# Large deviations: Growth of empirical average of iid non-negative random varialbes with infinite expectations?

Let $$X_1,X_2,X_3,...$$ be iid non-negative random variables with $$E[X_i]=\infty$$. I am looking for references on the growth in $$n$$ of the empirical average under assumptions on $$X_1.,..,X_n$$.

A more specific question is the following:

• Under the moment assumption $$E[X_i^p]=1$$ for some $$p\in (0,1)$$, what are deterministic $$a_{n,p}\to_{n\to\infty}\infty$$ such that $$\frac{1}{n}\sum_{i=1}^n X_i = O_P(a_{n,p})$$?

For $$p=1/2$$ a loose bound is given by applying the strong law of large numbers to $$\sqrt{X_i}$$ and then using $$\sum_i X_i \le (\sum_i \sqrt{X_i})^2$$; this gives $$a_{n,1/2}=n$$. I suspect much clever bounds exist.

(Above, $$W_n=O_P(a_n)$$ for $$a_n>0$$ if and only if $$\forall \epsilon>0, \exists K_\epsilon>0$$ such that $$P(|W_n|>K a_n)\le \epsilon$$.)

• You might look up some of the ones with nice scaling properties, like the inverse gaussian.
– mike
Apr 23, 2021 at 6:52

Here it is more convenient to consider the order of magnitude of $$S_n:=\sum_1^n X_i$$, rather than that of $$S_n/n$$.
Take any real $$c>0$$. Let $$x:=cn^{1/p}$$, $$Y_i:=X_i\,1(X_i, $$T_n:=\sum_1^n Y_i$$, $$M_n:=\max_1^n X_i$$. Then $$P(X_1\ge x)\le EX_1^p\,1(X_i\ge x)/x^p<<1/x^p\tag{1}$$ as $$n\to\infty$$, with $$a< meaning $$a/b\to0$$. Next, for any real $$A>0$$, $$EY_1=EX_i\,1(X_i $$EX_i\,1(A\le X_i So, $$\limsup_n EY_1/x^{1-p}\le EX_i^p \,1(A\le X_i for each real $$A$$. Letting now $$A\to\infty$$, we get $$EY_1< and hence $$ET_n< and $$P(T_n\ge x)\le ET_n/x<
So, $$P(S_n\ge x)\le P(S_n\ge x,M_n by (1) and (2). Thus, $$P(S_n\ge cn^{1/p})<<1,$$ for each real $$c>0$$; that is, $$S_n/n^{1/p}\to0$$ in probability; that is, $$S_n=o_P(n^{1/p})$$.
The exponent $$1/p$$ here is the correct (that is, the smallest) one. Indeed, take any real $$r>0$$ such that $$1/r<1/p$$, that is, $$r>p$$. Let $$X_1$$ be such that $$P(X_1\ge u)\sim u^{-r}$$ as $$u\to\infty$$ and $$EX_1^p=1$$. Then for $$y:=n^{1/r}$$ we have $$P(S_n\ge y)\ge P(M_n\ge y)=1-(1-P(X_1\ge y))^n \ge1-\exp\{-nP(X_1\ge y)\}\to1-\exp\{-1\}>0,$$ so that $$S_n\ne o_P(n^{1/r})$$.