Wasserstein penalization for time dependent measures
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I have time dependant measures, say $mu: [0,T] rightarrow mathcal{M}(Omega)$ and I'm looking to define a penalization in this space that would measure the amount of displacement in space through time.
My first option is to consider the Wasserstein metric. The simplest example I can think of is that for two time measurements $0leq t_1<t_2 leq T$, the following penalization can be considered
$$ P_2(mu) = W_2(mu(t_1), mu(t_2))$$
More generally, I could consider for $0 = t_0 < ldots < t_n = T$
$$ P_n(mu) = frac{1}{N} sum_{i=0}^{n-1} W_2(mu(t_i), mu(t_{i+1}))$$
So my question is, what happens when the limit $lim_{nrightarrow infty} P_n(mu)$ is considered? Does it has any name I could search in the literature?
optimal-transport
asked Nov 13 at 9:59
pancho
517210
add a comment |
up vote
0
down vote
favorite
I have time dependant measures, say $mu: [0,T] rightarrow mathcal{M}(Omega)$ and I'm looking to define a penalization in this space that would measure the amount of displacement in space through time.
My first option is to consider the Wasserstein metric. The simplest example I can think of is that for two time measurements $0leq t_1<t_2 leq T$, the following penalization can be considered
$$ P_2(mu) = W_2(mu(t_1), mu(t_2))$$
More generally, I could consider for $0 = t_0 < ldots < t_n = T$
$$ P_n(mu) = frac{1}{N} sum_{i=0}^{n-1} W_2(mu(t_i), mu(t_{i+1}))$$
So my question is, what happens when the limit $lim_{nrightarrow infty} P_n(mu)$ is considered? Does it has any name I could search in the literature?
optimal-transport
asked Nov 13 at 9:59
pancho
517210
1
You have to specify how you choose your partition to make this limit well-defined. Presumably you want that the mesh of the partition goes to zero, right? In this case, $n P_n(mu)$ converges to the length of $mu$ in the Wasserstein space. Thus $P_n(mu)$ converges to zero if $mu$ has finite length.
– MaoWao
Nov 14 at 10:17
Oh thanks, I didn't noticed that the closet the times are, the closer is it's wasserstein distance (It's obvious now). So indeed this converges to the arclength using the Wasserstein distance. Has this being studied somewhere?
– pancho
Nov 14 at 10:48
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have time dependant measures, say $mu: [0,T] rightarrow mathcal{M}(Omega)$ and I'm looking to define a penalization in this space that would measure the amount of displacement in space through time.
My first option is to consider the Wasserstein metric. The simplest example I can think of is that for two time measurements $0leq t_1<t_2 leq T$, the following penalization can be considered
$$ P_2(mu) = W_2(mu(t_1), mu(t_2))$$
More generally, I could consider for $0 = t_0 < ldots < t_n = T$
$$ P_n(mu) = frac{1}{N} sum_{i=0}^{n-1} W_2(mu(t_i), mu(t_{i+1}))$$
So my question is, what happens when the limit $lim_{nrightarrow infty} P_n(mu)$ is considered? Does it has any name I could search in the literature?
optimal-transport
asked Nov 13 at 9:59
pancho
517210
I have time dependant measures, say $mu: [0,T] rightarrow mathcal{M}(Omega)$ and I'm looking to define a penalization in this space that would measure the amount of displacement in space through time.
My first option is to consider the Wasserstein metric. The simplest example I can think of is that for two time measurements $0leq t_1<t_2 leq T$, the following penalization can be considered
$$ P_2(mu) = W_2(mu(t_1), mu(t_2))$$
More generally, I could consider for $0 = t_0 < ldots < t_n = T$
$$ P_n(mu) = frac{1}{N} sum_{i=0}^{n-1} W_2(mu(t_i), mu(t_{i+1}))$$
So my question is, what happens when the limit $lim_{nrightarrow infty} P_n(mu)$ is considered? Does it has any name I could search in the literature?
optimal-transport
optimal-transport
asked Nov 13 at 9:59
pancho
517210
asked Nov 13 at 9:59
pancho
517210
asked Nov 13 at 9:59
pancho
517210
asked Nov 13 at 9:59
pancho
517210
asked Nov 13 at 9:59
pancho
517210
517210
1
You have to specify how you choose your partition to make this limit well-defined. Presumably you want that the mesh of the partition goes to zero, right? In this case, $n P_n(mu)$ converges to the length of $mu$ in the Wasserstein space. Thus $P_n(mu)$ converges to zero if $mu$ has finite length.
– MaoWao
Nov 14 at 10:17
Oh thanks, I didn't noticed that the closet the times are, the closer is it's wasserstein distance (It's obvious now). So indeed this converges to the arclength using the Wasserstein distance. Has this being studied somewhere?
– pancho
Nov 14 at 10:48
add a comment |
1
You have to specify how you choose your partition to make this limit well-defined. Presumably you want that the mesh of the partition goes to zero, right? In this case, $n P_n(mu)$ converges to the length of $mu$ in the Wasserstein space. Thus $P_n(mu)$ converges to zero if $mu$ has finite length.
– MaoWao
Nov 14 at 10:17
Oh thanks, I didn't noticed that the closet the times are, the closer is it's wasserstein distance (It's obvious now). So indeed this converges to the arclength using the Wasserstein distance. Has this being studied somewhere?
– pancho
Nov 14 at 10:48
1
1
You have to specify how you choose your partition to make this limit well-defined. Presumably you want that the mesh of the partition goes to zero, right? In this case, $n P_n(mu)$ converges to the length of $mu$ in the Wasserstein space. Thus $P_n(mu)$ converges to zero if $mu$ has finite length.
– MaoWao
Nov 14 at 10:17
You have to specify how you choose your partition to make this limit well-defined. Presumably you want that the mesh of the partition goes to zero, right? In this case, $n P_n(mu)$ converges to the length of $mu$ in the Wasserstein space. Thus $P_n(mu)$ converges to zero if $mu$ has finite length.
– MaoWao
Nov 14 at 10:17
Oh thanks, I didn't noticed that the closet the times are, the closer is it's wasserstein distance (It's obvious now). So indeed this converges to the arclength using the Wasserstein distance. Has this being studied somewhere?
– pancho
Nov 14 at 10:48
Oh thanks, I didn't noticed that the closet the times are, the closer is it's wasserstein distance (It's obvious now). So indeed this converges to the arclength using the Wasserstein distance. Has this being studied somewhere?
– pancho
Nov 14 at 10:48
add a comment |
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You have to specify how you choose your partition to make this limit well-defined. Presumably you want that the mesh of the partition goes to zero, right? In this case, $n P_n(mu)$ converges to the length of $mu$ in the Wasserstein space. Thus $P_n(mu)$ converges to zero if $mu$ has finite length.
– MaoWao
Nov 14 at 10:17
Oh thanks, I didn't noticed that the closet the times are, the closer is it's wasserstein distance (It's obvious now). So indeed this converges to the arclength using the Wasserstein distance. Has this being studied somewhere?
– pancho
Nov 14 at 10:48