What is a martingale $(X_n)$ index by the finite set $(0,1,…,N)$?
In the book of Yor and Revuz : "Continuous martingale and Brownian motion" third edition, proposition 1.5 page 53 : it's written : If $(X_n)$ is an integrable submartingale indexed by the finite set $(0,1,...,N)$... what does it mean exactly ? First $(0,1,...,N)$ is not really a set, so what does he mean by that ?
probability definition
asked Dec 10 '18 at 20:40
NewMath
3318
|
show 1 more comment
In the book of Yor and Revuz : "Continuous martingale and Brownian motion" third edition, proposition 1.5 page 53 : it's written : If $(X_n)$ is an integrable submartingale indexed by the finite set $(0,1,...,N)$... what does it mean exactly ? First $(0,1,...,N)$ is not really a set, so what does he mean by that ?
probability definition
asked Dec 10 '18 at 20:40
NewMath
3318
1
If he said ${0,1,dots,N}$ would you be OK with the definition?
– GEdgar
Dec 10 '18 at 20:44
@GEdgar: If so, would it be just $(X_n)=(X_n)_{n=1,...,N}$ ?
– NewMath
Dec 10 '18 at 20:45
You missed the index $0$ in there.
– GEdgar
Dec 10 '18 at 20:47
@GEdgar yes, I mean $(X_n)_{n=0,...,N}$... but how can this be a martingale ? For example $mathbb E[X_mmid mathcal F_n]=X_n$ is still true if the set is finite ?
– NewMath
Dec 10 '18 at 20:52
Exactly: $mathbb E[X_mmid mathcal F_n]=X_n$ whenever $m ge n$.
– GEdgar
Dec 10 '18 at 20:54
|
show 1 more comment
In the book of Yor and Revuz : "Continuous martingale and Brownian motion" third edition, proposition 1.5 page 53 : it's written : If $(X_n)$ is an integrable submartingale indexed by the finite set $(0,1,...,N)$... what does it mean exactly ? First $(0,1,...,N)$ is not really a set, so what does he mean by that ?
probability definition
asked Dec 10 '18 at 20:40
NewMath
3318
In the book of Yor and Revuz : "Continuous martingale and Brownian motion" third edition, proposition 1.5 page 53 : it's written : If $(X_n)$ is an integrable submartingale indexed by the finite set $(0,1,...,N)$... what does it mean exactly ? First $(0,1,...,N)$ is not really a set, so what does he mean by that ?
probability definition
probability definition
asked Dec 10 '18 at 20:40
NewMath
3318
asked Dec 10 '18 at 20:40
NewMath
3318
asked Dec 10 '18 at 20:40
NewMath
3318
asked Dec 10 '18 at 20:40
NewMath
3318
asked Dec 10 '18 at 20:40
NewMath
3318
3318
1
If he said ${0,1,dots,N}$ would you be OK with the definition?
– GEdgar
Dec 10 '18 at 20:44
@GEdgar: If so, would it be just $(X_n)=(X_n)_{n=1,...,N}$ ?
– NewMath
Dec 10 '18 at 20:45
You missed the index $0$ in there.
– GEdgar
Dec 10 '18 at 20:47
@GEdgar yes, I mean $(X_n)_{n=0,...,N}$... but how can this be a martingale ? For example $mathbb E[X_mmid mathcal F_n]=X_n$ is still true if the set is finite ?
– NewMath
Dec 10 '18 at 20:52
Exactly: $mathbb E[X_mmid mathcal F_n]=X_n$ whenever $m ge n$.
– GEdgar
Dec 10 '18 at 20:54
|
show 1 more comment
1
If he said ${0,1,dots,N}$ would you be OK with the definition?
– GEdgar
Dec 10 '18 at 20:44
@GEdgar: If so, would it be just $(X_n)=(X_n)_{n=1,...,N}$ ?
– NewMath
Dec 10 '18 at 20:45
You missed the index $0$ in there.
– GEdgar
Dec 10 '18 at 20:47
@GEdgar yes, I mean $(X_n)_{n=0,...,N}$... but how can this be a martingale ? For example $mathbb E[X_mmid mathcal F_n]=X_n$ is still true if the set is finite ?
– NewMath
Dec 10 '18 at 20:52
Exactly: $mathbb E[X_mmid mathcal F_n]=X_n$ whenever $m ge n$.
– GEdgar
Dec 10 '18 at 20:54
1
1
If he said ${0,1,dots,N}$ would you be OK with the definition?
– GEdgar
Dec 10 '18 at 20:44
If he said ${0,1,dots,N}$ would you be OK with the definition?
– GEdgar
Dec 10 '18 at 20:44
@GEdgar: If so, would it be just $(X_n)=(X_n)_{n=1,...,N}$ ?
– NewMath
Dec 10 '18 at 20:45
@GEdgar: If so, would it be just $(X_n)=(X_n)_{n=1,...,N}$ ?
– NewMath
Dec 10 '18 at 20:45
You missed the index $0$ in there.
– GEdgar
Dec 10 '18 at 20:47
You missed the index $0$ in there.
– GEdgar
Dec 10 '18 at 20:47
@GEdgar yes, I mean $(X_n)_{n=0,...,N}$... but how can this be a martingale ? For example $mathbb E[X_mmid mathcal F_n]=X_n$ is still true if the set is finite ?
– NewMath
Dec 10 '18 at 20:52
@GEdgar yes, I mean $(X_n)_{n=0,...,N}$... but how can this be a martingale ? For example $mathbb E[X_mmid mathcal F_n]=X_n$ is still true if the set is finite ?
– NewMath
Dec 10 '18 at 20:52
Exactly: $mathbb E[X_mmid mathcal F_n]=X_n$ whenever $m ge n$.
– GEdgar
Dec 10 '18 at 20:54
Exactly: $mathbb E[X_mmid mathcal F_n]=X_n$ whenever $m ge n$.
– GEdgar
Dec 10 '18 at 20:54
|
show 1 more comment
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1
If he said ${0,1,dots,N}$ would you be OK with the definition?
– GEdgar
Dec 10 '18 at 20:44
@GEdgar: If so, would it be just $(X_n)=(X_n)_{n=1,...,N}$ ?
– NewMath
Dec 10 '18 at 20:45
You missed the index $0$ in there.
– GEdgar
Dec 10 '18 at 20:47
@GEdgar yes, I mean $(X_n)_{n=0,...,N}$... but how can this be a martingale ? For example $mathbb E[X_mmid mathcal F_n]=X_n$ is still true if the set is finite ?
– NewMath
Dec 10 '18 at 20:52
Exactly: $mathbb E[X_mmid mathcal F_n]=X_n$ whenever $m ge n$.
– GEdgar
Dec 10 '18 at 20:54