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Consider a set of i.i.d. (positive) random variables $\{X_i\}_{i=1}^N$. Each variable $X_i$ has a distribution with finite mean but infinite variance. In particular, if $P_{X_i}(x)$ is the P.D.F. of the random variable $X_i$ $P_{X_i}(x) \sim \frac{1}{x^{\alpha +1}}$ (with $1< \alpha <2$) for $x>\tilde{x}>0$ and $P_{X_i}(x) = 0$ otherwise.
If we consider the variable $W_N = \frac{(\sum_{i=1}^N (X_i - \langle X\rangle))}{N}$ (where $\langle X\rangle \equiv \int P_{X_i}(x) x dx$) , the variance of $W_N$ goes to 0 for $N \rightarrow \infty$ (for the law of large number)
I want to get the scaling exponent which determine the leading term of the variance's decrease rate of $W_N$.
In other words, for $N \rightarrow \infty$ $\int_{-\infty}^{\infty} w^2P_{W_N}(w) dw = aN^{-b} + o(N^{-b})$, with $a>0, b>0$. I want to get the value of $b$ (that will depend on $\alpha$ value) exponent (and possibly also of $a$).
Any idea?

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Assuming that $\langle X\rangle$ denotes the expectation of each of the iid $X_i$'s, the variance of $W_N$ is $\infty$. Indeed, $$Var\,W_N=\frac1{N^2}Var\sum_{i=1}^N X_i=\frac1{N^2}\,N\,Var\,X_1=\infty,$$ since $Var\,X_1=\infty$.

(Indeed, by the strong law of large numbers, $W_N\to0$ almost surely and hence in probability (as $N\to\infty$). However, in general (and in this particular case) this does not imply that $Var\,W_N=E(W_N^2)$ goes to $0$.)

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  • $\begingroup$ $E(X_i)<\infty$; i corrected the text of my question. There is no divergence in 0 (the random variable $X_i$ is positive). I don't understand how can the variance goes to infinity if the probability to find $|W_N| > \epsilon$ goes to 0 $\forall \epsilon >0 $ in the limit $N \rightarrow \infty$ $\endgroup$ Mar 21, 2021 at 14:59
  • $\begingroup$ @user1172131 : (i) The problem with an infinite mean/variance is never with what the distribution is near $0$; it is always with what the distribution is near $\pm\infty$. (ii) The convergence of a sequence $(Y_N)$ of random variables in probability to $0$ does not in general imply the convergence of $EY_N$ to $0$. E.g., suppose $P(Y_N=N^2)=1/N$ and $P(Y_N=0)=1-1/N$. Then for any real $\epsilon>0$ we have $P(|Y_N|>\epsilon)\le1/N\to0$ as $N\to\infty$, but $EY_N=N\to\infty$. $\endgroup$ Mar 21, 2021 at 15:48
  • $\begingroup$ So why did you wrote that $EX_i = \infty$? $EX_i = C\int_{\tilde{x}}^{\infty} \frac{x}{x^{1 + \alpha}} = C\int_{\tilde{x}}^{\infty} \frac{}{x^{\alpha}}$ (where C is the normalizzation factor). For $1<\alpha<2$ the integral is well defined $\endgroup$ Mar 21, 2021 at 15:52
  • $\begingroup$ @user1172131 : Oops! I have removed the comment at the end of the answer. $\endgroup$ Mar 21, 2021 at 17:03

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