Mixing
Wasserstein distance between hyperplane and cube
$begingroup$
Let $mu$ be the uniform measure on the cube $Q = [-1,1]^n$, and $nu$ be the uniform measure on the surface
$$
V = {(x_1,dots,x_n)in Q mid sum x_i = 0}.
$$
I am curious about Wasserstein distance between $mu$ and $nu$, using the $l^infty$ distance on $mathbb{R}^n$.
That is, I would like a bound on
$$
sup_{|f|_{Lip}=1} |mathbb{E}_mu f - mathbb{E}_nu f|.
$$
I expect that the distance is on the order of $n^{-1/2}$ based on the following two calculations. First, one guess is that the function $S(x)=frac{1}{n}|sum x_i|$ would be close to optimizing the above supremum, and
$$
|mathbb{E}_mu S - mathbb{E}_nu S| leq Cn^{-1/2}.
$$
The second heuristic follows from the fact that if $xin Q$ is sampled according to $mu$, then with high probability $|sum x_i| leq Cn^{1/2}$, so by nudging each of the coefficients by an amount $n^{-1/2}$ it should be possible to land on the plane. This gives a transport map from $mu$ to a measure supported on $V$ that has cost roughly $n^{-1/2}$. The problem is that the transported measure is not the same as $nu$, so this does not constitute a proof.
probability-theory measure-theory concentration-of-measure optimal-transport
asked Jan 26 at 20:12
felipehfelipeh
1,606619
$endgroup$
add a comment |
$begingroup$
Let $mu$ be the uniform measure on the cube $Q = [-1,1]^n$, and $nu$ be the uniform measure on the surface
$$
V = {(x_1,dots,x_n)in Q mid sum x_i = 0}.
$$
I am curious about Wasserstein distance between $mu$ and $nu$, using the $l^infty$ distance on $mathbb{R}^n$.
That is, I would like a bound on
$$
sup_{|f|_{Lip}=1} |mathbb{E}_mu f - mathbb{E}_nu f|.
$$
I expect that the distance is on the order of $n^{-1/2}$ based on the following two calculations. First, one guess is that the function $S(x)=frac{1}{n}|sum x_i|$ would be close to optimizing the above supremum, and
$$
|mathbb{E}_mu S - mathbb{E}_nu S| leq Cn^{-1/2}.
$$
The second heuristic follows from the fact that if $xin Q$ is sampled according to $mu$, then with high probability $|sum x_i| leq Cn^{1/2}$, so by nudging each of the coefficients by an amount $n^{-1/2}$ it should be possible to land on the plane. This gives a transport map from $mu$ to a measure supported on $V$ that has cost roughly $n^{-1/2}$. The problem is that the transported measure is not the same as $nu$, so this does not constitute a proof.
probability-theory measure-theory concentration-of-measure optimal-transport
asked Jan 26 at 20:12
felipehfelipeh
1,606619
$endgroup$
add a comment |
$begingroup$
Let $mu$ be the uniform measure on the cube $Q = [-1,1]^n$, and $nu$ be the uniform measure on the surface
$$
V = {(x_1,dots,x_n)in Q mid sum x_i = 0}.
$$
I am curious about Wasserstein distance between $mu$ and $nu$, using the $l^infty$ distance on $mathbb{R}^n$.
That is, I would like a bound on
$$
sup_{|f|_{Lip}=1} |mathbb{E}_mu f - mathbb{E}_nu f|.
$$
I expect that the distance is on the order of $n^{-1/2}$ based on the following two calculations. First, one guess is that the function $S(x)=frac{1}{n}|sum x_i|$ would be close to optimizing the above supremum, and
$$
|mathbb{E}_mu S - mathbb{E}_nu S| leq Cn^{-1/2}.
$$
The second heuristic follows from the fact that if $xin Q$ is sampled according to $mu$, then with high probability $|sum x_i| leq Cn^{1/2}$, so by nudging each of the coefficients by an amount $n^{-1/2}$ it should be possible to land on the plane. This gives a transport map from $mu$ to a measure supported on $V$ that has cost roughly $n^{-1/2}$. The problem is that the transported measure is not the same as $nu$, so this does not constitute a proof.
probability-theory measure-theory concentration-of-measure optimal-transport
asked Jan 26 at 20:12
felipehfelipeh
1,606619
$endgroup$
Let $mu$ be the uniform measure on the cube $Q = [-1,1]^n$, and $nu$ be the uniform measure on the surface
$$
V = {(x_1,dots,x_n)in Q mid sum x_i = 0}.
$$
I am curious about Wasserstein distance between $mu$ and $nu$, using the $l^infty$ distance on $mathbb{R}^n$.
That is, I would like a bound on
$$
sup_{|f|_{Lip}=1} |mathbb{E}_mu f - mathbb{E}_nu f|.
$$
I expect that the distance is on the order of $n^{-1/2}$ based on the following two calculations. First, one guess is that the function $S(x)=frac{1}{n}|sum x_i|$ would be close to optimizing the above supremum, and
$$
|mathbb{E}_mu S - mathbb{E}_nu S| leq Cn^{-1/2}.
$$
The second heuristic follows from the fact that if $xin Q$ is sampled according to $mu$, then with high probability $|sum x_i| leq Cn^{1/2}$, so by nudging each of the coefficients by an amount $n^{-1/2}$ it should be possible to land on the plane. This gives a transport map from $mu$ to a measure supported on $V$ that has cost roughly $n^{-1/2}$. The problem is that the transported measure is not the same as $nu$, so this does not constitute a proof.
probability-theory measure-theory concentration-of-measure optimal-transport
probability-theory measure-theory concentration-of-measure optimal-transport
asked Jan 26 at 20:12
felipehfelipeh
1,606619
asked Jan 26 at 20:12
felipehfelipeh
1,606619
asked Jan 26 at 20:12
felipehfelipeh
1,606619
asked Jan 26 at 20:12
felipehfelipeh
1,606619
asked Jan 26 at 20:12
felipehfelipeh
1,606619
1,606619
add a comment |
add a comment |
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