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$X$ and $Y$ are nonnegative rv's with distribution functions with reguraly varying distribution functions. They can be written as $$F_X(x)=L_X(x)x^{-\alpha}$$ and similarly for $Y$. $L_X$ and $L_Y$ are slowly varying function and $\alpha>0$. Why is limit $$ \lim_{x\to\infty}x^{-\alpha}\frac{L_X(x)L_Y(x)}{L_X(x)+L_Y(x) }=0? $$ Thanks!

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  • $\begingroup$ Can you specify what are $L_X(x)$ and $L_Y(y)$, kind of PDFs for some random variables? $\endgroup$ Aug 22, 2016 at 10:01
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    $\begingroup$ @josu, is now better? $\endgroup$
    – Kyoto
    Aug 22, 2016 at 11:03
  • $\begingroup$ I think that as your limit has a polynomial part value $x^{-\alpha}$, I think that it will go faster to infinity than your slowly varying functions, making the limit tend to zero always as $\alpha > 0$. However, it really depends on the nature of the "slowly varying functions", because that variation will imply the value of the limit. $\endgroup$ Aug 22, 2016 at 11:22

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Write your ratio as $$ {1\over {1\over L_X(x)x^{-\alpha}}+{1\over L_Y(x)x^{-\alpha}}} $$ and then let $x\to\infty$.

By the way, if $F_X$ is the cumulative distribution function of $X$ (that is, $F_X(x)=P[X\le x]$) then you must have in mind that $1-F_X(x)$ is regularly varying as $x\to \infty$.

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  • $\begingroup$ You are totaly right! Thanks! :-) $\endgroup$
    – Kyoto
    Aug 24, 2016 at 6:09
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Note that \begin{align*} L_X(x) = x^{\alpha} F_X(x). \end{align*} Then \begin{align*} x^{-\alpha}\frac{L_X(x)L_Y(x)}{L_X(x)+L_Y(x)} &= \frac{F_X(x)F_Y(x)}{F_X(x)+F_Y(x)} \rightarrow \frac{1}{2}. \end{align*}

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