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 $P(|W\_n|>K a\_n)\le \epsilon$.)



Large deviations: Growth of empirical average of iid non-negative random varialbes with infinite expectations?
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 $S\_n\ne o\_P(n^{1/r})$.


