Say I have a sequence of real independent functions $X_i$ on a probability space $(\Omega, P)$ such that $(X_i)_* P$ gives the probability distribution $d\alpha := \frac{cdx}{(1+x^2)\log(1+x^2)}$ (say for $|x| > 1$) where $c$ is a normalizing constant.

I want to solve an exercise which asks to show $\frac{S_n}{n} := \frac{X_1 + \dots + X_n}{n}$ converges in probability to a constant function. Or equivalently that the characteristic function of $\left(\frac{S_n}{n}\right)_*P$ converges to that of a probability supported on a point.

This boils down to checking the differentiability at $0$ of the characteristic function $\phi$ of $d\alpha$ which I'm having trouble with (Note, the function $x$ is not integrable with respect to $d\alpha$). I have

$$\frac{\phi(s) - 1}{s} = \frac{c}{s}\int\limits_{|x| > 1} \frac{e^{isx} - 1}{(1+x^2)\log(1+x^2)}dx $$

and the exercise would follow if I showed the limit as $s\to 0$ existed since the characteristic function of $\left(\frac{S_n}{n}\right)_*P$ is $\phi(\frac{\cdot}{n})^n$.

Can someone please help me evaluate this limit? Normally, pushing the limit inside the integral works but in this case it definitely will not :(

Edit: The original distribution was modified to have support away from $0$.

  • 1
    $\begingroup$ Surely no such distribution exists because $\int_0^\epsilon\frac{dx}{(1+x^2)\ln(1+x^2)}\sim\int_0^\epsilon\frac{dx}{x^2}=\infty$ for small $\epsilon>0$. $\endgroup$
    – J.G.
    Apr 5 '20 at 16:08
  • $\begingroup$ I see. I'm guessing here, but maybe the exercise wants us to assume that this density is supported away from a neighbourhood of $0$? Can anything be done in this case? For reference, this is Exercise 3.6 in Varadhan's book. I hope I've read correctly.. $\endgroup$ Apr 5 '20 at 16:33

Please refer to the book: W. Feller, An Introduction to Probability and Its Applications(2nd Ed.,1971), Th.VII.7.1, p235 and Th.XVII.2a, p565. In these theorems, the necessary and sufficient conditions of LLN for i.i.d.r.v's are given in terms of distribution function(also characteristic function).

  • $\begingroup$ Thanks for your answer. Can you please explain how the differentiation of the characteristic function argument goes? I understand the truncation argument. $\endgroup$ Apr 22 '20 at 11:44

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