Continuous square integrable martingales and family of stopping paths
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I have the following question and I'm not sure of the solution that I've been given:
"Show that for an element $Minmathcal{M}_c^2$, the family ${M_T}_{Tinmathcal{J}_a}$ is uniformly integrable for any $a>0$"
By definition I have that $mathcal{J}_a = {text{T a stopping time and} mathbb{P}(Tleq a)=1}$
The solution I have been given is very brief anf I do not understand:
$mathbb{E}[M_a|mathcal{F}_T]=M_T$
$mathbb{E}[M_a^2|mathcal{F}_T]geq M_T^2$
Then use something like Chebyshev
I am completely confused by this answer, could somebody shed some light on it? Thank you!
probability-theory stochastic-processes martingales uniform-integrability
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add a comment |
$begingroup$
I have the following question and I'm not sure of the solution that I've been given:
"Show that for an element $Minmathcal{M}_c^2$, the family ${M_T}_{Tinmathcal{J}_a}$ is uniformly integrable for any $a>0$"
By definition I have that $mathcal{J}_a = {text{T a stopping time and} mathbb{P}(Tleq a)=1}$
The solution I have been given is very brief anf I do not understand:
$mathbb{E}[M_a|mathcal{F}_T]=M_T$
$mathbb{E}[M_a^2|mathcal{F}_T]geq M_T^2$
Then use something like Chebyshev
I am completely confused by this answer, could somebody shed some light on it? Thank you!
probability-theory stochastic-processes martingales uniform-integrability
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What exactly do you not understand? The two statements follow from the optional stopping theorem and the fact that $(M_t)_t$ is a martingale and $(M_t^2)_t$ a submartingale.
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– saz
Apr 29 '18 at 5:25
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I am not aware of the optional stopping theorem, this was not in our notes. Could you explain what this theorem is?
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– R Thompson
Apr 29 '18 at 11:20
1
$begingroup$
Perhaps you know the optional sampling theorem? If not then give it a try and google it.
$endgroup$
– saz
Apr 29 '18 at 11:27
add a comment |
$begingroup$
I have the following question and I'm not sure of the solution that I've been given:
"Show that for an element $Minmathcal{M}_c^2$, the family ${M_T}_{Tinmathcal{J}_a}$ is uniformly integrable for any $a>0$"
By definition I have that $mathcal{J}_a = {text{T a stopping time and} mathbb{P}(Tleq a)=1}$
The solution I have been given is very brief anf I do not understand:
$mathbb{E}[M_a|mathcal{F}_T]=M_T$
$mathbb{E}[M_a^2|mathcal{F}_T]geq M_T^2$
Then use something like Chebyshev
I am completely confused by this answer, could somebody shed some light on it? Thank you!
probability-theory stochastic-processes martingales uniform-integrability
$endgroup$
I have the following question and I'm not sure of the solution that I've been given:
"Show that for an element $Minmathcal{M}_c^2$, the family ${M_T}_{Tinmathcal{J}_a}$ is uniformly integrable for any $a>0$"
By definition I have that $mathcal{J}_a = {text{T a stopping time and} mathbb{P}(Tleq a)=1}$
The solution I have been given is very brief anf I do not understand:
$mathbb{E}[M_a|mathcal{F}_T]=M_T$
$mathbb{E}[M_a^2|mathcal{F}_T]geq M_T^2$
Then use something like Chebyshev
I am completely confused by this answer, could somebody shed some light on it? Thank you!
probability-theory stochastic-processes martingales uniform-integrability
probability-theory stochastic-processes martingales uniform-integrability
asked Apr 28 '18 at 18:25
R ThompsonR Thompson
526
526
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What exactly do you not understand? The two statements follow from the optional stopping theorem and the fact that $(M_t)_t$ is a martingale and $(M_t^2)_t$ a submartingale.
$endgroup$
– saz
Apr 29 '18 at 5:25
$begingroup$
I am not aware of the optional stopping theorem, this was not in our notes. Could you explain what this theorem is?
$endgroup$
– R Thompson
Apr 29 '18 at 11:20
1
$begingroup$
Perhaps you know the optional sampling theorem? If not then give it a try and google it.
$endgroup$
– saz
Apr 29 '18 at 11:27
add a comment |
$begingroup$
What exactly do you not understand? The two statements follow from the optional stopping theorem and the fact that $(M_t)_t$ is a martingale and $(M_t^2)_t$ a submartingale.
$endgroup$
– saz
Apr 29 '18 at 5:25
$begingroup$
I am not aware of the optional stopping theorem, this was not in our notes. Could you explain what this theorem is?
$endgroup$
– R Thompson
Apr 29 '18 at 11:20
1
$begingroup$
Perhaps you know the optional sampling theorem? If not then give it a try and google it.
$endgroup$
– saz
Apr 29 '18 at 11:27
$begingroup$
What exactly do you not understand? The two statements follow from the optional stopping theorem and the fact that $(M_t)_t$ is a martingale and $(M_t^2)_t$ a submartingale.
$endgroup$
– saz
Apr 29 '18 at 5:25
$begingroup$
What exactly do you not understand? The two statements follow from the optional stopping theorem and the fact that $(M_t)_t$ is a martingale and $(M_t^2)_t$ a submartingale.
$endgroup$
– saz
Apr 29 '18 at 5:25
$begingroup$
I am not aware of the optional stopping theorem, this was not in our notes. Could you explain what this theorem is?
$endgroup$
– R Thompson
Apr 29 '18 at 11:20
$begingroup$
I am not aware of the optional stopping theorem, this was not in our notes. Could you explain what this theorem is?
$endgroup$
– R Thompson
Apr 29 '18 at 11:20
1
1
$begingroup$
Perhaps you know the optional sampling theorem? If not then give it a try and google it.
$endgroup$
– saz
Apr 29 '18 at 11:27
$begingroup$
Perhaps you know the optional sampling theorem? If not then give it a try and google it.
$endgroup$
– saz
Apr 29 '18 at 11:27
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let us admit that $$tag{*}mathbb{E}[M_a|mathcal{F}_T]=M_T.$$ Then
$$
leftlvert M_Trightrvertleqslant mathbb{E}left[leftlvert M_arightrvert|mathcal{F}_Tright]
$$
and you probably now that for each non-negative integrable random variable $Y$ on a probability space $(Omega,mathcal F,Pr)$, the family $left{mathbb Eleft[Ymid mathcal Gright],mathcal Gmbox{ is a sub sigma-algebra of }mathcal Fright}$ is uniformly integrable.
In order to prove (*), you can try to start with the case where $T$ has finitely many values, and then look at $T_n=2^nlfloor 2^{-n}Trfloor$.
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add a comment |
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1 Answer
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1 Answer
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active
oldest
votes
active
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$begingroup$
Let us admit that $$tag{*}mathbb{E}[M_a|mathcal{F}_T]=M_T.$$ Then
$$
leftlvert M_Trightrvertleqslant mathbb{E}left[leftlvert M_arightrvert|mathcal{F}_Tright]
$$
and you probably now that for each non-negative integrable random variable $Y$ on a probability space $(Omega,mathcal F,Pr)$, the family $left{mathbb Eleft[Ymid mathcal Gright],mathcal Gmbox{ is a sub sigma-algebra of }mathcal Fright}$ is uniformly integrable.
In order to prove (*), you can try to start with the case where $T$ has finitely many values, and then look at $T_n=2^nlfloor 2^{-n}Trfloor$.
$endgroup$
add a comment |
$begingroup$
Let us admit that $$tag{*}mathbb{E}[M_a|mathcal{F}_T]=M_T.$$ Then
$$
leftlvert M_Trightrvertleqslant mathbb{E}left[leftlvert M_arightrvert|mathcal{F}_Tright]
$$
and you probably now that for each non-negative integrable random variable $Y$ on a probability space $(Omega,mathcal F,Pr)$, the family $left{mathbb Eleft[Ymid mathcal Gright],mathcal Gmbox{ is a sub sigma-algebra of }mathcal Fright}$ is uniformly integrable.
In order to prove (*), you can try to start with the case where $T$ has finitely many values, and then look at $T_n=2^nlfloor 2^{-n}Trfloor$.
$endgroup$
add a comment |
$begingroup$
Let us admit that $$tag{*}mathbb{E}[M_a|mathcal{F}_T]=M_T.$$ Then
$$
leftlvert M_Trightrvertleqslant mathbb{E}left[leftlvert M_arightrvert|mathcal{F}_Tright]
$$
and you probably now that for each non-negative integrable random variable $Y$ on a probability space $(Omega,mathcal F,Pr)$, the family $left{mathbb Eleft[Ymid mathcal Gright],mathcal Gmbox{ is a sub sigma-algebra of }mathcal Fright}$ is uniformly integrable.
In order to prove (*), you can try to start with the case where $T$ has finitely many values, and then look at $T_n=2^nlfloor 2^{-n}Trfloor$.
$endgroup$
Let us admit that $$tag{*}mathbb{E}[M_a|mathcal{F}_T]=M_T.$$ Then
$$
leftlvert M_Trightrvertleqslant mathbb{E}left[leftlvert M_arightrvert|mathcal{F}_Tright]
$$
and you probably now that for each non-negative integrable random variable $Y$ on a probability space $(Omega,mathcal F,Pr)$, the family $left{mathbb Eleft[Ymid mathcal Gright],mathcal Gmbox{ is a sub sigma-algebra of }mathcal Fright}$ is uniformly integrable.
In order to prove (*), you can try to start with the case where $T$ has finitely many values, and then look at $T_n=2^nlfloor 2^{-n}Trfloor$.
answered Dec 12 '18 at 10:47
Davide GiraudoDavide Giraudo
126k16150261
126k16150261
add a comment |
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$begingroup$
What exactly do you not understand? The two statements follow from the optional stopping theorem and the fact that $(M_t)_t$ is a martingale and $(M_t^2)_t$ a submartingale.
$endgroup$
– saz
Apr 29 '18 at 5:25
$begingroup$
I am not aware of the optional stopping theorem, this was not in our notes. Could you explain what this theorem is?
$endgroup$
– R Thompson
Apr 29 '18 at 11:20
1
$begingroup$
Perhaps you know the optional sampling theorem? If not then give it a try and google it.
$endgroup$
– saz
Apr 29 '18 at 11:27