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Let $X_i$ is power-law-distributed random variable $f(x)=C_0x^{-k}$ where $1

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

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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).2012-05-14
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    Assuming that what you do not understand is *how to prove it* (and not *its meaning*), please see edited answer.2012-05-14
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    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.2012-05-15
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    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 $c_iN^{k_i-1}$.2012-05-15
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    @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)2012-08-21