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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?


Thoughts. Regarding (1): The limiting condition says that $ h(y) =\alpha+\beta y $ with $\beta>0$ in$ 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?

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?


Thoughts. Regarding (1): The limiting condition says that $ h(y) =\alpha+\beta y $ with $\beta>0$ 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?

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?


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?

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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)<\infty$$\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?


Thoughts. Regarding (1): SupposeThe limiting condition says that $ h $ is$ h(y) =\alpha+\beta y $ with increasing$\beta>0$ in $y $the limit. Then, I think it holds that $$ (B)\quad \lim_{y \rightarrow \infty} \Pr(Y \geq y) = \lim_{y \rightarrow \infty} \Pr(Y \geq \kappa h(y)) = 0. $$ In this setting with $h$ increasing, AssumptionAssumption $(A)$ 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). If I am correct, what is an example of function $h$ such that $(B)$ holds but $(A)$ does not?

When $h$ is not increasing, in general, we may find that $\lim_{y \rightarrow \infty} \Pr(Y \geq \kappa h(y))$ does not exist. However, does statement (A) still imply that $\lim_{y \rightarrow \infty} \Pr(Y \geq \kappa h(y))$ exists and must be equal to zero?

Regarding (2): Somehow, this result suggests that if $(A)$ holds, then $h$ can be anything (can shift, shift and rescale, or apply nonlinear transformations to $y$ (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?

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 $\lim_{y\rightarrow \infty} h'(y)<\infty$. 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?


Thoughts. Regarding (1): Suppose that $ h $ is increasing in $y $. Then, $$ (B)\quad \lim_{y \rightarrow \infty} \Pr(Y \geq y) = \lim_{y \rightarrow \infty} \Pr(Y \geq \kappa h(y)) = 0. $$ In this setting with $h$ increasing, 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). If I am correct, what is an example of function $h$ such that $(B)$ holds but $(A)$ does not?

When $h$ is not increasing, in general, we may find that $\lim_{y \rightarrow \infty} \Pr(Y \geq \kappa h(y))$ does not exist. However, does statement (A) still imply that $\lim_{y \rightarrow \infty} \Pr(Y \geq \kappa h(y))$ exists and must be equal to zero?

Regarding (2): Somehow, this result suggests that if $(A)$ holds, then $h$ can be anything (can shift, rescale, or apply nonlinear transformations to $y$), 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?

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?


Thoughts. Regarding (1): The limiting condition says that $ h(y) =\alpha+\beta y $ with $\beta>0$ 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?

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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 $\lim_{y\rightarrow \infty} h'(y)<\infty$. 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?


Thoughts. Regarding (1): Suppose that $ h $ is increasing in $y $. Then, $$ (B)\quad \lim_{y \rightarrow \infty} \Pr(Y \geq y) = \lim_{y \rightarrow \infty} \Pr(Y \geq \kappa h(y)) = 0. $$ In this setting with $h$ increasing, 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). If I am correct, what is an example of function $h$ such that $(B)$ holds but $(A)$ does not?

When $h$ is not monotone increasing, in general, we may find that $\lim_{y \rightarrow \infty} \Pr(Y \geq \kappa h(y))$ does not exist. However, does statement (A) still imply that $\lim_{y \rightarrow \infty} \Pr(Y \geq \kappa h(y))$ exists and must be equal to zero?

Regarding (2): Somehow, this result suggests that if $(A)$ holds, then $h$ can be anything (can shift, rescale, or apply nonlinear transformations to $y$), 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?

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 $\lim_{y\rightarrow \infty} h'(y)<\infty$. 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?


Thoughts. Regarding (1): Suppose that $ h $ is increasing in $y $. Then, $$ (B)\quad \lim_{y \rightarrow \infty} \Pr(Y \geq y) = \lim_{y \rightarrow \infty} \Pr(Y \geq \kappa h(y)) = 0. $$ In this setting with $h$ increasing, 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). If I am correct, what is an example of function $h$ such that $(B)$ holds but $(A)$ does not?

When $h$ is not monotone increasing, in general, we may find that $\lim_{y \rightarrow \infty} \Pr(Y \geq \kappa h(y))$ does not exist. However, does statement (A) still imply that $\lim_{y \rightarrow \infty} \Pr(Y \geq \kappa h(y))$ exists and must be equal to zero?

Regarding (2): Somehow, this result suggests that if $(A)$ holds, then $h$ can be anything (can shift, rescale, or apply nonlinear transformations to $y$), 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?

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 $\lim_{y\rightarrow \infty} h'(y)<\infty$. 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?


Thoughts. Regarding (1): Suppose that $ h $ is increasing in $y $. Then, $$ (B)\quad \lim_{y \rightarrow \infty} \Pr(Y \geq y) = \lim_{y \rightarrow \infty} \Pr(Y \geq \kappa h(y)) = 0. $$ In this setting with $h$ increasing, 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). If I am correct, what is an example of function $h$ such that $(B)$ holds but $(A)$ does not?

When $h$ is not increasing, in general, we may find that $\lim_{y \rightarrow \infty} \Pr(Y \geq \kappa h(y))$ does not exist. However, does statement (A) still imply that $\lim_{y \rightarrow \infty} \Pr(Y \geq \kappa h(y))$ exists and must be equal to zero?

Regarding (2): Somehow, this result suggests that if $(A)$ holds, then $h$ can be anything (can shift, rescale, or apply nonlinear transformations to $y$), 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?

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