Can someone provide a proof for this theorem that was used in the proof of law of large number?

Let $$X_1,X_2,X_3,...$$ be $$i.i.d.$$ with finite mean $$\mu$$.

Then let $$Y_i=X_i1_{\{|X_i|

There will be only finitely many terms such that $$Y_i\neq X_i$$

While that lecture notes did not provide proof, it hinted it is related to Borel–Cantelli.

What is the proof for the statement

• What is $k$? After all, when $k=0$ (or, more generally, when $\Pr(|X_i|\lt k)=0$) the statement evidently can be false, so there must be some kind of restriction on $k.$ – whuber Jun 9 at 13:53
• Sorry, that is a mistype, there is no k, there is only i – Preston Lui Jun 9 at 14:08
• You can find the proof in any strong law proofs of which there’s a ton in literature and textbooks, e.g. see p 57 here math.mit.edu/~sheffield/2016175/Lecture6.pdf – Aksakal Jun 9 at 14:21
• @Aksakal Indeed, and this is the first lemma used for the general proof, which I don't know how to prove – Preston Lui Jun 9 at 15:19

If you are going to apply the Borel-Cantelli lemma, that means you expect to show that the sum of all the chances of the events $$Y_i\ne X_i$$ is finite. Somehow this must be derivable from the assumptions--and about the only useful assumption available is that the common distribution has finite mean. What is the connection between these statements?

Notice that $$Y_i \ne X_i$$ is equivalent to $$|X_i| \ge i.$$ There's no problem working with $$|X_i|$$ instead of $$X_i$$ because

1. The independence of the $$X_i$$ implies the independence of the $$|X_i|.$$

2. Because the $$X_i$$ are identically distributed, the $$|X_i|$$ are identically distributed.

3. The mean of $$X_i$$ is defined and finite if and only if the mean of $$|X_i|$$ is defined and finite.

What's the connection between the mean of a variable and the chances that it's large? The answer lies in the "tail probability expectation formula" (see the reference or consult Expectation of a function of a random variable from CDF),

$$\mathbb{E}(|X_j|) = \int_0^\infty \Pr(|X_j| \gt x)\,\mathrm{d}x$$

for any $$j.$$

To relate this to $$\Pr(|X_i|\ge i)$$ we can break this integral into pieces at the integers $$i=1,2,3,\ldots$$ and underestimate its argument a little in each piece because the $$X_i$$ are identically distributed. Specifically, for any $$x\lt i,$$

$$\Pr(|X_j| \gt x) \ge \Pr(|X_j| \ge i) = \Pr(X_i \ge i).$$

This is the key step in the following derivation, which begins with observation $$(3)$$ that $$|X_j|$$ has finite expectation:

\eqalign{ \infty \gt \int_0^\infty \Pr(|X_j| \gt x)\,\mathrm{d}x &= \sum_{i=1}^\infty \int_{i-1}^i \Pr(|X_j| \gt x)\,\mathrm{d}x \\ &\ge\sum_{i=1}^\infty \int_{i-1}^i \Pr(|X_i| \ge i)\,\mathrm{d}x \\ &= \sum_{i=1}^\infty \Pr(|X_i| \ge i)\,\int_{i-1}^i \mathrm{d}x \\ &= \sum_{i=1}^\infty \Pr(|X_i| \ge i). }

That's the condition for applying Borel-Cantelli, QED.

Reference

Ambrose Lo, Demystifying the Integrated Tail Probability Expectation Formula. The American Statistician Volume 73, Number 4, November 2019, pp 367-374.

• Thanks, to be precise so it also works for discrete RV, the inequality can also be rewritten as $\geq \sum_{i=1}^{+\infty}Pr(|X_i|\geq i)$ – Preston Lui Jun 9 at 15:57
• Thank you. I have taken care of that in an edited version of the derivation. – whuber Jun 9 at 16:28