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Iosif Pinelis
Iosif Pinelis
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$\newcommand{\N}{\mathbb N}$Let $P$ be the set of all probability mass functions on $\N_0:=\{0\}\cup\N$, where $\N:=\{1,2,\dots\}$. Let $P_{>0}$ denote the set of all $q=(q_0,q_1,\dots)\in P$ such that $q_n>0$ for all $n\in\N_0$.

A previous MO post asked the following question:

Is there a function $f\colon[0,\infty)\to[0,\infty)$ with $f(0)=0$ such that for all $p=(p_0,p_1,\dots)\in P$ and all $q=(q_0,q_1,\dots)\in P_{>0}$ we have \begin{equation} W_1(p,q)\le f(\chi^2(p,q)), \end{equation} where $W_1(p,q)$ is the Wasserstein distance of order $1$ between $p$ and $q$, and \begin{equation} \chi^2(p,q):=\sum_{n\in\N_0}\frac{(p_n-q_n)^2}{q_n}, \end{equation} the chi-squared "distance" between $p$ and $q$?

The answer to this question is negative, even with the later added condition that the means for $p$ and $q$ are the same.

However, as will be shown below, the answer becomes positive if $f$ is allowed to depend on $q$ or, more specifically, just on the second moment \begin{equation} m_2(q):=\sum_{n\in\N}n^2 q_n \end{equation} of $q$.

$\newcommand{\N}{\mathbb N}$Let $P$ be the set of probability mass functions on $\N_0:=\{0\}\cup\N$, where $\N:=\{1,2,\dots\}$. Let $P_{>0}$ denote the set of all $q=(q_0,q_1,\dots)\in P$ such that $q_n>0$ for all $n\in\N_0$.

A previous MO post asked the following question:

Is there a function $f\colon[0,\infty)\to[0,\infty)$ with $f(0)=0$ such that for all $p=(p_0,p_1,\dots)\in P$ and all $q=(q_0,q_1,\dots)\in P_{>0}$ we have \begin{equation} W_1(p,q)\le f(\chi^2(p,q)), \end{equation} where $W_1(p,q)$ is the Wasserstein distance of order $1$ between $p$ and $q$, and \begin{equation} \chi^2(p,q):=\sum_{n\in\N_0}\frac{(p_n-q_n)^2}{q_n}, \end{equation} the chi-squared "distance" between $p$ and $q$?

The answer to this question is negative, even with the later added condition that the means for $p$ and $q$ are the same.

However, as will be shown below, the answer becomes positive if $f$ is allowed to depend on $q$ or, more specifically, just on the second moment \begin{equation} m_2(q):=\sum_{n\in\N}n^2 q_n \end{equation} of $q$.

$\newcommand{\N}{\mathbb N}$Let $P$ be the set of all probability mass functions on $\N_0:=\{0\}\cup\N$, where $\N:=\{1,2,\dots\}$. Let $P_{>0}$ denote the set of all $q=(q_0,q_1,\dots)\in P$ such that $q_n>0$ for all $n\in\N_0$.

A previous MO post asked the following question:

Is there a function $f\colon[0,\infty)\to[0,\infty)$ with $f(0)=0$ such that for all $p=(p_0,p_1,\dots)\in P$ and all $q=(q_0,q_1,\dots)\in P_{>0}$ we have \begin{equation} W_1(p,q)\le f(\chi^2(p,q)), \end{equation} where $W_1(p,q)$ is the Wasserstein distance of order $1$ between $p$ and $q$, and \begin{equation} \chi^2(p,q):=\sum_{n\in\N_0}\frac{(p_n-q_n)^2}{q_n}, \end{equation} the chi-squared "distance" between $p$ and $q$?

The answer to this question is negative, even with the later added condition that the means for $p$ and $q$ are the same.

However, as will be shown below, the answer becomes positive if $f$ is allowed to depend on $q$ or, more specifically, just on the second moment \begin{equation} m_2(q):=\sum_{n\in\N}n^2 q_n \end{equation} of $q$.

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Iosif Pinelis
Iosif Pinelis
  • 127.7k
  • 8
  • 107
  • 229

An inequality involving the Wasserstein distance and chi-squared distance

$\newcommand{\N}{\mathbb N}$Let $P$ be the set of probability mass functions on $\N_0:=\{0\}\cup\N$, where $\N:=\{1,2,\dots\}$. Let $P_{>0}$ denote the set of all $q=(q_0,q_1,\dots)\in P$ such that $q_n>0$ for all $n\in\N_0$.

A previous MO post asked the following question:

Is there a function $f\colon[0,\infty)\to[0,\infty)$ with $f(0)=0$ such that for all $p=(p_0,p_1,\dots)\in P$ and all $q=(q_0,q_1,\dots)\in P_{>0}$ we have \begin{equation} W_1(p,q)\le f(\chi^2(p,q)), \end{equation} where $W_1(p,q)$ is the Wasserstein distance of order $1$ between $p$ and $q$, and \begin{equation} \chi^2(p,q):=\sum_{n\in\N_0}\frac{(p_n-q_n)^2}{q_n}, \end{equation} the chi-squared "distance" between $p$ and $q$?

The answer to this question is negative, even with the later added condition that the means for $p$ and $q$ are the same.

However, as will be shown below, the answer becomes positive if $f$ is allowed to depend on $q$ or, more specifically, just on the second moment \begin{equation} m_2(q):=\sum_{n\in\N}n^2 q_n \end{equation} of $q$.