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.


2 Answers 2


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 CDF $$\lim_{u \to \infty} P(X_n > u) = 0$$

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

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

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

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


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? $\endgroup$ Apr 24, 2013 at 15:19

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