For cdf $F(x)$ ad empirical cdf $F_n(x)$, show that $|F^{-1}big(F_n(xi_p)big)-hatxi_p|overset{a.s}to0$












2












$begingroup$


Suppose $X_1,cdots,X_n$ are i.i.d. continuous random variables from distribution with cdf $F(x)$.Let $F_n(x)$
be a random variable defined by
$$F_n(x)=frac{1}nsum_{i=1}^nI{X_ile x}.$$
Define the $p$th quantile for cdf $F(x)$ as
$$xi_pequiv F^{-1}(p)=inf{x:F(x)ge p},$$
and the $p$th quantile for empirical cdf $F_n(x)$ as
$$hatxi_pequiv F_n^{-1}(p)=inf{x:F_n(x)ge p}.$$
Show that
$$Big|F^{-1}big(F_n(xi_p)big)-hatxi_pBig|overset{a.s}to0$$




How can we prove the result? By LLN, we have $F_n(x)overset{a.s}to F(x)$. Can we write the equation as
$$Big|F^{-1}big(F_n(xi_p)big)-hatxi_pBig|=Big|F^{-1}big(F_n(xi_p)big)-F_n^{-1}big(F(xi_p)big)Big|,$$
and get the result directly?






probability probability-theory probability-distributions quantile







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    For a fixed $x$, $F_n(x) = frac1n sum_{k=1}^n 1_{X_k le x}$ is a random variable with binomial distribution $Pr(F_n(x) le l/n)= sum_{m=0}^{lfloor l rfloor} {n choose m} F(x)^m (1-F(x))^m$. So $mathbb{E}[F_n(x)]= F(x)$ and $mathbb{E}[(F_n(x)-F(x))^2]= frac1n F(x)(1-F(x))to 0$. Whence $F_n(x) to F(x)$ almost surely and since $F^{-1}$ is continuous then $F^{-1}(F_n(x)) to F^{-1}(F(x))$ a.s.
    $endgroup$
    – reuns
    Dec 29 '18 at 7:38






  • 1




    $begingroup$
    @reuns The continuity of $F(x)$ does not make $F^{-1}$ continuous. Consider uniform distribution on the union of intervals $(0,1)$ and $(2,3)$ and you will get $F^{-1}(0.5)=1$, $F^{-1}(0.5+)=2$ (right limit).
    $endgroup$
    – NCh
    Dec 29 '18 at 15:15










  • $begingroup$
    Not so easy. If $F^{-1}$ is discontinuous only at finitely many point it is easier.
    $endgroup$
    – reuns
    Jan 8 at 11:27












  • $begingroup$
    @reuns Thank you for your kind comments!
    $endgroup$
    – J.Mike
    Jan 8 at 17:59
















2












$begingroup$


Suppose $X_1,cdots,X_n$ are i.i.d. continuous random variables from distribution with cdf $F(x)$.Let $F_n(x)$
be a random variable defined by
$$F_n(x)=frac{1}nsum_{i=1}^nI{X_ile x}.$$
Define the $p$th quantile for cdf $F(x)$ as
$$xi_pequiv F^{-1}(p)=inf{x:F(x)ge p},$$
and the $p$th quantile for empirical cdf $F_n(x)$ as
$$hatxi_pequiv F_n^{-1}(p)=inf{x:F_n(x)ge p}.$$
Show that
$$Big|F^{-1}big(F_n(xi_p)big)-hatxi_pBig|overset{a.s}to0$$




How can we prove the result? By LLN, we have $F_n(x)overset{a.s}to F(x)$. Can we write the equation as
$$Big|F^{-1}big(F_n(xi_p)big)-hatxi_pBig|=Big|F^{-1}big(F_n(xi_p)big)-F_n^{-1}big(F(xi_p)big)Big|,$$
and get the result directly?






probability probability-theory probability-distributions quantile







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    For a fixed $x$, $F_n(x) = frac1n sum_{k=1}^n 1_{X_k le x}$ is a random variable with binomial distribution $Pr(F_n(x) le l/n)= sum_{m=0}^{lfloor l rfloor} {n choose m} F(x)^m (1-F(x))^m$. So $mathbb{E}[F_n(x)]= F(x)$ and $mathbb{E}[(F_n(x)-F(x))^2]= frac1n F(x)(1-F(x))to 0$. Whence $F_n(x) to F(x)$ almost surely and since $F^{-1}$ is continuous then $F^{-1}(F_n(x)) to F^{-1}(F(x))$ a.s.
    $endgroup$
    – reuns
    Dec 29 '18 at 7:38






  • 1




    $begingroup$
    @reuns The continuity of $F(x)$ does not make $F^{-1}$ continuous. Consider uniform distribution on the union of intervals $(0,1)$ and $(2,3)$ and you will get $F^{-1}(0.5)=1$, $F^{-1}(0.5+)=2$ (right limit).
    $endgroup$
    – NCh
    Dec 29 '18 at 15:15










  • $begingroup$
    Not so easy. If $F^{-1}$ is discontinuous only at finitely many point it is easier.
    $endgroup$
    – reuns
    Jan 8 at 11:27












  • $begingroup$
    @reuns Thank you for your kind comments!
    $endgroup$
    – J.Mike
    Jan 8 at 17:59














2












2








2





$begingroup$


Suppose $X_1,cdots,X_n$ are i.i.d. continuous random variables from distribution with cdf $F(x)$.Let $F_n(x)$
be a random variable defined by
$$F_n(x)=frac{1}nsum_{i=1}^nI{X_ile x}.$$
Define the $p$th quantile for cdf $F(x)$ as
$$xi_pequiv F^{-1}(p)=inf{x:F(x)ge p},$$
and the $p$th quantile for empirical cdf $F_n(x)$ as
$$hatxi_pequiv F_n^{-1}(p)=inf{x:F_n(x)ge p}.$$
Show that
$$Big|F^{-1}big(F_n(xi_p)big)-hatxi_pBig|overset{a.s}to0$$




How can we prove the result? By LLN, we have $F_n(x)overset{a.s}to F(x)$. Can we write the equation as
$$Big|F^{-1}big(F_n(xi_p)big)-hatxi_pBig|=Big|F^{-1}big(F_n(xi_p)big)-F_n^{-1}big(F(xi_p)big)Big|,$$
and get the result directly?






probability probability-theory probability-distributions quantile







share|cite|improve this question











$endgroup$




Suppose $X_1,cdots,X_n$ are i.i.d. continuous random variables from distribution with cdf $F(x)$.Let $F_n(x)$
be a random variable defined by
$$F_n(x)=frac{1}nsum_{i=1}^nI{X_ile x}.$$
Define the $p$th quantile for cdf $F(x)$ as
$$xi_pequiv F^{-1}(p)=inf{x:F(x)ge p},$$
and the $p$th quantile for empirical cdf $F_n(x)$ as
$$hatxi_pequiv F_n^{-1}(p)=inf{x:F_n(x)ge p}.$$
Show that
$$Big|F^{-1}big(F_n(xi_p)big)-hatxi_pBig|overset{a.s}to0$$




How can we prove the result? By LLN, we have $F_n(x)overset{a.s}to F(x)$. Can we write the equation as
$$Big|F^{-1}big(F_n(xi_p)big)-hatxi_pBig|=Big|F^{-1}big(F_n(xi_p)big)-F_n^{-1}big(F(xi_p)big)Big|,$$
and get the result directly?






probability probability-theory probability-distributions quantile




probability probability-theory probability-distributions quantile






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 8 at 10:00









TZakrevskiy

20.1k12354




20.1k12354










asked Dec 29 '18 at 6:36









J.MikeJ.Mike

336110




336110








  • 1




    $begingroup$
    For a fixed $x$, $F_n(x) = frac1n sum_{k=1}^n 1_{X_k le x}$ is a random variable with binomial distribution $Pr(F_n(x) le l/n)= sum_{m=0}^{lfloor l rfloor} {n choose m} F(x)^m (1-F(x))^m$. So $mathbb{E}[F_n(x)]= F(x)$ and $mathbb{E}[(F_n(x)-F(x))^2]= frac1n F(x)(1-F(x))to 0$. Whence $F_n(x) to F(x)$ almost surely and since $F^{-1}$ is continuous then $F^{-1}(F_n(x)) to F^{-1}(F(x))$ a.s.
    $endgroup$
    – reuns
    Dec 29 '18 at 7:38






  • 1




    $begingroup$
    @reuns The continuity of $F(x)$ does not make $F^{-1}$ continuous. Consider uniform distribution on the union of intervals $(0,1)$ and $(2,3)$ and you will get $F^{-1}(0.5)=1$, $F^{-1}(0.5+)=2$ (right limit).
    $endgroup$
    – NCh
    Dec 29 '18 at 15:15










  • $begingroup$
    Not so easy. If $F^{-1}$ is discontinuous only at finitely many point it is easier.
    $endgroup$
    – reuns
    Jan 8 at 11:27












  • $begingroup$
    @reuns Thank you for your kind comments!
    $endgroup$
    – J.Mike
    Jan 8 at 17:59














  • 1




    $begingroup$
    For a fixed $x$, $F_n(x) = frac1n sum_{k=1}^n 1_{X_k le x}$ is a random variable with binomial distribution $Pr(F_n(x) le l/n)= sum_{m=0}^{lfloor l rfloor} {n choose m} F(x)^m (1-F(x))^m$. So $mathbb{E}[F_n(x)]= F(x)$ and $mathbb{E}[(F_n(x)-F(x))^2]= frac1n F(x)(1-F(x))to 0$. Whence $F_n(x) to F(x)$ almost surely and since $F^{-1}$ is continuous then $F^{-1}(F_n(x)) to F^{-1}(F(x))$ a.s.
    $endgroup$
    – reuns
    Dec 29 '18 at 7:38






  • 1




    $begingroup$
    @reuns The continuity of $F(x)$ does not make $F^{-1}$ continuous. Consider uniform distribution on the union of intervals $(0,1)$ and $(2,3)$ and you will get $F^{-1}(0.5)=1$, $F^{-1}(0.5+)=2$ (right limit).
    $endgroup$
    – NCh
    Dec 29 '18 at 15:15










  • $begingroup$
    Not so easy. If $F^{-1}$ is discontinuous only at finitely many point it is easier.
    $endgroup$
    – reuns
    Jan 8 at 11:27












  • $begingroup$
    @reuns Thank you for your kind comments!
    $endgroup$
    – J.Mike
    Jan 8 at 17:59








1




1




$begingroup$
For a fixed $x$, $F_n(x) = frac1n sum_{k=1}^n 1_{X_k le x}$ is a random variable with binomial distribution $Pr(F_n(x) le l/n)= sum_{m=0}^{lfloor l rfloor} {n choose m} F(x)^m (1-F(x))^m$. So $mathbb{E}[F_n(x)]= F(x)$ and $mathbb{E}[(F_n(x)-F(x))^2]= frac1n F(x)(1-F(x))to 0$. Whence $F_n(x) to F(x)$ almost surely and since $F^{-1}$ is continuous then $F^{-1}(F_n(x)) to F^{-1}(F(x))$ a.s.
$endgroup$
– reuns
Dec 29 '18 at 7:38




$begingroup$
For a fixed $x$, $F_n(x) = frac1n sum_{k=1}^n 1_{X_k le x}$ is a random variable with binomial distribution $Pr(F_n(x) le l/n)= sum_{m=0}^{lfloor l rfloor} {n choose m} F(x)^m (1-F(x))^m$. So $mathbb{E}[F_n(x)]= F(x)$ and $mathbb{E}[(F_n(x)-F(x))^2]= frac1n F(x)(1-F(x))to 0$. Whence $F_n(x) to F(x)$ almost surely and since $F^{-1}$ is continuous then $F^{-1}(F_n(x)) to F^{-1}(F(x))$ a.s.
$endgroup$
– reuns
Dec 29 '18 at 7:38




1




1




$begingroup$
@reuns The continuity of $F(x)$ does not make $F^{-1}$ continuous. Consider uniform distribution on the union of intervals $(0,1)$ and $(2,3)$ and you will get $F^{-1}(0.5)=1$, $F^{-1}(0.5+)=2$ (right limit).
$endgroup$
– NCh
Dec 29 '18 at 15:15




$begingroup$
@reuns The continuity of $F(x)$ does not make $F^{-1}$ continuous. Consider uniform distribution on the union of intervals $(0,1)$ and $(2,3)$ and you will get $F^{-1}(0.5)=1$, $F^{-1}(0.5+)=2$ (right limit).
$endgroup$
– NCh
Dec 29 '18 at 15:15












$begingroup$
Not so easy. If $F^{-1}$ is discontinuous only at finitely many point it is easier.
$endgroup$
– reuns
Jan 8 at 11:27






$begingroup$
Not so easy. If $F^{-1}$ is discontinuous only at finitely many point it is easier.
$endgroup$
– reuns
Jan 8 at 11:27














$begingroup$
@reuns Thank you for your kind comments!
$endgroup$
– J.Mike
Jan 8 at 17:59




$begingroup$
@reuns Thank you for your kind comments!
$endgroup$
– J.Mike
Jan 8 at 17:59










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