3
$\begingroup$

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?

Improve this question
$\endgroup$
7
  • $\begingroup$ Should that second $\overline{F}_1$ in the first line be an $\overline{F}_2$? $\endgroup$
    – jbowman
    Commented May 28, 2012 at 14:49
  • $\begingroup$ Yes, you are correct! $\endgroup$
    – Chris
    Commented May 28, 2012 at 15:08
  • $\begingroup$ I assume the $o(1)$ is taken as $x \to \infty$? $\endgroup$
    – cardinal
    Commented May 28, 2012 at 15:11
  • $\begingroup$ What is $\bar{F}$ compared to $F$? $\endgroup$
    – gui11aume
    Commented May 28, 2012 at 15:22
  • 1
    $\begingroup$ You should carefully study the result and the proof you obtain. In particular, you should try to understand why the restriction $\delta < 1/2$ is completely superfluous and why the slowly-varying condition need only be satisfied by one of the two $\overline F_i$ in order for the result to hold. $\endgroup$
    – cardinal
    Commented May 28, 2012 at 20:27

1 Answer 1

Reset to default
5
$\begingroup$

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.

Improve this answer
$\endgroup$
8
  • $\begingroup$ 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: $0<\delta<\frac{1}{2}$. $\endgroup$
    – Chris
    Commented May 28, 2012 at 20:15
  • $\begingroup$ 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.) $\endgroup$
    – cardinal
    Commented May 28, 2012 at 20:17
  • $\begingroup$ (+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. $\endgroup$
    – cardinal
    Commented May 28, 2012 at 20:46
  • $\begingroup$ Thanks @cardinal. You did most of the job ;-) I just wrote it down. $\endgroup$
    – gui11aume
    Commented May 28, 2012 at 20:47
  • $\begingroup$ Nice job. (And, I didn't do anything, I don't think!) :-) $\endgroup$
    – cardinal
    Commented May 28, 2012 at 20:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.