how to understand results of WLLN with iid condition and without iid condition

Question

Consider independent $r.v.$ $X_1,X_2,...$

WLLN: If there is a constant $p\in[1,2]$ s.t.

$$\sum_{i=1}^\infty\frac{E|X_i|^p}{i^p}<\infty$$

then$$\frac{1}{n}\sum_{i=1}^n(X_i-EX_i)\rightarrow_p0$$

Consider $iid$ $r.v.$ $X_1,X_2,...$

WLLN: the necessary and sufficient condition for $$\frac{1}{n}\sum_{i=1}^nX_i-E(X_1I_{\{|X_1|\le n\}})\rightarrow_p0$$is that$$nP(|X_1|>n)\rightarrow0$$

My question is:

we can treat "$iid$ WLLN" as a special case of "independent WLLN", then why do we have an indicator function in the $iid$ case while we don't have an indicator function in "independent WLLN"? isn't it means we have different WLLN results?

Answer 1

Yes, the results are a bit different.

• in the "independent WLLN", the random variables $X_i$ do not necessarily have the same distribution, but the involved random variables should have a finite moment of order $p$, hence should be in particular integrable.
• In the "i.i.d. WLLN", as the name indicates, the random variables $X_i$ should have the same distribution. The assumption on the tail of $\lvert X_1\rvert$ is weaker than integrability (the tail could for example behave as $1/(n \ln n)$ hence we are not allowed to write $\mathbb E[X_1]$ in this context, while $\mathbb E\left[X_1 \mathbf{1}\{\lvert X_1\rvert\leqslant n\right]$ does make sense.

In order words, these two version of the weak law of large numbers cover different situations.