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Let $X_1,X_2$ be independent distributed with cdfs $F_1(x),F_2(x)$, so that $$\overline F_i(x) := 1 - F_i(x) = x^{-\alpha}L_i(x),\ \alpha \geq 0 \>,$$ where $L_i(x)$ is a slowly varying function, that is, it satisfies $L(tx)/L(x) \to 1$ as $x \to \infty$ for all $t > 0$.

For $0<\delta<\frac{1}{2}$ I have the following line:

$\overline{F_{1}}((1-\delta)x)+\overline{F_{2}}((1-\delta)x)+\overline{F_{1}}(\delta x)\overline{F_{2}}(\delta x) =\left(\overline{F_{1}}((1-\delta)x)+\overline{F_{2}}((1-\delta)x)\right)\left(1+o(1)\right).$

Can someone explain why we have $o(1)$ as $x \rightarrow \infty$ in the second line of the equation?

cardinal
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Chris
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1 Answers1

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Thanks @cardinal for the definition.

Simplifying on both sides of the $=$ sign, we get

$$ \frac{\bar{F}_1(\delta x) \bar{F}_2(\delta x)}{\bar{F}_1((1-\delta) x) + \bar{F}_2((1-\delta) x)} = o(1), $$

which means that it tends to 0 (as $x \rightarrow \infty$ since you specified it).

Is your question about the meaning of $o(1)$ or how to prove this inequality? If it is the first case, the small $o$ Landau notation $f(x) = o(g(x))$, $x\to a$ sometimes said "$f$ is negligible compared to $g$ in the neighborhood of $a$", means that the function $f$ is such that

$$ \lim_{x\to a} \frac{f(x)}{g(x)} = 0, $$

so $o(1)$ means that $f$ tends to 0.

If your question is actually the second one, I feel we miss some information, like can $\delta$ be greater than 1 (in which case the limit is trivially true)?

EDIT: with the new assumptions you can prove it as follows:

$$ \frac{\bar{F}_1(\delta x) \bar{F}_2(\delta x)}{\bar{F}_1((1-\delta) x) + \bar{F}_2((1-\delta) x)} = \frac{(\delta x)^{-2\alpha}L_1(\delta x)L_2(\delta x)}{((1 - \delta)x)^{-\alpha} (L_1((1-\delta) x) + L_2((1-\delta) x))}. $$

Dividing numerator and denominator by $L_1(x)L_2(x)$ and taking the limit, the whole things is equivalent to

EDIT:

$$\frac{(\delta x)^{-2\alpha}}{((1 - \delta)x)^{-\alpha}(\frac{1}{L_2(x)} + \frac{1}{L_1(x)})}.$$

By hypothesis $L_i(x) = o(x^\alpha)$, in the limit $\frac{1}{L_i(x)} > x^{-\alpha}$ so the denominator is bounded and the ration tends to 0.

mpiktas
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gui11aume
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  • Thank you very much for your answer. Indeed, I wanted a proof. I did not post all the information I have, because I didn't think that it was necessary. I am sorry for that. I am updating the post now: $0 – Chris May 28 '12 at 20:15
  • There are two clear situations in which the result obviously follows. (**1**) $\delta \geq 1/2$ and, what I suspect to be the likely implicit assumption here given the OP's previous questions, (**2**) $\bar F_i = x^{-\alpha} L_i$ for $i=1,2$ where $L_i$ is slowly varying. (Actually, those with a keen eye will note that only *one* of $\bar F_1$ and $\bar F_2$ need satisfy this condition.) – cardinal May 28 '12 at 20:17
  • (+1) I get the feeling the OP's proof is being proved for him on this site in somewhat random order. For example, this seems to be essentially one side of the argument one would need to show the $n=2$ case from [one of his other questions](http://stats.stackexchange.com/questions/29052). – cardinal May 28 '12 at 20:46
  • Thanks @cardinal. You did most of the job ;-) I just wrote it down. – gui11aume May 28 '12 at 20:47
  • Nice job. (And, I didn't do *anything*, I don't think!) :-) – cardinal May 28 '12 at 20:49
  • Asking the right question, giving the right hint and doing the right edits... that's a rare talent. I mean it :D – gui11aume May 28 '12 at 20:51
  • Great job! One question, when you divide the denominator by $L_1(x)L_2(x)$ you have: $\frac{(L_1((1−δ)x)+L_2((1−δ)x))}{L_1(x)L_2(x)}$. Why is this 2? – Chris May 29 '12 at 02:59
  • Oops, seems I went a bit too fast. Thanks for spotting this. I edited the proof. – gui11aume May 29 '12 at 08:18