The law of large numbers for diverging moments [closed]

Question

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.

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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 $C_n\to\infty$.

Answer 1

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 $n\to\infty$.

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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), 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.

Answer 2

See theorem 1 in Erickson's 1973 paper.