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

Question

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 $P(|W_n|>K a_n)\le \epsilon$.)

Answer 1

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 $x:=cn^{1/p}$, $Y_i:=X_i\,1(X_i<x)$, $T_n:=\sum_1^n Y_i$, $M_n:=\max_1^n X_i$. 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 $a<<b$ meaning $a/b\to0$. Next, for any real $A>0$, $$EY_1=EX_i\,1(X_i<A)+EX_i\,1(A\le X_i<x),$$ $$EX_i\,1(A\le X_i<x)\le EX_i^p \,1(A\le X_i<x)x^{1-p}.$$ So, $$\limsup_n EY_1/x^{1-p}\le EX_i^p \,1(A\le X_i<x)\le EX_i^p \,1(A\le X_i)$$ for each real $A$. Letting now $A\to\infty$, we get $EY_1<<x^{1-p}$ and hence $ET_n<<nx^{1-p}$ and $$P(T_n\ge x)\le ET_n/x<<n/x^p=1/c^p.\tag{2}$$

So, $$P(S_n\ge x)\le P(S_n\ge x,M_n<x)+P(M_n\ge x) \le P(T_n\ge x)+nP(X_1\ge x)<<1,$$ 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 $S_n\ne o_P(n^{1/r})$.