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Let $X_i$ is power-law-distributed random variable $f(x)=C_0x^{-k}$ where $1<k_i\le3$. What is the exponent $k$ of the variable $$ X=\sum_{i=1}^N X_i \ ? $$

My doubt come from the fact that $X$ as a sum of i.i.d has to tend to a $\alpha$-stable distribution. The generic exponent $\alpha$ of a generic $\alpha$-stable distribution can lay only in the range $(0,2]$, that imply $1<k\le 3$. But if we try to use the rule of Fourier transform for the sum of i.i.d. random variables (namely the convolution is the product of the Fourier transforms) as power law we can get an arbitrary big exponent $k$ (isn't it?). So at some point my reasoning is wrong. I guess that the mistake is in the convolution of the power law distributions.

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Assume for simplicity that $\bar c_i/x^{k_i-1}\leqslant\mathrm P(X_i\geqslant x)\leqslant c_i/x^{k_i-1}$ when $x\to\infty$ and that each random variable $X_i$ is almost surely nonnegative. Then, for every $1\leqslant j\leqslant N$, $$ [X_j\geqslant x]\subseteq[X\geqslant x]\subseteq\bigcup_{i=1}^N[X_i\geqslant x/N]. $$ This implies that $$ \max\limits_{i=1}^N\mathrm P(X_i\geqslant x)\leqslant\mathrm P(X\geqslant x)\leqslant\sum_{i=1}^N\mathrm P(X_i\geqslant x/N), $$ hence $X$ has exponent $k=\min\limits_{i=1}^Nk_i$ in the sense that, when $x\to+\infty$, $$ \bar C_N/x^{k-1}\leqslant\mathrm P(X\geqslant x)\leqslant C_N/x^{k-1}. $$ The result you mention about $\alpha$-stable distribution concerns the regime where $N\to\infty$ and one rescales $X$, hence the constants $C_N$ and $\bar C_N$ come into play and modify the exponent $k$.

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I don't understand the meaning of the "main" inequality (the third row from the top). – emanuele May 14 '12 at 15:57
Assuming that what you do not understand is how to prove it (and not its meaning), please see edited answer. – Did May 14 '12 at 16:36
I took some time to think about your proof, but i still don't understand it. I am sorry but, i can't guess why you get last inequality from the previous. Actually i understand that must be as you stated, but i don't understand the last logical steps. – emanuele May 15 '12 at 8:02
The LHS is at least $\bar c_i/x^{k_i-1}$ for each $i$, hence at least $\bar C_N/x^{k-1}$ where $k$ may be any $k_i$, for example their minimum, and $\bar C_N$ is $\bar c_i$ for this $k_i$. // Each term in the RHS is at most $c_i/(x/N)^{k_i-1}\leqslant c_iN^{k_i-1}/x^{k-1}$ for $x\geqslant1$. Hence the whole RHS is at most $C_N/x^k$ where $C_N$ is the sum over $i$ of $c_iN^{k_i-1}$. – Did May 15 '12 at 8:39
@did this time I looked at the question and the first two lines of the answer and I guessed that you gave the answer. I was right. You are estimatable) – Seyhmus Güngören Aug 21 '12 at 10:05

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