Rate of convergence in distribution












1














Imagine a sequence of random variables ${X_n}_{n=1}^infty$ in [0,1] converging pointwise (i.e, sure convergence) to the random variable $X$, with convergence factor $lambda$:
$$
|X_n(omega)-X(omega)|=O(lambda^n),quadtext{for all }omegainOmega=[0,1]
$$
It is also clear that sure convergence implies convergence in distribution:
$X_noverset{mathcal D}{rightarrow}X$.



Question: can the convergence factor of $,overset{mathcal D}{rightarrow},$ be different than $lambda$? My rough intuition is that convergence in distribution could potentially be orders of magnitude faster, being convergence in distribution less demanding than sure convergence.



I found out that the Wasserstein metric $W_1$ metrizes weak convergence in our particular case (because our space is bounded in $mathbb R$), so $W_1$ can be used in this context as a metric to analyze weak convergence. Moreover, since we are in $mathbb R$, $W_1$ can be written as:



$$
W_1(P_n,P)=int_0^1|F_n(x)-F(x)|dx,
$$



where ${P_n}_{n=1}^infty$ and $P$ are corresponding probability measures.



So my question is whether there is a possibility to obtain something like



$$
W_1(P_n,P)=o(lambda^n)
$$



meaning that convergence in distribution is much faster than sure convergence in this context.



I have thought of using the transformation



$$
W_1(P_n,P)=int_0^1|F^{-1}_n(y)-F^{-1}(y)|dy,
$$
by defining $F^{-1}$ suitably, because this formula involves distances in our domain $[0,1]$, which are bounded by $lambda^n$.



Thanks.






convergence weak-convergence







share|cite|improve this question
























  • What exactly do you mean by $|X_n(omega)-X(omega)|=O(lambda^n)$..? Clearly, $|X_n(omega)-X(omega)| leq C lambda^n$ but in general $C$ will depend on $omega$. Do you assume that $C$ can be chosen independently of $omega$...?
    – saz
    Nov 22 '18 at 9:12










  • You're right. As you suggest, $C$ may depend on $omega$. But it can be bounded in our case because $omegain[0,1]$. So $C$ is a constant independent of $omega$.
    – Coralio
    Nov 22 '18 at 10:13












  • Note that $C(omega)<infty$ for each $omega in [0,1]$ does, in general, not imply $sup_{omega in [0,1]} C(omega)<infty$.
    – saz
    Nov 22 '18 at 10:50






  • 1




    Our $C$ is fixed, finite, independent of $omega$. Thanks
    – Coralio
    Nov 22 '18 at 11:11












  • thanks for the clarification.
    – saz
    Nov 22 '18 at 11:22
















1














Imagine a sequence of random variables ${X_n}_{n=1}^infty$ in [0,1] converging pointwise (i.e, sure convergence) to the random variable $X$, with convergence factor $lambda$:
$$
|X_n(omega)-X(omega)|=O(lambda^n),quadtext{for all }omegainOmega=[0,1]
$$
It is also clear that sure convergence implies convergence in distribution:
$X_noverset{mathcal D}{rightarrow}X$.



Question: can the convergence factor of $,overset{mathcal D}{rightarrow},$ be different than $lambda$? My rough intuition is that convergence in distribution could potentially be orders of magnitude faster, being convergence in distribution less demanding than sure convergence.



I found out that the Wasserstein metric $W_1$ metrizes weak convergence in our particular case (because our space is bounded in $mathbb R$), so $W_1$ can be used in this context as a metric to analyze weak convergence. Moreover, since we are in $mathbb R$, $W_1$ can be written as:



$$
W_1(P_n,P)=int_0^1|F_n(x)-F(x)|dx,
$$



where ${P_n}_{n=1}^infty$ and $P$ are corresponding probability measures.



So my question is whether there is a possibility to obtain something like



$$
W_1(P_n,P)=o(lambda^n)
$$



meaning that convergence in distribution is much faster than sure convergence in this context.



I have thought of using the transformation



$$
W_1(P_n,P)=int_0^1|F^{-1}_n(y)-F^{-1}(y)|dy,
$$
by defining $F^{-1}$ suitably, because this formula involves distances in our domain $[0,1]$, which are bounded by $lambda^n$.



Thanks.






convergence weak-convergence







share|cite|improve this question
























  • What exactly do you mean by $|X_n(omega)-X(omega)|=O(lambda^n)$..? Clearly, $|X_n(omega)-X(omega)| leq C lambda^n$ but in general $C$ will depend on $omega$. Do you assume that $C$ can be chosen independently of $omega$...?
    – saz
    Nov 22 '18 at 9:12










  • You're right. As you suggest, $C$ may depend on $omega$. But it can be bounded in our case because $omegain[0,1]$. So $C$ is a constant independent of $omega$.
    – Coralio
    Nov 22 '18 at 10:13












  • Note that $C(omega)<infty$ for each $omega in [0,1]$ does, in general, not imply $sup_{omega in [0,1]} C(omega)<infty$.
    – saz
    Nov 22 '18 at 10:50






  • 1




    Our $C$ is fixed, finite, independent of $omega$. Thanks
    – Coralio
    Nov 22 '18 at 11:11












  • thanks for the clarification.
    – saz
    Nov 22 '18 at 11:22














1












1








1







Imagine a sequence of random variables ${X_n}_{n=1}^infty$ in [0,1] converging pointwise (i.e, sure convergence) to the random variable $X$, with convergence factor $lambda$:
$$
|X_n(omega)-X(omega)|=O(lambda^n),quadtext{for all }omegainOmega=[0,1]
$$
It is also clear that sure convergence implies convergence in distribution:
$X_noverset{mathcal D}{rightarrow}X$.



Question: can the convergence factor of $,overset{mathcal D}{rightarrow},$ be different than $lambda$? My rough intuition is that convergence in distribution could potentially be orders of magnitude faster, being convergence in distribution less demanding than sure convergence.



I found out that the Wasserstein metric $W_1$ metrizes weak convergence in our particular case (because our space is bounded in $mathbb R$), so $W_1$ can be used in this context as a metric to analyze weak convergence. Moreover, since we are in $mathbb R$, $W_1$ can be written as:



$$
W_1(P_n,P)=int_0^1|F_n(x)-F(x)|dx,
$$



where ${P_n}_{n=1}^infty$ and $P$ are corresponding probability measures.



So my question is whether there is a possibility to obtain something like



$$
W_1(P_n,P)=o(lambda^n)
$$



meaning that convergence in distribution is much faster than sure convergence in this context.



I have thought of using the transformation



$$
W_1(P_n,P)=int_0^1|F^{-1}_n(y)-F^{-1}(y)|dy,
$$
by defining $F^{-1}$ suitably, because this formula involves distances in our domain $[0,1]$, which are bounded by $lambda^n$.



Thanks.






convergence weak-convergence







share|cite|improve this question















Imagine a sequence of random variables ${X_n}_{n=1}^infty$ in [0,1] converging pointwise (i.e, sure convergence) to the random variable $X$, with convergence factor $lambda$:
$$
|X_n(omega)-X(omega)|=O(lambda^n),quadtext{for all }omegainOmega=[0,1]
$$
It is also clear that sure convergence implies convergence in distribution:
$X_noverset{mathcal D}{rightarrow}X$.



Question: can the convergence factor of $,overset{mathcal D}{rightarrow},$ be different than $lambda$? My rough intuition is that convergence in distribution could potentially be orders of magnitude faster, being convergence in distribution less demanding than sure convergence.



I found out that the Wasserstein metric $W_1$ metrizes weak convergence in our particular case (because our space is bounded in $mathbb R$), so $W_1$ can be used in this context as a metric to analyze weak convergence. Moreover, since we are in $mathbb R$, $W_1$ can be written as:



$$
W_1(P_n,P)=int_0^1|F_n(x)-F(x)|dx,
$$



where ${P_n}_{n=1}^infty$ and $P$ are corresponding probability measures.



So my question is whether there is a possibility to obtain something like



$$
W_1(P_n,P)=o(lambda^n)
$$



meaning that convergence in distribution is much faster than sure convergence in this context.



I have thought of using the transformation



$$
W_1(P_n,P)=int_0^1|F^{-1}_n(y)-F^{-1}(y)|dy,
$$
by defining $F^{-1}$ suitably, because this formula involves distances in our domain $[0,1]$, which are bounded by $lambda^n$.



Thanks.






convergence weak-convergence




convergence weak-convergence






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 22 '18 at 12:31







Coralio

















asked Nov 22 '18 at 0:07









CoralioCoralio

62




62












  • What exactly do you mean by $|X_n(omega)-X(omega)|=O(lambda^n)$..? Clearly, $|X_n(omega)-X(omega)| leq C lambda^n$ but in general $C$ will depend on $omega$. Do you assume that $C$ can be chosen independently of $omega$...?
    – saz
    Nov 22 '18 at 9:12










  • You're right. As you suggest, $C$ may depend on $omega$. But it can be bounded in our case because $omegain[0,1]$. So $C$ is a constant independent of $omega$.
    – Coralio
    Nov 22 '18 at 10:13












  • Note that $C(omega)<infty$ for each $omega in [0,1]$ does, in general, not imply $sup_{omega in [0,1]} C(omega)<infty$.
    – saz
    Nov 22 '18 at 10:50






  • 1




    Our $C$ is fixed, finite, independent of $omega$. Thanks
    – Coralio
    Nov 22 '18 at 11:11












  • thanks for the clarification.
    – saz
    Nov 22 '18 at 11:22


















  • What exactly do you mean by $|X_n(omega)-X(omega)|=O(lambda^n)$..? Clearly, $|X_n(omega)-X(omega)| leq C lambda^n$ but in general $C$ will depend on $omega$. Do you assume that $C$ can be chosen independently of $omega$...?
    – saz
    Nov 22 '18 at 9:12










  • You're right. As you suggest, $C$ may depend on $omega$. But it can be bounded in our case because $omegain[0,1]$. So $C$ is a constant independent of $omega$.
    – Coralio
    Nov 22 '18 at 10:13












  • Note that $C(omega)<infty$ for each $omega in [0,1]$ does, in general, not imply $sup_{omega in [0,1]} C(omega)<infty$.
    – saz
    Nov 22 '18 at 10:50






  • 1




    Our $C$ is fixed, finite, independent of $omega$. Thanks
    – Coralio
    Nov 22 '18 at 11:11












  • thanks for the clarification.
    – saz
    Nov 22 '18 at 11:22
















What exactly do you mean by $|X_n(omega)-X(omega)|=O(lambda^n)$..? Clearly, $|X_n(omega)-X(omega)| leq C lambda^n$ but in general $C$ will depend on $omega$. Do you assume that $C$ can be chosen independently of $omega$...?
– saz
Nov 22 '18 at 9:12




What exactly do you mean by $|X_n(omega)-X(omega)|=O(lambda^n)$..? Clearly, $|X_n(omega)-X(omega)| leq C lambda^n$ but in general $C$ will depend on $omega$. Do you assume that $C$ can be chosen independently of $omega$...?
– saz
Nov 22 '18 at 9:12












You're right. As you suggest, $C$ may depend on $omega$. But it can be bounded in our case because $omegain[0,1]$. So $C$ is a constant independent of $omega$.
– Coralio
Nov 22 '18 at 10:13






You're right. As you suggest, $C$ may depend on $omega$. But it can be bounded in our case because $omegain[0,1]$. So $C$ is a constant independent of $omega$.
– Coralio
Nov 22 '18 at 10:13














Note that $C(omega)<infty$ for each $omega in [0,1]$ does, in general, not imply $sup_{omega in [0,1]} C(omega)<infty$.
– saz
Nov 22 '18 at 10:50




Note that $C(omega)<infty$ for each $omega in [0,1]$ does, in general, not imply $sup_{omega in [0,1]} C(omega)<infty$.
– saz
Nov 22 '18 at 10:50




1




1




Our $C$ is fixed, finite, independent of $omega$. Thanks
– Coralio
Nov 22 '18 at 11:11






Our $C$ is fixed, finite, independent of $omega$. Thanks
– Coralio
Nov 22 '18 at 11:11














thanks for the clarification.
– saz
Nov 22 '18 at 11:22




thanks for the clarification.
– saz
Nov 22 '18 at 11:22










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