# Rate of variance's decrease for the mean's distribution of infinite variance i.i.d. random variables

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?

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 $$0$$.)
• $E(X_i)<\infty$; 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 $N \rightarrow \infty$ Mar 21, 2021 at 14:59
• @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 $EY_N=N\to\infty$. Mar 21, 2021 at 15:48
• 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 Mar 21, 2021 at 15:52