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Is the topology of weak convergence of probability measures first-countable?
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Let $S$ be a separable metric space, and $mu,mu_1,mu_2,ldots$ be Borel probability measures on $S$.
We know that $mu_n to mu$ weakly if and only if $pi(mu_n,mu) to 0$ where $pi$ is the Prokhorov metric.
However, this (a priori) doesn't imply that the metric topology induced by $pi$ is the same as the topology of weak convergence of probability measures (which is the smallest topology such that $mu mapsto int f ,dmu$ is continuous for all continuous bounded $f:S to mathbbR$), since the topology of weak convergence of probability measures might not be first-countable.
So my question is: is the topology of weak convergence of probability measures on $S$ first-countable?
Thank you in advance.
general-topology functional-analysis probability-theory weak-convergence first-countable
edited Jul 19 at 8:48
Davide Giraudo
121k15147249
asked Jul 15 at 5:40
Wooyoung Chin
24519
 |Â
show 1 more comment
up vote
2
down vote
favorite
Let $S$ be a separable metric space, and $mu,mu_1,mu_2,ldots$ be Borel probability measures on $S$.
We know that $mu_n to mu$ weakly if and only if $pi(mu_n,mu) to 0$ where $pi$ is the Prokhorov metric.
However, this (a priori) doesn't imply that the metric topology induced by $pi$ is the same as the topology of weak convergence of probability measures (which is the smallest topology such that $mu mapsto int f ,dmu$ is continuous for all continuous bounded $f:S to mathbbR$), since the topology of weak convergence of probability measures might not be first-countable.
So my question is: is the topology of weak convergence of probability measures on $S$ first-countable?
Thank you in advance.
general-topology functional-analysis probability-theory weak-convergence first-countable
edited Jul 19 at 8:48
Davide Giraudo
121k15147249
asked Jul 15 at 5:40
Wooyoung Chin
24519
See en.wikipedia.org/wiki/L%C3%A9vy%E2%80%93Prokhorov_metric
– Kavi Rama Murthy
Jul 15 at 11:55
1
@KaviRamaMurthy That Wikipedia articles references the book by Billingsley. The proof that the Prokhorov metric metrizes weak convergence in that book is incomplete exactly because it does not show the weak topology is first countable.
– Michael Greinecker♦
Jul 15 at 12:00
Billingsley's book clearly states that if $S$ is separable then the space $Z$ of Borel probability measures on it with the topology of weak convergence is metrizable (and separable). This is in line14 of page 239. If you show this fact it follows as a consequence that weak convergence topology is first countable.
– Kavi Rama Murthy
Jul 15 at 12:15
1
I probably have the first edition. The statement I am quoting appears exactly four lines above the heading "Prohorov's Theorem"
– Kavi Rama Murthy
Jul 15 at 12:27
1
@KaviRamaMurthy I just looked at the first edition and you are right. This seems to be one of the cases where earlier editions are better.
– Michael Greinecker♦
Jul 15 at 12:38
 |Â
show 1 more comment
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $S$ be a separable metric space, and $mu,mu_1,mu_2,ldots$ be Borel probability measures on $S$.
We know that $mu_n to mu$ weakly if and only if $pi(mu_n,mu) to 0$ where $pi$ is the Prokhorov metric.
However, this (a priori) doesn't imply that the metric topology induced by $pi$ is the same as the topology of weak convergence of probability measures (which is the smallest topology such that $mu mapsto int f ,dmu$ is continuous for all continuous bounded $f:S to mathbbR$), since the topology of weak convergence of probability measures might not be first-countable.
So my question is: is the topology of weak convergence of probability measures on $S$ first-countable?
Thank you in advance.
general-topology functional-analysis probability-theory weak-convergence first-countable
edited Jul 19 at 8:48
Davide Giraudo
121k15147249
asked Jul 15 at 5:40
Wooyoung Chin
24519
Let $S$ be a separable metric space, and $mu,mu_1,mu_2,ldots$ be Borel probability measures on $S$.
We know that $mu_n to mu$ weakly if and only if $pi(mu_n,mu) to 0$ where $pi$ is the Prokhorov metric.
However, this (a priori) doesn't imply that the metric topology induced by $pi$ is the same as the topology of weak convergence of probability measures (which is the smallest topology such that $mu mapsto int f ,dmu$ is continuous for all continuous bounded $f:S to mathbbR$), since the topology of weak convergence of probability measures might not be first-countable.
So my question is: is the topology of weak convergence of probability measures on $S$ first-countable?
Thank you in advance.
general-topology functional-analysis probability-theory weak-convergence first-countable
edited Jul 19 at 8:48
Davide Giraudo
121k15147249
asked Jul 15 at 5:40
Wooyoung Chin
24519
edited Jul 19 at 8:48
Davide Giraudo
121k15147249
edited Jul 19 at 8:48
Davide Giraudo
121k15147249
edited Jul 19 at 8:48
Davide Giraudo
121k15147249
121k15147249
asked Jul 15 at 5:40
Wooyoung Chin
24519
asked Jul 15 at 5:40
Wooyoung Chin
24519
asked Jul 15 at 5:40
Wooyoung Chin
24519
24519
See en.wikipedia.org/wiki/L%C3%A9vy%E2%80%93Prokhorov_metric
– Kavi Rama Murthy
Jul 15 at 11:55
1
@KaviRamaMurthy That Wikipedia articles references the book by Billingsley. The proof that the Prokhorov metric metrizes weak convergence in that book is incomplete exactly because it does not show the weak topology is first countable.
– Michael Greinecker♦
Jul 15 at 12:00
Billingsley's book clearly states that if $S$ is separable then the space $Z$ of Borel probability measures on it with the topology of weak convergence is metrizable (and separable). This is in line14 of page 239. If you show this fact it follows as a consequence that weak convergence topology is first countable.
– Kavi Rama Murthy
Jul 15 at 12:15
1
I probably have the first edition. The statement I am quoting appears exactly four lines above the heading "Prohorov's Theorem"
– Kavi Rama Murthy
Jul 15 at 12:27
1
@KaviRamaMurthy I just looked at the first edition and you are right. This seems to be one of the cases where earlier editions are better.
– Michael Greinecker♦
Jul 15 at 12:38
 |Â
show 1 more comment
See en.wikipedia.org/wiki/L%C3%A9vy%E2%80%93Prokhorov_metric
– Kavi Rama Murthy
Jul 15 at 11:55
1
@KaviRamaMurthy That Wikipedia articles references the book by Billingsley. The proof that the Prokhorov metric metrizes weak convergence in that book is incomplete exactly because it does not show the weak topology is first countable.
– Michael Greinecker♦
Jul 15 at 12:00
Billingsley's book clearly states that if $S$ is separable then the space $Z$ of Borel probability measures on it with the topology of weak convergence is metrizable (and separable). This is in line14 of page 239. If you show this fact it follows as a consequence that weak convergence topology is first countable.
– Kavi Rama Murthy
Jul 15 at 12:15
1
I probably have the first edition. The statement I am quoting appears exactly four lines above the heading "Prohorov's Theorem"
– Kavi Rama Murthy
Jul 15 at 12:27
1
@KaviRamaMurthy I just looked at the first edition and you are right. This seems to be one of the cases where earlier editions are better.
– Michael Greinecker♦
Jul 15 at 12:38
See en.wikipedia.org/wiki/L%C3%A9vy%E2%80%93Prokhorov_metric
– Kavi Rama Murthy
Jul 15 at 11:55
See en.wikipedia.org/wiki/L%C3%A9vy%E2%80%93Prokhorov_metric
– Kavi Rama Murthy
Jul 15 at 11:55
1
1
@KaviRamaMurthy That Wikipedia articles references the book by Billingsley. The proof that the Prokhorov metric metrizes weak convergence in that book is incomplete exactly because it does not show the weak topology is first countable.
– Michael Greinecker♦
Jul 15 at 12:00
@KaviRamaMurthy That Wikipedia articles references the book by Billingsley. The proof that the Prokhorov metric metrizes weak convergence in that book is incomplete exactly because it does not show the weak topology is first countable.
– Michael Greinecker♦
Jul 15 at 12:00
Billingsley's book clearly states that if $S$ is separable then the space $Z$ of Borel probability measures on it with the topology of weak convergence is metrizable (and separable). This is in line14 of page 239. If you show this fact it follows as a consequence that weak convergence topology is first countable.
– Kavi Rama Murthy
Jul 15 at 12:15
Billingsley's book clearly states that if $S$ is separable then the space $Z$ of Borel probability measures on it with the topology of weak convergence is metrizable (and separable). This is in line14 of page 239. If you show this fact it follows as a consequence that weak convergence topology is first countable.
– Kavi Rama Murthy
Jul 15 at 12:15
1
1
I probably have the first edition. The statement I am quoting appears exactly four lines above the heading "Prohorov's Theorem"
– Kavi Rama Murthy
Jul 15 at 12:27
I probably have the first edition. The statement I am quoting appears exactly four lines above the heading "Prohorov's Theorem"
– Kavi Rama Murthy
Jul 15 at 12:27
1
1
@KaviRamaMurthy I just looked at the first edition and you are right. This seems to be one of the cases where earlier editions are better.
– Michael Greinecker♦
Jul 15 at 12:38
@KaviRamaMurthy I just looked at the first edition and you are right. This seems to be one of the cases where earlier editions are better.
– Michael Greinecker♦
Jul 15 at 12:38
 |Â
show 1 more comment
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See en.wikipedia.org/wiki/L%C3%A9vy%E2%80%93Prokhorov_metric
– Kavi Rama Murthy
Jul 15 at 11:55
1
@KaviRamaMurthy That Wikipedia articles references the book by Billingsley. The proof that the Prokhorov metric metrizes weak convergence in that book is incomplete exactly because it does not show the weak topology is first countable.
– Michael Greinecker♦
Jul 15 at 12:00
Billingsley's book clearly states that if $S$ is separable then the space $Z$ of Borel probability measures on it with the topology of weak convergence is metrizable (and separable). This is in line14 of page 239. If you show this fact it follows as a consequence that weak convergence topology is first countable.
– Kavi Rama Murthy
Jul 15 at 12:15
1
I probably have the first edition. The statement I am quoting appears exactly four lines above the heading "Prohorov's Theorem"
– Kavi Rama Murthy
Jul 15 at 12:27
1
@KaviRamaMurthy I just looked at the first edition and you are right. This seems to be one of the cases where earlier editions are better.
– Michael Greinecker♦
Jul 15 at 12:38