A closed (?) subset of the set of probability measures. Is my reasoning correct?
$begingroup$
Let $f:mathbb{R}^{d} rightarrow [0,infty) $ continuous function and denote by $P$ the space of probability measures on $mathbb{R}^{d}$ and by $P_R ,=, {nu in P , | , int f d nu leqslant R , } $ for $R>0$.
We equip both spaces with the topology of weak convergence of probability measures.
I am trying to see if $P_R$ is a closed subset of $P$.
My attempt, which shows that it is indeed closed subset, is the following :
Let ${mu_n} subseteq P_R $ such that $mu_n rightarrow mu in P $. I will show that $mu in P_R$.
${mu_n} subseteq P_R ,,Rightarrow ,, sup limits_{n} int f d mu_n , leqslant R < infty ,, Rightarrow ,, {mu_n}$ tight sequence in $P_R$ $Rightarrow$ sequentially compact in $P_R$ by Prokhorov's theorem $Rightarrow$ there exists subsequence ${mu_{k_n}}_n subseteq P_R$ which converges in $P_R$. So by uniqueness of the limit $mu in P_R$.
What do you think ? Is it correct ?
probability-theory proof-verification
asked Dec 7 '18 at 20:31
vl.athvl.ath
879
$endgroup$
add a comment |
$begingroup$
Let $f:mathbb{R}^{d} rightarrow [0,infty) $ continuous function and denote by $P$ the space of probability measures on $mathbb{R}^{d}$ and by $P_R ,=, {nu in P , | , int f d nu leqslant R , } $ for $R>0$.
We equip both spaces with the topology of weak convergence of probability measures.
I am trying to see if $P_R$ is a closed subset of $P$.
My attempt, which shows that it is indeed closed subset, is the following :
Let ${mu_n} subseteq P_R $ such that $mu_n rightarrow mu in P $. I will show that $mu in P_R$.
${mu_n} subseteq P_R ,,Rightarrow ,, sup limits_{n} int f d mu_n , leqslant R < infty ,, Rightarrow ,, {mu_n}$ tight sequence in $P_R$ $Rightarrow$ sequentially compact in $P_R$ by Prokhorov's theorem $Rightarrow$ there exists subsequence ${mu_{k_n}}_n subseteq P_R$ which converges in $P_R$. So by uniqueness of the limit $mu in P_R$.
What do you think ? Is it correct ?
probability-theory proof-verification
asked Dec 7 '18 at 20:31
vl.athvl.ath
879
$endgroup$
$begingroup$
Obviously false if you have strict inequality in the definition of $P_R$. (Take degenerate measures)
$endgroup$
– Kavi Rama Murthy
Dec 8 '18 at 0:03
$begingroup$
No I meant less or equal in the definition of P_R.. sorry typo
$endgroup$
– vl.ath
Dec 8 '18 at 9:38
add a comment |
$begingroup$
Let $f:mathbb{R}^{d} rightarrow [0,infty) $ continuous function and denote by $P$ the space of probability measures on $mathbb{R}^{d}$ and by $P_R ,=, {nu in P , | , int f d nu leqslant R , } $ for $R>0$.
We equip both spaces with the topology of weak convergence of probability measures.
I am trying to see if $P_R$ is a closed subset of $P$.
My attempt, which shows that it is indeed closed subset, is the following :
Let ${mu_n} subseteq P_R $ such that $mu_n rightarrow mu in P $. I will show that $mu in P_R$.
${mu_n} subseteq P_R ,,Rightarrow ,, sup limits_{n} int f d mu_n , leqslant R < infty ,, Rightarrow ,, {mu_n}$ tight sequence in $P_R$ $Rightarrow$ sequentially compact in $P_R$ by Prokhorov's theorem $Rightarrow$ there exists subsequence ${mu_{k_n}}_n subseteq P_R$ which converges in $P_R$. So by uniqueness of the limit $mu in P_R$.
What do you think ? Is it correct ?
probability-theory proof-verification
asked Dec 7 '18 at 20:31
vl.athvl.ath
879
$endgroup$
Let $f:mathbb{R}^{d} rightarrow [0,infty) $ continuous function and denote by $P$ the space of probability measures on $mathbb{R}^{d}$ and by $P_R ,=, {nu in P , | , int f d nu leqslant R , } $ for $R>0$.
We equip both spaces with the topology of weak convergence of probability measures.
I am trying to see if $P_R$ is a closed subset of $P$.
My attempt, which shows that it is indeed closed subset, is the following :
Let ${mu_n} subseteq P_R $ such that $mu_n rightarrow mu in P $. I will show that $mu in P_R$.
${mu_n} subseteq P_R ,,Rightarrow ,, sup limits_{n} int f d mu_n , leqslant R < infty ,, Rightarrow ,, {mu_n}$ tight sequence in $P_R$ $Rightarrow$ sequentially compact in $P_R$ by Prokhorov's theorem $Rightarrow$ there exists subsequence ${mu_{k_n}}_n subseteq P_R$ which converges in $P_R$. So by uniqueness of the limit $mu in P_R$.
What do you think ? Is it correct ?
probability-theory proof-verification
probability-theory proof-verification
asked Dec 7 '18 at 20:31
vl.athvl.ath
879
asked Dec 7 '18 at 20:31
vl.athvl.ath
879
edited Dec 8 '18 at 15:06
vl.ath
asked Dec 7 '18 at 20:31
vl.athvl.ath
879
asked Dec 7 '18 at 20:31
vl.athvl.ath
879
asked Dec 7 '18 at 20:31
vl.athvl.ath
879
879
$begingroup$
Obviously false if you have strict inequality in the definition of $P_R$. (Take degenerate measures)
$endgroup$
– Kavi Rama Murthy
Dec 8 '18 at 0:03
$begingroup$
No I meant less or equal in the definition of P_R.. sorry typo
$endgroup$
– vl.ath
Dec 8 '18 at 9:38
add a comment |
$begingroup$
Obviously false if you have strict inequality in the definition of $P_R$. (Take degenerate measures)
$endgroup$
– Kavi Rama Murthy
Dec 8 '18 at 0:03
$begingroup$
No I meant less or equal in the definition of P_R.. sorry typo
$endgroup$
– vl.ath
Dec 8 '18 at 9:38
$begingroup$
Obviously false if you have strict inequality in the definition of $P_R$. (Take degenerate measures)
$endgroup$
– Kavi Rama Murthy
Dec 8 '18 at 0:03
$begingroup$
Obviously false if you have strict inequality in the definition of $P_R$. (Take degenerate measures)
$endgroup$
– Kavi Rama Murthy
Dec 8 '18 at 0:03
$begingroup$
No I meant less or equal in the definition of P_R.. sorry typo
$endgroup$
– vl.ath
Dec 8 '18 at 9:38
$begingroup$
No I meant less or equal in the definition of P_R.. sorry typo
$endgroup$
– vl.ath
Dec 8 '18 at 9:38
add a comment |
1 Answer
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Boundedness of the integrals for one particular $f$ does not give you tightness: take $f equiv 0$, for example. Hence your argument is invalid. I suppose you meant $P_R={nu: int f d nu leq R}$. Suppose $nu_j to nu$ weakly. If $N$ is a positive integer and $f_N=min {f,N}$ then $f_N$ is a bounded continuous function. Hence $int f_N d nu =lim_{j to infty} int f_N dnu_j leq lim_{j to infty} int f dnu_j leq R$. Now let $N to infty$ and apply Monotone Convefrgence Theorem.
answered Dec 8 '18 at 0:08
Kavi Rama MurthyKavi Rama Murthy
66.9k53067
$endgroup$
$begingroup$
I have one more question. Based on this math.stackexchange.com/questions/1142631/… , if I also assume that my fuction f goes to $infty$ as $|x|rightarrow infty$, then is my attempt correct ?
$endgroup$
– vl.ath
Dec 8 '18 at 15:09
1
$begingroup$
@vl.ath In that case tightness is valid but still your argument is circular. You are already assuming that $P_R$ is closed when you say that the limiting measure $mu in P_R$.
$endgroup$
– Kavi Rama Murthy
Dec 8 '18 at 23:17
add a comment |
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1 Answer
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$begingroup$
Boundedness of the integrals for one particular $f$ does not give you tightness: take $f equiv 0$, for example. Hence your argument is invalid. I suppose you meant $P_R={nu: int f d nu leq R}$. Suppose $nu_j to nu$ weakly. If $N$ is a positive integer and $f_N=min {f,N}$ then $f_N$ is a bounded continuous function. Hence $int f_N d nu =lim_{j to infty} int f_N dnu_j leq lim_{j to infty} int f dnu_j leq R$. Now let $N to infty$ and apply Monotone Convefrgence Theorem.
answered Dec 8 '18 at 0:08
Kavi Rama MurthyKavi Rama Murthy
66.9k53067
$endgroup$
$begingroup$
I have one more question. Based on this math.stackexchange.com/questions/1142631/… , if I also assume that my fuction f goes to $infty$ as $|x|rightarrow infty$, then is my attempt correct ?
$endgroup$
– vl.ath
Dec 8 '18 at 15:09
1
$begingroup$
@vl.ath In that case tightness is valid but still your argument is circular. You are already assuming that $P_R$ is closed when you say that the limiting measure $mu in P_R$.
$endgroup$
– Kavi Rama Murthy
Dec 8 '18 at 23:17
add a comment |
$begingroup$
Boundedness of the integrals for one particular $f$ does not give you tightness: take $f equiv 0$, for example. Hence your argument is invalid. I suppose you meant $P_R={nu: int f d nu leq R}$. Suppose $nu_j to nu$ weakly. If $N$ is a positive integer and $f_N=min {f,N}$ then $f_N$ is a bounded continuous function. Hence $int f_N d nu =lim_{j to infty} int f_N dnu_j leq lim_{j to infty} int f dnu_j leq R$. Now let $N to infty$ and apply Monotone Convefrgence Theorem.
answered Dec 8 '18 at 0:08
Kavi Rama MurthyKavi Rama Murthy
66.9k53067
$endgroup$
$begingroup$
I have one more question. Based on this math.stackexchange.com/questions/1142631/… , if I also assume that my fuction f goes to $infty$ as $|x|rightarrow infty$, then is my attempt correct ?
$endgroup$
– vl.ath
Dec 8 '18 at 15:09
1
$begingroup$
@vl.ath In that case tightness is valid but still your argument is circular. You are already assuming that $P_R$ is closed when you say that the limiting measure $mu in P_R$.
$endgroup$
– Kavi Rama Murthy
Dec 8 '18 at 23:17
add a comment |
$begingroup$
Boundedness of the integrals for one particular $f$ does not give you tightness: take $f equiv 0$, for example. Hence your argument is invalid. I suppose you meant $P_R={nu: int f d nu leq R}$. Suppose $nu_j to nu$ weakly. If $N$ is a positive integer and $f_N=min {f,N}$ then $f_N$ is a bounded continuous function. Hence $int f_N d nu =lim_{j to infty} int f_N dnu_j leq lim_{j to infty} int f dnu_j leq R$. Now let $N to infty$ and apply Monotone Convefrgence Theorem.
answered Dec 8 '18 at 0:08
Kavi Rama MurthyKavi Rama Murthy
66.9k53067
$endgroup$
Boundedness of the integrals for one particular $f$ does not give you tightness: take $f equiv 0$, for example. Hence your argument is invalid. I suppose you meant $P_R={nu: int f d nu leq R}$. Suppose $nu_j to nu$ weakly. If $N$ is a positive integer and $f_N=min {f,N}$ then $f_N$ is a bounded continuous function. Hence $int f_N d nu =lim_{j to infty} int f_N dnu_j leq lim_{j to infty} int f dnu_j leq R$. Now let $N to infty$ and apply Monotone Convefrgence Theorem.
answered Dec 8 '18 at 0:08
Kavi Rama MurthyKavi Rama Murthy
66.9k53067
answered Dec 8 '18 at 0:08
Kavi Rama MurthyKavi Rama Murthy
66.9k53067
answered Dec 8 '18 at 0:08
Kavi Rama MurthyKavi Rama Murthy
66.9k53067
answered Dec 8 '18 at 0:08
Kavi Rama MurthyKavi Rama Murthy
66.9k53067
66.9k53067
$begingroup$
I have one more question. Based on this math.stackexchange.com/questions/1142631/… , if I also assume that my fuction f goes to $infty$ as $|x|rightarrow infty$, then is my attempt correct ?
$endgroup$
– vl.ath
Dec 8 '18 at 15:09
1
$begingroup$
@vl.ath In that case tightness is valid but still your argument is circular. You are already assuming that $P_R$ is closed when you say that the limiting measure $mu in P_R$.
$endgroup$
– Kavi Rama Murthy
Dec 8 '18 at 23:17
add a comment |
$begingroup$
I have one more question. Based on this math.stackexchange.com/questions/1142631/… , if I also assume that my fuction f goes to $infty$ as $|x|rightarrow infty$, then is my attempt correct ?
$endgroup$
– vl.ath
Dec 8 '18 at 15:09
1
$begingroup$
@vl.ath In that case tightness is valid but still your argument is circular. You are already assuming that $P_R$ is closed when you say that the limiting measure $mu in P_R$.
$endgroup$
– Kavi Rama Murthy
Dec 8 '18 at 23:17
$begingroup$
I have one more question. Based on this math.stackexchange.com/questions/1142631/… , if I also assume that my fuction f goes to $infty$ as $|x|rightarrow infty$, then is my attempt correct ?
$endgroup$
– vl.ath
Dec 8 '18 at 15:09
$begingroup$
I have one more question. Based on this math.stackexchange.com/questions/1142631/… , if I also assume that my fuction f goes to $infty$ as $|x|rightarrow infty$, then is my attempt correct ?
$endgroup$
– vl.ath
Dec 8 '18 at 15:09
1
1
$begingroup$
@vl.ath In that case tightness is valid but still your argument is circular. You are already assuming that $P_R$ is closed when you say that the limiting measure $mu in P_R$.
$endgroup$
– Kavi Rama Murthy
Dec 8 '18 at 23:17
$begingroup$
@vl.ath In that case tightness is valid but still your argument is circular. You are already assuming that $P_R$ is closed when you say that the limiting measure $mu in P_R$.
$endgroup$
– Kavi Rama Murthy
Dec 8 '18 at 23:17
add a comment |
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$begingroup$
Obviously false if you have strict inequality in the definition of $P_R$. (Take degenerate measures)
$endgroup$
– Kavi Rama Murthy
Dec 8 '18 at 0:03
$begingroup$
No I meant less or equal in the definition of P_R.. sorry typo
$endgroup$
– vl.ath
Dec 8 '18 at 9:38