Interpretation of an asymptotic result in probability

Question

A result in asymptotic theory says the following: Let $Y$ be a real random variable with full support and $E(|Y|)<\infty$. Assume that: $$ (A)\quad \lim_{y\rightarrow \infty} \frac{\Pr(Y\geq y)}{\Pr(Y\geq \kappa h(y))}=1 $$ for some $\kappa>0$ and function $h$ such that $\lim_{y\rightarrow \infty} h'(y)=1 $. Then, $\kappa=1$.

Could you help me understand:(1) How to interpret $(A)$; (2) How to interpret the result that $\kappa$ must be 1; (3) Why would you expect this result to intuitively hold or not hold?

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Thoughts. Regarding (1): The limiting condition says that $ h(y) =\alpha+ y $ in the limit. Then, I think it holds that $$ (B)\quad \lim_{y \rightarrow \infty} \Pr(Y \geq y) = \lim_{y \rightarrow \infty} \Pr(Y \geq \kappa h(y)) = 0. $$ Assumption $(A)$ is stronger than $(B)$, because I think $(A)$ says that $\Pr(Y \geq y)$ and $\Pr(Y \geq \kappa h(y))$ go to zero at the same rate (but I am not sure here about this).

Regarding (2): Somehow, this result suggests that if $(A)$ holds, then $h$ can shift and rescale $y$ (in the limit), but the multiplicative constant $\kappa$ in front of $h$ must be one. Is this a correct interpretation of the result?

Regarding (3): I have zero intuition. If my interpretation of the result is correct (see above), why can $h$ be anything and $\kappa$ must instead be one?

Answer 1

$\newcommand\ka\kappa$Intuition for this result is as follows.

The condition on $h'$ implies that $h(y)=(1+o(1))y$ (as $y\to\infty$). So, for the tail function $T$ given by $T(y):=P(Y\ge y)$ we have $$T(y)=(1+o(1))T(\ka h(y)) =(1+o(1))T((\ka+o(1))y) \tag{1}\label{1}.$$ So, if $\ka\in(0,\infty)\setminus\{1\}$, then $T(y)$ cannot decrease to $0$ faster than $y^{-p}$, for any real $p>0$. So, then we can expect that $$E\max(0,Y)^p=\int_0^\infty dy\,py^{p-1}T(y)=\infty$$ for any real $p>0$, which would contradict the condition $E|Y|<\infty$.

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This intuition is easy to convert to a formal proof, as follows. Moreover, instead of your condition on $h'$, it is enough to assume its less restrictive consequence that $h(y)=(1+o(1))y$ (as $y\to\infty$). Furthermore, instead of your condition $E|Y|<\infty$, it is enough to assume its less restrictive consequence that $E\max(0,Y)^p<\infty$ for some real $p>0$.

Indeed, without loss of generality $\ka>1$. (If $\ka\in(0,1)$, rewrite \eqref{1} as $T(y)=(1+o(1))T((1/\ka+o(1))y))$. Take now any $k\in(1,\ka)$. Then \eqref{1} and the full support condition imply that $$T(ky)\ge k^{-p} T(y)>0$$ for some real $y_0>0$ and all real $y\ge y_0$. So, $$y\in[k^n y_0,k^{n+1}y_0)\implies T(y)\ge T(k^{n+1}y_0)\ge k^{-(n+1)p} T(y_0).$$ So, $$E\max(0,Y)^p=\int_0^\infty dy\,py^{p-1}T(y) \\ \ge\sum_{n=0}^\infty\int_{k^n y_0}^{k^{n+1}y_0} dy\,py^{p-1}k^{-(n+1)p} T(y_0) =\sum_{n=0}^\infty(1-k^{-p})y_0^p T(y_0) =\infty,$$ which contradicts the condition $E\max(0,Y)^p<\infty$.