Convergence of measure

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Let $X_1,dots,X_n$ be iid random variables with mean $theta$ and variance 1. Let $F_n$ denote the distribution of $sqrtn(barX-theta)$, where $barX=frac1nsum_i=1^nX_i$. Let $Phi$ denote the standard normal distribution. Are there sufficient conditions on the distribution of the $X_i$s so that



$$sup_g left| fracint g(x) dF_n(x)int g(x) dPhi(x) - 1 right| to 0,$$



where the supremum is over all bounded functions on the interval $[0,1]$ (excluding the case $g=0$ $Phi$-a.s.). Convergence in total variation seems not strong enough.




probability-theory measure-theory




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  • $int g , d Phi$ may be $0$.
    – Kavi Rama Murthy
    Aug 22 at 8:01










  • Right, thanks! I'll edited the post.
    – Danijel
    Aug 22 at 8:04










  • Even excluding the case $g=0$ almost surely, the integral in the denominator may be $0$.
    – Did
    Aug 22 at 11:22










  • Really? Isn't it equivalent for non-negative functions?
    – Danijel
    Aug 22 at 11:41















up vote
4
down vote

favorite












Let $X_1,dots,X_n$ be iid random variables with mean $theta$ and variance 1. Let $F_n$ denote the distribution of $sqrtn(barX-theta)$, where $barX=frac1nsum_i=1^nX_i$. Let $Phi$ denote the standard normal distribution. Are there sufficient conditions on the distribution of the $X_i$s so that



$$sup_g left| fracint g(x) dF_n(x)int g(x) dPhi(x) - 1 right| to 0,$$



where the supremum is over all bounded functions on the interval $[0,1]$ (excluding the case $g=0$ $Phi$-a.s.). Convergence in total variation seems not strong enough.




probability-theory measure-theory




share|cite|improve this question






















  • $int g , d Phi$ may be $0$.
    – Kavi Rama Murthy
    Aug 22 at 8:01










  • Right, thanks! I'll edited the post.
    – Danijel
    Aug 22 at 8:04










  • Even excluding the case $g=0$ almost surely, the integral in the denominator may be $0$.
    – Did
    Aug 22 at 11:22










  • Really? Isn't it equivalent for non-negative functions?
    – Danijel
    Aug 22 at 11:41













up vote
4
down vote

favorite









up vote
4
down vote

favorite











Let $X_1,dots,X_n$ be iid random variables with mean $theta$ and variance 1. Let $F_n$ denote the distribution of $sqrtn(barX-theta)$, where $barX=frac1nsum_i=1^nX_i$. Let $Phi$ denote the standard normal distribution. Are there sufficient conditions on the distribution of the $X_i$s so that



$$sup_g left| fracint g(x) dF_n(x)int g(x) dPhi(x) - 1 right| to 0,$$



where the supremum is over all bounded functions on the interval $[0,1]$ (excluding the case $g=0$ $Phi$-a.s.). Convergence in total variation seems not strong enough.




probability-theory measure-theory




share|cite|improve this question














Let $X_1,dots,X_n$ be iid random variables with mean $theta$ and variance 1. Let $F_n$ denote the distribution of $sqrtn(barX-theta)$, where $barX=frac1nsum_i=1^nX_i$. Let $Phi$ denote the standard normal distribution. Are there sufficient conditions on the distribution of the $X_i$s so that



$$sup_g left| fracint g(x) dF_n(x)int g(x) dPhi(x) - 1 right| to 0,$$



where the supremum is over all bounded functions on the interval $[0,1]$ (excluding the case $g=0$ $Phi$-a.s.). Convergence in total variation seems not strong enough.





probability-theory measure-theory





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 22 at 9:57

























asked Aug 22 at 7:52









Danijel

414




414











  • $int g , d Phi$ may be $0$.
    – Kavi Rama Murthy
    Aug 22 at 8:01










  • Right, thanks! I'll edited the post.
    – Danijel
    Aug 22 at 8:04










  • Even excluding the case $g=0$ almost surely, the integral in the denominator may be $0$.
    – Did
    Aug 22 at 11:22










  • Really? Isn't it equivalent for non-negative functions?
    – Danijel
    Aug 22 at 11:41

















  • $int g , d Phi$ may be $0$.
    – Kavi Rama Murthy
    Aug 22 at 8:01










  • Right, thanks! I'll edited the post.
    – Danijel
    Aug 22 at 8:04










  • Even excluding the case $g=0$ almost surely, the integral in the denominator may be $0$.
    – Did
    Aug 22 at 11:22










  • Really? Isn't it equivalent for non-negative functions?
    – Danijel
    Aug 22 at 11:41
















$int g , d Phi$ may be $0$.
– Kavi Rama Murthy
Aug 22 at 8:01




$int g , d Phi$ may be $0$.
– Kavi Rama Murthy
Aug 22 at 8:01












Right, thanks! I'll edited the post.
– Danijel
Aug 22 at 8:04




Right, thanks! I'll edited the post.
– Danijel
Aug 22 at 8:04












Even excluding the case $g=0$ almost surely, the integral in the denominator may be $0$.
– Did
Aug 22 at 11:22




Even excluding the case $g=0$ almost surely, the integral in the denominator may be $0$.
– Did
Aug 22 at 11:22












Really? Isn't it equivalent for non-negative functions?
– Danijel
Aug 22 at 11:41





Really? Isn't it equivalent for non-negative functions?
– Danijel
Aug 22 at 11:41
















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