Show that the limit inferior of the sum of third moments of mutually independent Gaussians tends to negative infinity Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)An (possible) application of the Central Limit TheoremConvergence in Law implies a.s. convergence (Donsker-like statement)Definition of gaussian processUpper bound for a multivariate continuous random variableIf $Φ$ is the cdf of the standard normal distribution and $φ_i:ℝ→ℝ$ is linear, are the 1st and 2nd derivative of $x↦Φ(φ_1(x))+e^xΦ(φ_2(x))$ bounded?If $X_i∼fλ$, $Z∼mathcal N(0,I_d)$ and $Y=X+ell d^-αZ$ with $α<1/2$, then $liminf_d→∞text Eleft[1∧prod_i=1^dfracf(Y_i)f(X_i)right]=0$Strong law of large numbers for a scaled sequence of normally distributed random variablesWhich version of the central limit theorem do we need to apply here?If $text Pleft[|X_n|>n^-alpharight]to0$ as $ntoinfty$ for some $alpha>0$, does $(X_n)_ninmathbb N$ converge in probability?If $(W_i)_iinmathbb N$ obeys the strong law of large numbers, what can we say about $liminf_dtoinftyfrac1d^2alphasum_i=1^dW_i$?

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



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)An (possible) application of the Central Limit TheoremConvergence in Law implies a.s. convergence (Donsker-like statement)Definition of gaussian processUpper bound for a multivariate continuous random variableIf $Φ$ is the cdf of the standard normal distribution and $φ_i:ℝ→ℝ$ is linear, are the 1st and 2nd derivative of $x↦Φ(φ_1(x))+e^xΦ(φ_2(x))$ bounded?If $X_i∼fλ$, $Z∼mathcal N(0,I_d)$ and $Y=X+ell d^-αZ$ with $α<1/2$, then $liminf_d→∞text Eleft[1∧prod_i=1^dfracf(Y_i)f(X_i)right]=0$Strong law of large numbers for a scaled sequence of normally distributed random variablesWhich version of the central limit theorem do we need to apply here?If $text Pleft[|X_n|>n^-alpharight]to0$ as $ntoinfty$ for some $alpha>0$, does $(X_n)_ninmathbb N$ converge in probability?If $(W_i)_iinmathbb N$ obeys the strong law of large numbers, what can we say about $liminf_dtoinftyfrac1d^2alphasum_i=1^dW_i$?










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 $sigma^2$.










share|cite|improve this question











$endgroup$











  • $begingroup$
    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
















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 $sigma^2$.










share|cite|improve this question











$endgroup$











  • $begingroup$
    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














0












0








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 $sigma^2$.










share|cite|improve this question











$endgroup$




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















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Apr 6 at 10:23







0xbadf00d

















asked Mar 26 at 18:07









0xbadf00d0xbadf00d

1,68241534




1,68241534











  • $begingroup$
    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

















  • $begingroup$
    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
















$begingroup$
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





$begingroup$
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











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