Mixing
Prove that supermartingale with specific characteristics is a martingale
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Prove that if $ (X_{n}, mathcal{F}_n)_{n=0}^infty $ is a supermartingale such as
$ EX_{n} = EX_{0} $ for all $ n $ then $ (X_{n}, mathcal{F}_n)_{n=0}^infty $ is a martingale
Is it enough to say that if $ EX_{n} = EX_{0} $ for all $ n $ then $ EX_{n} = const $ for all $ n $ and use a prove of theory saying that that every supermartingale $X_{t}$ for which the map $tmapsto E(X_{t})$ is constant is already a martingale?
probability-theory martingales
asked Jan 9 at 8:18
PabloPablo
1086
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add a comment |
$begingroup$
Prove that if $ (X_{n}, mathcal{F}_n)_{n=0}^infty $ is a supermartingale such as
$ EX_{n} = EX_{0} $ for all $ n $ then $ (X_{n}, mathcal{F}_n)_{n=0}^infty $ is a martingale
Is it enough to say that if $ EX_{n} = EX_{0} $ for all $ n $ then $ EX_{n} = const $ for all $ n $ and use a prove of theory saying that that every supermartingale $X_{t}$ for which the map $tmapsto E(X_{t})$ is constant is already a martingale?
probability-theory martingales
asked Jan 9 at 8:18
PabloPablo
1086
$endgroup$
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Yes. I can post a proof later if no one else does.
$endgroup$
– Calculon
Jan 9 at 8:21
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Thank you i know the proof
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– Pablo
Jan 9 at 8:24
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If you know the proof, then why pose this question...?
$endgroup$
– saz
Jan 9 at 8:31
add a comment |
$begingroup$
Prove that if $ (X_{n}, mathcal{F}_n)_{n=0}^infty $ is a supermartingale such as
$ EX_{n} = EX_{0} $ for all $ n $ then $ (X_{n}, mathcal{F}_n)_{n=0}^infty $ is a martingale
Is it enough to say that if $ EX_{n} = EX_{0} $ for all $ n $ then $ EX_{n} = const $ for all $ n $ and use a prove of theory saying that that every supermartingale $X_{t}$ for which the map $tmapsto E(X_{t})$ is constant is already a martingale?
probability-theory martingales
asked Jan 9 at 8:18
PabloPablo
1086
$endgroup$
Prove that if $ (X_{n}, mathcal{F}_n)_{n=0}^infty $ is a supermartingale such as
$ EX_{n} = EX_{0} $ for all $ n $ then $ (X_{n}, mathcal{F}_n)_{n=0}^infty $ is a martingale
Is it enough to say that if $ EX_{n} = EX_{0} $ for all $ n $ then $ EX_{n} = const $ for all $ n $ and use a prove of theory saying that that every supermartingale $X_{t}$ for which the map $tmapsto E(X_{t})$ is constant is already a martingale?
probability-theory martingales
probability-theory martingales
asked Jan 9 at 8:18
PabloPablo
1086
asked Jan 9 at 8:18
PabloPablo
1086
asked Jan 9 at 8:18
PabloPablo
1086
asked Jan 9 at 8:18
PabloPablo
1086
asked Jan 9 at 8:18
PabloPablo
1086
1086
$begingroup$
Yes. I can post a proof later if no one else does.
$endgroup$
– Calculon
Jan 9 at 8:21
$begingroup$
Thank you i know the proof
$endgroup$
– Pablo
Jan 9 at 8:24
$begingroup$
If you know the proof, then why pose this question...?
$endgroup$
– saz
Jan 9 at 8:31
add a comment |
$begingroup$
Yes. I can post a proof later if no one else does.
$endgroup$
– Calculon
Jan 9 at 8:21
$begingroup$
Thank you i know the proof
$endgroup$
– Pablo
Jan 9 at 8:24
$begingroup$
If you know the proof, then why pose this question...?
$endgroup$
– saz
Jan 9 at 8:31
$begingroup$
Yes. I can post a proof later if no one else does.
$endgroup$
– Calculon
Jan 9 at 8:21
$begingroup$
Yes. I can post a proof later if no one else does.
$endgroup$
– Calculon
Jan 9 at 8:21
$begingroup$
Thank you i know the proof
$endgroup$
– Pablo
Jan 9 at 8:24
$begingroup$
Thank you i know the proof
$endgroup$
– Pablo
Jan 9 at 8:24
$begingroup$
If you know the proof, then why pose this question...?
$endgroup$
– saz
Jan 9 at 8:31
$begingroup$
If you know the proof, then why pose this question...?
$endgroup$
– saz
Jan 9 at 8:31
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$E(X_{n+1} |mathcal F_n) geq X_n$. But $E (E(X_{n+1} |mathcal F_n) - X_n) =EX_{n=1} -EX_n=EX_0-EX_0=0$. If a non-negative random variable has zero mean it is $0$ almost surely. Hence $E(X_{n+1} |mathcal F_n) = X_n$.
answered Jan 9 at 8:26
Kavi Rama MurthyKavi Rama Murthy
57.5k42160
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
$E(X_{n+1} |mathcal F_n) geq X_n$. But $E (E(X_{n+1} |mathcal F_n) - X_n) =EX_{n=1} -EX_n=EX_0-EX_0=0$. If a non-negative random variable has zero mean it is $0$ almost surely. Hence $E(X_{n+1} |mathcal F_n) = X_n$.
answered Jan 9 at 8:26
Kavi Rama MurthyKavi Rama Murthy
57.5k42160
$endgroup$
add a comment |
$begingroup$
$E(X_{n+1} |mathcal F_n) geq X_n$. But $E (E(X_{n+1} |mathcal F_n) - X_n) =EX_{n=1} -EX_n=EX_0-EX_0=0$. If a non-negative random variable has zero mean it is $0$ almost surely. Hence $E(X_{n+1} |mathcal F_n) = X_n$.
answered Jan 9 at 8:26
Kavi Rama MurthyKavi Rama Murthy
57.5k42160
$endgroup$
add a comment |
$begingroup$
$E(X_{n+1} |mathcal F_n) geq X_n$. But $E (E(X_{n+1} |mathcal F_n) - X_n) =EX_{n=1} -EX_n=EX_0-EX_0=0$. If a non-negative random variable has zero mean it is $0$ almost surely. Hence $E(X_{n+1} |mathcal F_n) = X_n$.
answered Jan 9 at 8:26
Kavi Rama MurthyKavi Rama Murthy
57.5k42160
$endgroup$
$E(X_{n+1} |mathcal F_n) geq X_n$. But $E (E(X_{n+1} |mathcal F_n) - X_n) =EX_{n=1} -EX_n=EX_0-EX_0=0$. If a non-negative random variable has zero mean it is $0$ almost surely. Hence $E(X_{n+1} |mathcal F_n) = X_n$.
answered Jan 9 at 8:26
Kavi Rama MurthyKavi Rama Murthy
57.5k42160
answered Jan 9 at 8:26
Kavi Rama MurthyKavi Rama Murthy
57.5k42160
answered Jan 9 at 8:26
Kavi Rama MurthyKavi Rama Murthy
57.5k42160
answered Jan 9 at 8:26
Kavi Rama MurthyKavi Rama Murthy
57.5k42160
57.5k42160
add a comment |
add a comment |
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$begingroup$
Yes. I can post a proof later if no one else does.
$endgroup$
– Calculon
Jan 9 at 8:21
$begingroup$
Thank you i know the proof
$endgroup$
– Pablo
Jan 9 at 8:24
$begingroup$
If you know the proof, then why pose this question...?
$endgroup$
– saz
Jan 9 at 8:31