I have a random variable $X$ whose first and second moments are given as
$$
E[X] \propto C\_n^{1-a},\quad E[X^2] \propto C\_n^{2-a}\quad (0 < a < 1),
$$
where $C\_n$ satisfies
$$
\lim\_{n\to\infty} C\_n = \infty, \quad \lim\_{n\to\infty} \frac{C\_n^a}{n} = 0.
$$
I want to see the limit of a ratio, 
$$
R\_n = \frac{\sum\_{i=1}^n X\_i^2}{(\sum\_{i=1}^n X\_i)^2},
$$
where $X\_1,\dots, X\_n$ is i.i.d. samples of $X$. If the first and second moment is finite, one can easily see that
$$
\lim\_{n\to\infty} R\_n = \lim\_{n\to\infty} \frac{1}{n} \frac{\sum\_{i=1}^n X\_i^2/n}{(\sum\_{i=1}^n X\_i)^2/n^2} = 0,
$$
by the law of large numbers. But in this case, $C\_n$ also goes to infinity and as far as I know, the law of large numbers does not hold for infinite moments. At first, I forgot this fact and just thought as follows,
$$
\lim\_{n\to\infty} R\_n \propto \lim\_{n\to\infty} \frac{1}{n} \frac{C\_n^{2-a}}{C\_n^{2-2a}}
= \lim\_{n\to\infty} \frac{C\_n^a}{n} = 0,
$$
by the condition given to $C\_n$. I empirically checked whether $R\_n$ converges to zero as $n\to\infty$, and it indeed goes to zero. Is there any clear proof, or if it is wrong, can anybody explain why it is wrong?


Thanks in advance for the answers.




---


Sorry for the confusion. What Iosif Pinelis interpreted is right. To be more specific, I have a r.v. $X\_n$ constrained on an interval $[0, C\_n]$, and its first and second moments
are given as


$$
EX\_n = \frac{C\_n^{1-a}}{1-a} - \gamma(1-a, C\_n),\quad EX\_n^2 = \frac{C\_n^{2-a}}{2-a} - \gamma(2-a, C\_n),
$$
where $\gamma(\cdot,\cdot)$ is a lower incomplete gamma function. I compute the ratio $R\_n$ by drawing $n$ i.i.d. samples $X\_{n,1}, \dots X\_{n,n}$ of $X\_n$. For a fixed $n$ (and thus $C\_n$), there is no problem, but I want to know how $R\_n$ behaves as $n\to\infty$, so $C\_n\to\infty$. 



The law of large numbers for diverging moments
I interpret the question as follows (cf. the comment by **Nate Eldredge**).  

For each natural $n$, let $X\_n,X\_{n,1},\dots,X\_{n,n}$ be independent identically distributed (i.i.d.) random variables (r.v.'s) such that 
$$
EX\_n \bowtie C\_n^{1-a},\quad EX\_n^2\bowtie C\_n^{2-a}\quad (0 < a < 1),
$$
where $C\_n$ satisfies
$$(1)\qquad
\lim\_{n\to\infty} C\_n = \infty, \quad \lim\_{n\to\infty} \frac{C\_n^a}{n} = 0, 
$$
and $a\bowtie b$ means that $a\triangleleft b\triangleleft a$ and $a\triangleleft b$ means $a=O(b)$. 
Let  

$$
R\_n := \frac{\sum\_{i=1}^n X\_{n,i}^2}{(\sum\_{i=1}^n X\_{n,i})^2}. 
$$
Show that $R\_n\to0$; here and elsewhere, the convergence of r.v.'s is in probability, as $n\to\infty$.




---


The conclusion $R\_n\to0$ is indeed true and can be proved as follows, actually under the more general condition
$$(2)\qquad 
EX\_n \triangleright C\_n^{1-a},\quad EX\_n^2 \triangleleft C\_n^{2-a}\quad (0 < a < 1),
$$
where $a\triangleright b$ means $b\triangleleft a$. 


For each natural $n$, let $V\_n,V\_{n,1},\dots,V\_{n,n}$ be i.i.d. r.v.'s. 
By the well-known necessary and sufficient condition for the (weak) law of large numbers (LLN) (see e.g. Theorem 3 of Ch. IX of [Petrov](http://rads.stackoverflow.com/amzn/click/3642658113)), for $\sum\_{i=1}^n V\_{n,i}\to0$ it is sufficient that for each real $\tau>0$ 


(i) $nP(|V\_n|\ge\tau)\to0$, 


(ii) $nEV\_n^2\,I\{|V\_n|<\tau\}\to0$, and 


(iii) $nEV\_n\,I\{|V\_n|<\tau\}\to0$, where $I$ denotes the indicator. 


Let now $Y\_{n,i}:=X\_{n,i}^2/B\_n$ and $Y\_n:=X\_n^2/B\_n$, where $B\_n:=\rho\_n n C\_n^{2-a}$ and the $\rho\_n$'s are any positive real numbers such that $\rho\_n\to\infty$ and $\rho\_n C\_n^a/n\to0$; by (1), such $\rho\_n$'s exist. Then, by (2), for each real $\tau>0$,  

$$nP(|Y\_n|\ge\tau)\le nE|Y\_n|/\tau=nEX\_n^2/(\tau B\_n)\triangleleft nC\_n^{2-a}/B\_n=1/\rho\_n\to0,$$ 
$$nEY\_n^2\,I\{|Y\_n|<\tau\}\le\tau nE|Y\_n|\triangleleft1/\rho\_n\to0\ \text{(cf. the previous line)},$$
$$\big|nEY\_n\,I\{|Y\_n|<\tau\}\big|\le nE|Y\_n|\triangleleft1/\rho\_n\to0.$$ 
So, by the above sufficient condition for the LLN, 
$$(3)\qquad \sum\_{i=1}^n X\_{n,i}^2/B\_n=\sum\_{i=1}^n Y\_{n,i}\to0.
$$


Next, let $Z\_{n,i}:=(X\_{n,i}-EX\_{n,i})/E\_n$ and $Z\_n:=(X\_n-EX\_n)/E\_n$, where $E\_n:=nC\_n^{1-a}$. Then, by (2) and (1), for each real $\tau>0$,  

$$nP(|Z\_n|\ge\tau)\le nEZ\_n^2/\tau^2\triangleleft nEX\_n^2/E\_n^2\triangleleft nC\_n^{2-a}/E\_n^2=C\_n^a/n\to0,
$$ 
$$nEZ\_n^2\,I\{|Z\_n|<\tau\}\le nEZ\_n^2\to0\ \text{(cf. the previous line)},$$
$$\big|nEZ\_n\,I\{|Z\_n|<\tau\}\big|=\big|nEZ\_n\,I\{|Z\_n|\ge\tau\}\big|\le nEZ\_n^2/\tau\to0.$$ 
So, by the same sufficient condition for the LLN, $\sum\_{i=1}^n Z\_{n,i}\to0$; that is, 
$\sum\_{i=1}^n X\_{n,i}/E\_n-nEX\_n/E\_n\to0.
$
On the other hand, by (2), $nEX\_n/E\_n\triangleright1$. So, $\sum\_{i=1}^n X\_{n,i}/E\_n\triangleright1$ and hence 
$$\Big(\sum\_{i=1}^n X\_{n,i}\Big)^2\triangleright E\_n^2=n^2C\_n^{2-2a}. 
$$
Comparing this with (3), one concludes that 
$$R\_n<<\frac{B\_n}{E\_n^2}=\rho\_n C\_n^a/n\to0, 
$$
by the choice of $\rho\_n$. 
This completes the proof. 


