I was working out some problems from Rick Durrett's Probability theory and Examples (2010 edition), when I came across a very unusual question(reproduced here ad-verbatim):

If $X_n$ is ANY sequence of random variables, there are constants $c_n \to \infty$ so that $$\frac{X_n}{c_n} \to 0 \quad \mbox{a.s}$$

You will find it in Chapter 2: Laws of Large Numbers, under the section on Borel Cantelli Lemma. What makes this question unusual is that there are NO assumptions made on the random variables.

My attempt: I rephrased the question equivalently as:

If $X_n$ is ANY sequence of random variables, there are constants $a_n \to 0$ so that we need to show $$a_nX_n \to 0 \quad \mbox{a.s}$$

Then I showed that it was sufficient to assume $a_n > 0$ and $X_n \geq 0$. To see this, since the limit is going to , the sign of the constants do not matter, hence positivity of $a_n$ can be assumed w.l.o.g. As for an arbitrary r.v, any r.v can be written as:

$$X_n = X_n^+ - X_n^-$$

where $X_n^+ = \max(X_n,0)$ and $X_n^- = \max(-X_n,0)$

Suppose we prove the result for non-negative random variables, then say we have $$b_n X_n^+ \to 0 \quad c_n X_n^- \to 0 \quad \mbox{a.s}$$

Then pick $a_n = \min(b_n,c_n)$. Then this will ensure $$a_nX_n \to 0 \quad \mbox{a.s}$$

For the non-negative case, I was able to prove the result for simple functions: If $X_n$ is simple, and $X_n = \sum_{k=1}^n s_k 1_{A_k}$, take $a_n = \frac{1}{2^{n} \sum_{k=1}^n s_k}$ would work, but only if all functions were simple.

Now the tough part. Handling the non -ve measurable function case. I had difficulty here.

Hence my question is:

I'd like a hint/answer (preferably a hint) on how to solve this particular case. Right now I am looking at manipulating the lemma that every non-negative measurable function can be approximated by a monotone sequence of simple functions.

Thank you.

Note: I use measurable functions and random variables interchangeably. But note that the space is a probability space. Additionally, I didn't find any similar question (I typed convergence random variables). If it has been answered, kindly provide the link.


Hint: Consider $\xi_n(c) := P(|X_n| \geq c)$ and choose $c_n$ such that $\xi_n(2^{-n}c_n)$ is very small.

  • $\begingroup$ Can this be done even if $X_n$ is not integrable? $\endgroup$ – Gautam Shenoy Apr 24 '13 at 15:19

Its very interesting. Based on Thomas' reply, let me respond:

Let us consider the case where $X_n \geq 0$. Now, we know this property of CCDF $$\lim_{u \to \infty} P(X_n > u) = 0$$

So given $n \geq 1, m \geq 1 \quad\exists \delta_{n,m} > 0$ s.t $$ u \geq \delta_{n,m} \Rightarrow P(X_n > u) \leq \frac{1}{2^m}$$

Now pick $c_{m,n}$ such that $c_{m,n} > \delta_{n,m}2^m$ and substitute above to get $$P\left(\frac{X_n}{c_{m,n}} > \frac{1}{2^m}\right) \leq \frac{1}{2^m}$$

Now arrange the $c_{m,n}$ in a grid and pick the diagonal sequence i.e. let $c_n = c_{n,n}$. Now apply the Borel Cantelli Lemma to conclude:

$\forall \epsilon > 0$, pick m s.t $2^{-m} < \epsilon$ and

$$\sum_{n=1}^{\infty}P\left(\frac{X_n}{c_n} > \epsilon\right) \leq \sum_{n=1}^{\infty}P\left(\frac{X_n}{c_n} > \frac{1}{2^m}\right) $$ $$ = \sum_{n=1}^{m}P\left(\frac{X_n}{c_n} > \frac{1}{2^m}\right) + \sum_{n=m+1}^{\infty}P\left(\frac{X_n}{c_n} > \frac{1}{2^m}\right)$$ $$ = \sum_{n=1}^{m}P\left(\frac{X_n}{c_n} > \frac{1}{2^m}\right) + \sum_{n=m+1}^{\infty}P\left(\frac{X_n}{c_n} > \frac{1}{2^n}\right)$$

$$ \leq \sum_{n=1}^{m}P\left(\frac{X_n}{c_n} > \frac{1}{2^m}\right) + \sum_{n=m+1}^{\infty}\frac{1}{2^n} < \infty$$ and thus

$$\Rightarrow P\left(\frac{X_n}{c_n}\to 0 \right) =1$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.