X_n/n$ does not converge to $0$ almost surely; it gives also an explicit expression for a set where the convergence does not hold true. $\endgroup$
    – saz
    Commented May 24, 2015 at 7:26
  • $\begingroup$ Shouldn't it be more precise if you defined your $A_n$ as $\{|X_n/n|>0.5\}$? If not $0.5$ then anything smaller than $1$? Because $P(A_n)=0$ for all $n$ in your case. $\endgroup$ Commented May 24, 2015 at 7:28
  • $\begingroup$ Okay, so the extra thing that you have shown is that $A$ is our desired set . I will try to keep this technique in mind. Thanks!! $\endgroup$ Commented May 24, 2015 at 7:30
  • $\begingroup$ @LandonCarter What do you mean by "more precise"? If you want to make the set as big as possible, then you can define $$A_n^k := \left\{ \left| \frac{X_n}{n} \right|>\frac{1}{k} \right\}$$ and set $$A:= \bigcup_{k \in \mathbb{N}} \limsup_{n \to \infty} A_n^k.$$ $\endgroup$
    – saz
    Commented May 24, 2015 at 7:31

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