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Show that the limit inferior of the sum of third moments of mutually independent Gaussians tends to negative infinity

0

$begingroup$

Let $dinmathbb N$, $sigma:=ell d^-alpha$ for some $ell>0$ and $alpha>0$, $Y$ be a Gaussian $mathbb R^d$-valued random variable with mean $0$ and covariance matrix $sigma^2I_d$$^1$, $Z$ be a $mathbb R^d$-valued random variable, $fin C^3(mathbb R)$ with $f>0$ and $g:=ln f$. Assume $g'$ is Lipschitz continuous and $g'''$ is bounded. Now let $$s_d:=sum_i=1^dg'''(Z_i)Y_i^3.$$

Are we able to show that $limsup_dtoinftys_dle0$ in a suitable mode of convergence?

I've figured out the following: As $$operatorname Eleft[|s_d|right]le2sqrtfrac2pileft|g'''right|_infty dsigmatag1,$$ we've got $L^1$-convergence of $s_d$ to $0$ as $dtoinfty$ as long as $alpha>1/3$. But what if $alphale1/3$?

---

$^1$ i.e. $Y_1,ldots,Y_d$ are mutually independent Gaussian random variables with mean $0$ and variacne $sigma^2$.

probability-theory normal-distribution weak-convergence limsup-and-liminf probability-limit-theorems

share|cite|improve this question

edited Apr 6 at 10:23

0xbadf00d

asked Mar 26 at 18:07

0xbadf00d0xbadf00d

1,68241534

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  By $Y_i^3$, do you mean $|Y_i|^3$? And why not replace $g'''$ with $h$, and replace the conditions on $f$ and $g$ with $h$ being continuous and bounded? And where does the result for $alpha>1/3$ come from?
  $endgroup$
  – Matt F.
  Apr 10 at 20:26
  
  

  
  
  
  

add a comment | 

0

$begingroup$

Let $dinmathbb N$, $sigma:=ell d^-alpha$ for some $ell>0$ and $alpha>0$, $Y$ be a Gaussian $mathbb R^d$-valued random variable with mean $0$ and covariance matrix $sigma^2I_d$$^1$, $Z$ be a $mathbb R^d$-valued random variable, $fin C^3(mathbb R)$ with $f>0$ and $g:=ln f$. Assume $g'$ is Lipschitz continuous and $g'''$ is bounded. Now let $$s_d:=sum_i=1^dg'''(Z_i)Y_i^3.$$

Are we able to show that $limsup_dtoinftys_dle0$ in a suitable mode of convergence?

I've figured out the following: As $$operatorname Eleft[|s_d|right]le2sqrtfrac2pileft|g'''right|_infty dsigmatag1,$$ we've got $L^1$-convergence of $s_d$ to $0$ as $dtoinfty$ as long as $alpha>1/3$. But what if $alphale1/3$?

---

$^1$ i.e. $Y_1,ldots,Y_d$ are mutually independent Gaussian random variables with mean $0$ and variacne $sigma^2$.

probability-theory normal-distribution weak-convergence limsup-and-liminf probability-limit-theorems

share|cite|improve this question

edited Apr 6 at 10:23

0xbadf00d

asked Mar 26 at 18:07

0xbadf00d0xbadf00d

1,68241534

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  By $Y_i^3$, do you mean $|Y_i|^3$? And why not replace $g'''$ with $h$, and replace the conditions on $f$ and $g$ with $h$ being continuous and bounded? And where does the result for $alpha>1/3$ come from?
  $endgroup$
  – Matt F.
  Apr 10 at 20:26
  
  

  
  
  
  

add a comment | 

$begingroup$

Let $dinmathbb N$, $sigma:=ell d^-alpha$ for some $ell>0$ and $alpha>0$, $Y$ be a Gaussian $mathbb R^d$-valued random variable with mean $0$ and covariance matrix $sigma^2I_d$$^1$, $Z$ be a $mathbb R^d$-valued random variable, $fin C^3(mathbb R)$ with $f>0$ and $g:=ln f$. Assume $g'$ is Lipschitz continuous and $g'''$ is bounded. Now let $$s_d:=sum_i=1^dg'''(Z_i)Y_i^3.$$

Are we able to show that $limsup_dtoinftys_dle0$ in a suitable mode of convergence?

I've figured out the following: As $$operatorname Eleft[|s_d|right]le2sqrtfrac2pileft|g'''right|_infty dsigmatag1,$$ we've got $L^1$-convergence of $s_d$ to $0$ as $dtoinfty$ as long as $alpha>1/3$. But what if $alphale1/3$?

---

$^1$ i.e. $Y_1,ldots,Y_d$ are mutually independent Gaussian random variables with mean $0$ and variacne $sigma^2$.

probability-theory normal-distribution weak-convergence limsup-and-liminf probability-limit-theorems

share|cite|improve this question

edited Apr 6 at 10:23

0xbadf00d

asked Mar 26 at 18:07

0xbadf00d0xbadf00d

1,68241534

$endgroup$

share|cite|improve this question

edited Apr 6 at 10:23

0xbadf00d

asked Mar 26 at 18:07

0xbadf00d0xbadf00d

1,68241534

share|cite|improve this question

edited Apr 6 at 10:23

0xbadf00d

asked Mar 26 at 18:07

0xbadf00d0xbadf00d

1,68241534

asked Mar 26 at 18:07

0xbadf00d0xbadf00d

1,68241534

asked Mar 26 at 18:07

0xbadf00d0xbadf00d

1,68241534