Meaning of finite expectation for an absolute value of random variable

Question

I am trying to understand the proof for the following theorem:

Given a sequence of independent identically distributed random variables $X_n$ with common mean $\mu = \mathbb{E}[X_n]$ $\forall n$ and $\mathbb{E}[|X_1|] < \infty$, then the Weak Law of Large Numbers holds:

$$\lim_{n\rightarrow \infty}\dfrac{S_n}{n}\xrightarrow{\mathbb{P}}\mu$$

Now my question is, what is the intuition of $\mathbb{E}[|X_1|]<\infty$? If I understood correctly, this theorem is an alternate for the WLLN, without the assumption of existence of variance. If $\mathbb{E}[X_1]=\mu$, how does that not imply that $\mathbb{E}[|X_1|]<\infty$?

Can you provide an example such that $\mathbb{E}[X]=\mu$ and $\mathbb{E}[|X|]=\infty$ as a counterexample?

Answer 1

As remarked in another solution, the mean $\mathbf E[X]$ is only defined when $\mathbf E[|X|]$ is defined.

The version of the weak law that you state above, requiring only $\mathbf E[|X|] < \infty$ is in fact the standard defition of the weak law of large numbers.

So why do we often see the weak law stated with the condition $\text{Var}(X) < \infty$. The answer is that this version of the weak law is significantly easier to prove.

Proof of WL under finite variance.

If $\text{Var}(X_1) = \sigma^2 < \infty$, and if we denote $\overline X_n = n^{-1} \sum_{i=1}^n X_n$, then we can use Chebyshev's Inequality to note

$$ \mathbf P \left[ | \overline X_n - \mu | > \epsilon \right] \leq \frac{\sigma^2}{n \epsilon^2},$$ which converges to $0$ for any fixed $\epsilon$.

In the absence of finite variance.

We do not prove this in detail, but rather sketch the idea.

The idea is to define truncated variables $Y = X$ if $|X| < C$ and $Y = 0$ otherwise. The random variables $Y$ neccessarily have finite variance, and hence satisfy a weak law of large numbers (by the above).

A technical proof is then required to relate the weak law for this truncated sequence to the full weak law. A proof is given here.

Answer 2

The way I learned things, it's part of the definition that in order for the mean $E(X)$ to exist we must have $E(|X|)<\infty$ (which, under the hood, is just the usual definition of Lebesgue integrability).

This is the standard definition, I think. It's possible that by modifying the definition of integral you can define the expectation without having $E(|X|)<\infty.$ For instance, if you make it principal parts integral, then a centered Cauchy distribution with PDF proportional to $\frac{1}{1+x^2}$ would have "mean zero," but this is not standard, nor is it useful, as far as I know.

So I would say it's just put in that definition for emphasis. Otherwise, people might erroneously think the example above has mean zero since it's symmetric and try to apply the LLN to it.