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3

1

$begingroup$

Imagine it holds that

$$limlimits_t to 0F(tx)/F(t)=x^alpha,$$

where $F$ is the cdf of $X$ respectively $Y$ and $alpha>0$. Does it then also hold that

$$limlimits_t to 0F_X+Y(tx)/F_X+Y(t)=x^beta,$$

for some $beta>0$?

It is a well known fact that this is true if $X$ and $Y$ are regularly varying at infinity(Then X+Y is regularly varying at infinity as well with the same $alpha$), however it is not possible to easily adapt this result to this problem

probability-theory probability-distributions

share|cite|improve this question

edited Mar 13 at 21:06

Cettt

1,890622

asked Mar 12 at 11:10

user299124

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I could not prove it, but it holds for simple distributions like the exponential distribution at least, so you might be right.
  $endgroup$
  – Martin
  Mar 12 at 20:50
  
  

  
  
  
  

add a comment | 

3

1

$begingroup$

Imagine it holds that

$$limlimits_t to 0F(tx)/F(t)=x^alpha,$$

where $F$ is the cdf of $X$ respectively $Y$ and $alpha>0$. Does it then also hold that

$$limlimits_t to 0F_X+Y(tx)/F_X+Y(t)=x^beta,$$

for some $beta>0$?

It is a well known fact that this is true if $X$ and $Y$ are regularly varying at infinity(Then X+Y is regularly varying at infinity as well with the same $alpha$), however it is not possible to easily adapt this result to this problem

probability-theory probability-distributions

share|cite|improve this question

edited Mar 13 at 21:06

Cettt

1,890622

asked Mar 12 at 11:10

user299124

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I could not prove it, but it holds for simple distributions like the exponential distribution at least, so you might be right.
  $endgroup$
  – Martin
  Mar 12 at 20:50
  
  

  
  
  
  

add a comment | 

$begingroup$

Imagine it holds that

$$limlimits_t to 0F(tx)/F(t)=x^alpha,$$

where $F$ is the cdf of $X$ respectively $Y$ and $alpha>0$. Does it then also hold that

$$limlimits_t to 0F_X+Y(tx)/F_X+Y(t)=x^beta,$$

for some $beta>0$?

It is a well known fact that this is true if $X$ and $Y$ are regularly varying at infinity(Then X+Y is regularly varying at infinity as well with the same $alpha$), however it is not possible to easily adapt this result to this problem

probability-theory probability-distributions

share|cite|improve this question

edited Mar 13 at 21:06

Cettt

1,890622

asked Mar 12 at 11:10

user299124

$endgroup$

share|cite|improve this question

edited Mar 13 at 21:06

Cettt

1,890622

asked Mar 12 at 11:10

user299124

share|cite|improve this question

edited Mar 13 at 21:06

Cettt

1,890622

asked Mar 12 at 11:10

user299124

edited Mar 13 at 21:06

Cettt

1,890622

edited Mar 13 at 21:06

Cettt

1,890622

1 Answer
1

active

oldest

votes

0

$begingroup$

Yes, there is a result at zero, however, it is very different from the one at infinity you mentioned. Namely, the indices add up:

If $X_i$, $i=1,2$, are independent with cdfs $F_X_i$, which are regulatly varying at zero with indices $alpha_i$, $i=1,2$, then $F_X_1 +X_2$ is regularly varying at zero of index $alpha_1+alpha_2$.

Proof follows from the corresponding Tauberian theorem (see e.g. Bingham, Goldie, Teugels, Theorem 1.7.1'):

A non-decreasing function $Ucolon (0,infty)to (0,infty)$, whose Laplace-Stieltjes transform $hat U(s) = int_0^infty e^-st dU(s)$ is finite for large $s$, is regularly varying at zero of index $rhoge 0$ iff $hat U(s)$ is regularly varying on infinity of index $-rho$.

That said, $hat F_X_i$ are regularly varying at infinity with indices $-alpha_i$, $i=1,2$, so $hat F_X_1+X_2 = hat F_X_1hat F_X_2$ is regularly varying at infinity of index $-alpha_1-alpha_2$. Hence, $F_X_1+X_2$ is regularly varying at zero of index $alpha_1+alpha_2$.

share|cite|improve this answer

answered Mar 13 at 10:20

zhorasterzhoraster

15.9k21853

$endgroup$

add a comment | 

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0

$begingroup$

Yes, there is a result at zero, however, it is very different from the one at infinity you mentioned. Namely, the indices add up:

If $X_i$, $i=1,2$, are independent with cdfs $F_X_i$, which are regulatly varying at zero with indices $alpha_i$, $i=1,2$, then $F_X_1 +X_2$ is regularly varying at zero of index $alpha_1+alpha_2$.

Proof follows from the corresponding Tauberian theorem (see e.g. Bingham, Goldie, Teugels, Theorem 1.7.1'):

A non-decreasing function $Ucolon (0,infty)to (0,infty)$, whose Laplace-Stieltjes transform $hat U(s) = int_0^infty e^-st dU(s)$ is finite for large $s$, is regularly varying at zero of index $rhoge 0$ iff $hat U(s)$ is regularly varying on infinity of index $-rho$.

That said, $hat F_X_i$ are regularly varying at infinity with indices $-alpha_i$, $i=1,2$, so $hat F_X_1+X_2 = hat F_X_1hat F_X_2$ is regularly varying at infinity of index $-alpha_1-alpha_2$. Hence, $F_X_1+X_2$ is regularly varying at zero of index $alpha_1+alpha_2$.

share|cite|improve this answer

answered Mar 13 at 10:20

zhorasterzhoraster

15.9k21853

$endgroup$

add a comment | 

0

$begingroup$

Yes, there is a result at zero, however, it is very different from the one at infinity you mentioned. Namely, the indices add up:

If $X_i$, $i=1,2$, are independent with cdfs $F_X_i$, which are regulatly varying at zero with indices $alpha_i$, $i=1,2$, then $F_X_1 +X_2$ is regularly varying at zero of index $alpha_1+alpha_2$.

Proof follows from the corresponding Tauberian theorem (see e.g. Bingham, Goldie, Teugels, Theorem 1.7.1'):

A non-decreasing function $Ucolon (0,infty)to (0,infty)$, whose Laplace-Stieltjes transform $hat U(s) = int_0^infty e^-st dU(s)$ is finite for large $s$, is regularly varying at zero of index $rhoge 0$ iff $hat U(s)$ is regularly varying on infinity of index $-rho$.

That said, $hat F_X_i$ are regularly varying at infinity with indices $-alpha_i$, $i=1,2$, so $hat F_X_1+X_2 = hat F_X_1hat F_X_2$ is regularly varying at infinity of index $-alpha_1-alpha_2$. Hence, $F_X_1+X_2$ is regularly varying at zero of index $alpha_1+alpha_2$.

share|cite|improve this answer

answered Mar 13 at 10:20

zhorasterzhoraster

15.9k21853

$endgroup$

add a comment | 

$begingroup$

Yes, there is a result at zero, however, it is very different from the one at infinity you mentioned. Namely, the indices add up:

If $X_i$, $i=1,2$, are independent with cdfs $F_X_i$, which are regulatly varying at zero with indices $alpha_i$, $i=1,2$, then $F_X_1 +X_2$ is regularly varying at zero of index $alpha_1+alpha_2$.

Proof follows from the corresponding Tauberian theorem (see e.g. Bingham, Goldie, Teugels, Theorem 1.7.1'):

A non-decreasing function $Ucolon (0,infty)to (0,infty)$, whose Laplace-Stieltjes transform $hat U(s) = int_0^infty e^-st dU(s)$ is finite for large $s$, is regularly varying at zero of index $rhoge 0$ iff $hat U(s)$ is regularly varying on infinity of index $-rho$.

That said, $hat F_X_i$ are regularly varying at infinity with indices $-alpha_i$, $i=1,2$, so $hat F_X_1+X_2 = hat F_X_1hat F_X_2$ is regularly varying at infinity of index $-alpha_1-alpha_2$. Hence, $F_X_1+X_2$ is regularly varying at zero of index $alpha_1+alpha_2$.

share|cite|improve this answer

answered Mar 13 at 10:20

zhorasterzhoraster

15.9k21853

$endgroup$

share|cite|improve this answer

answered Mar 13 at 10:20

zhorasterzhoraster

15.9k21853

answered Mar 13 at 10:20

zhorasterzhoraster

15.9k21853

answered Mar 13 at 10:20

zhorasterzhoraster

15.9k21853