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?


1 Answer 1


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 $0$.)

  • $\begingroup$ $E(X_i)<\infty$; 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.