How to prove a probability measure is relatively compact？

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$mu_alpha$ is a series of probability measure corresponding to random variables $X_alpha$. If $exists r>0,M>0,s.t.forall alpha,mathbb E|X_alpha|^r<M,$ then $mu_alpha$ is relatively compact.



I know that if $X_alpha$ is defined in a complete separable space, then I just need to prove $mu_alpha$ is tight based on 'prohorov theorem' and it will be easier .



But generally, how can I get this statement?



Thanks for your help.






probability sequences-and-series measure-theory







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  • 1




    I really don't think you're going to get any general answer. I mean, write down that expected value and see if you can bound it... or... try other things...
    – Jack M
    Sep 2 at 9:19











  • @ Jack M I mean if $X_alpha$ is defined in a general metric space, how can I prove this statement
    – HQR
    Sep 2 at 9:31










  • You are asking, why does a uniform moment bound imply tightness?
    – kimchi lover
    Sep 2 at 12:00














up vote
0
down vote

favorite












$mu_alpha$ is a series of probability measure corresponding to random variables $X_alpha$. If $exists r>0,M>0,s.t.forall alpha,mathbb E|X_alpha|^r<M,$ then $mu_alpha$ is relatively compact.



I know that if $X_alpha$ is defined in a complete separable space, then I just need to prove $mu_alpha$ is tight based on 'prohorov theorem' and it will be easier .



But generally, how can I get this statement?



Thanks for your help.






probability sequences-and-series measure-theory







share|cite|improve this question



















  • 1




    I really don't think you're going to get any general answer. I mean, write down that expected value and see if you can bound it... or... try other things...
    – Jack M
    Sep 2 at 9:19











  • @ Jack M I mean if $X_alpha$ is defined in a general metric space, how can I prove this statement
    – HQR
    Sep 2 at 9:31










  • You are asking, why does a uniform moment bound imply tightness?
    – kimchi lover
    Sep 2 at 12:00












up vote
0
down vote

favorite









up vote
0
down vote

favorite











$mu_alpha$ is a series of probability measure corresponding to random variables $X_alpha$. If $exists r>0,M>0,s.t.forall alpha,mathbb E|X_alpha|^r<M,$ then $mu_alpha$ is relatively compact.



I know that if $X_alpha$ is defined in a complete separable space, then I just need to prove $mu_alpha$ is tight based on 'prohorov theorem' and it will be easier .



But generally, how can I get this statement?



Thanks for your help.






probability sequences-and-series measure-theory







share|cite|improve this question















$mu_alpha$ is a series of probability measure corresponding to random variables $X_alpha$. If $exists r>0,M>0,s.t.forall alpha,mathbb E|X_alpha|^r<M,$ then $mu_alpha$ is relatively compact.



I know that if $X_alpha$ is defined in a complete separable space, then I just need to prove $mu_alpha$ is tight based on 'prohorov theorem' and it will be easier .



But generally, how can I get this statement?



Thanks for your help.






probability sequences-and-series measure-theory




probability sequences-and-series measure-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Sep 2 at 9:33

























asked Sep 2 at 9:02









HQR

329113




329113







  • 1




    I really don't think you're going to get any general answer. I mean, write down that expected value and see if you can bound it... or... try other things...
    – Jack M
    Sep 2 at 9:19











  • @ Jack M I mean if $X_alpha$ is defined in a general metric space, how can I prove this statement
    – HQR
    Sep 2 at 9:31










  • You are asking, why does a uniform moment bound imply tightness?
    – kimchi lover
    Sep 2 at 12:00












  • 1




    I really don't think you're going to get any general answer. I mean, write down that expected value and see if you can bound it... or... try other things...
    – Jack M
    Sep 2 at 9:19











  • @ Jack M I mean if $X_alpha$ is defined in a general metric space, how can I prove this statement
    – HQR
    Sep 2 at 9:31










  • You are asking, why does a uniform moment bound imply tightness?
    – kimchi lover
    Sep 2 at 12:00







1




1




I really don't think you're going to get any general answer. I mean, write down that expected value and see if you can bound it... or... try other things...
– Jack M
Sep 2 at 9:19





I really don't think you're going to get any general answer. I mean, write down that expected value and see if you can bound it... or... try other things...
– Jack M
Sep 2 at 9:19













@ Jack M I mean if $X_alpha$ is defined in a general metric space, how can I prove this statement
– HQR
Sep 2 at 9:31




@ Jack M I mean if $X_alpha$ is defined in a general metric space, how can I prove this statement
– HQR
Sep 2 at 9:31












You are asking, why does a uniform moment bound imply tightness?
– kimchi lover
Sep 2 at 12:00




You are asking, why does a uniform moment bound imply tightness?
– kimchi lover
Sep 2 at 12:00










1 Answer
1






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If I understand the question correctly $mu_alpha$ is a family of Borel probability measures on $mathbb R$ with $sup _alpha int |x|^r, dmu (x) <infty$ and you want to show that $mu_alpha$ is tight. The domain of $X_alpha$'s does not come i not the picture at all. (You are probably saying that the basic probability space is a complete separable metric space with some given Borel probability measure but this has no realtion to the tightness question). Since $mu_alpha x: leq frac 1 A^r sup _alpha int |x|^r, dmu_alpha (x)$ we see that $mu_alpha$ is tight. PS: if the random variables take values in a metric space then $E|X_alpha|^r$ does not even make sense, so I am assuming that $X_alpha$'s are real valued.






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  • Thanks very much.Maybe you are right.
    – HQR
    Sep 3 at 1:17










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote



accepted










If I understand the question correctly $mu_alpha$ is a family of Borel probability measures on $mathbb R$ with $sup _alpha int |x|^r, dmu (x) <infty$ and you want to show that $mu_alpha$ is tight. The domain of $X_alpha$'s does not come i not the picture at all. (You are probably saying that the basic probability space is a complete separable metric space with some given Borel probability measure but this has no realtion to the tightness question). Since $mu_alpha x: leq frac 1 A^r sup _alpha int |x|^r, dmu_alpha (x)$ we see that $mu_alpha$ is tight. PS: if the random variables take values in a metric space then $E|X_alpha|^r$ does not even make sense, so I am assuming that $X_alpha$'s are real valued.






share|cite|improve this answer




















  • Thanks very much.Maybe you are right.
    – HQR
    Sep 3 at 1:17














up vote
2
down vote



accepted










If I understand the question correctly $mu_alpha$ is a family of Borel probability measures on $mathbb R$ with $sup _alpha int |x|^r, dmu (x) <infty$ and you want to show that $mu_alpha$ is tight. The domain of $X_alpha$'s does not come i not the picture at all. (You are probably saying that the basic probability space is a complete separable metric space with some given Borel probability measure but this has no realtion to the tightness question). Since $mu_alpha x: leq frac 1 A^r sup _alpha int |x|^r, dmu_alpha (x)$ we see that $mu_alpha$ is tight. PS: if the random variables take values in a metric space then $E|X_alpha|^r$ does not even make sense, so I am assuming that $X_alpha$'s are real valued.






share|cite|improve this answer




















  • Thanks very much.Maybe you are right.
    – HQR
    Sep 3 at 1:17












up vote
2
down vote



accepted







up vote
2
down vote



accepted






If I understand the question correctly $mu_alpha$ is a family of Borel probability measures on $mathbb R$ with $sup _alpha int |x|^r, dmu (x) <infty$ and you want to show that $mu_alpha$ is tight. The domain of $X_alpha$'s does not come i not the picture at all. (You are probably saying that the basic probability space is a complete separable metric space with some given Borel probability measure but this has no realtion to the tightness question). Since $mu_alpha x: leq frac 1 A^r sup _alpha int |x|^r, dmu_alpha (x)$ we see that $mu_alpha$ is tight. PS: if the random variables take values in a metric space then $E|X_alpha|^r$ does not even make sense, so I am assuming that $X_alpha$'s are real valued.






share|cite|improve this answer












If I understand the question correctly $mu_alpha$ is a family of Borel probability measures on $mathbb R$ with $sup _alpha int |x|^r, dmu (x) <infty$ and you want to show that $mu_alpha$ is tight. The domain of $X_alpha$'s does not come i not the picture at all. (You are probably saying that the basic probability space is a complete separable metric space with some given Borel probability measure but this has no realtion to the tightness question). Since $mu_alpha x: leq frac 1 A^r sup _alpha int |x|^r, dmu_alpha (x)$ we see that $mu_alpha$ is tight. PS: if the random variables take values in a metric space then $E|X_alpha|^r$ does not even make sense, so I am assuming that $X_alpha$'s are real valued.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Sep 2 at 12:25









Kavi Rama Murthy

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  • Thanks very much.Maybe you are right.
    – HQR
    Sep 3 at 1:17
















  • Thanks very much.Maybe you are right.
    – HQR
    Sep 3 at 1:17















Thanks very much.Maybe you are right.
– HQR
Sep 3 at 1:17




Thanks very much.Maybe you are right.
– HQR
Sep 3 at 1:17

















 

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