Mixing
Where is an error in my deduction? (question about martingales)
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Suppose we have a filtration ${mathcal{F_{t}},tgeq 0}$ and a stochastic process ${ X_{t},tgeq 0}$ which is adapted to this filtration and also integrable. All we need for this process to be a martingale is to hold $ E[X_{t} | mathcal{F_{s}} ] = X_{s}$ for every $t > s$. But suppose we take a trivial filtration i.e. $mathcal{F_{t}}=mathcal{F}$ for every $tgeq 0$
where $mathcal{F}$ is some sigma-algebra in our probability space. Then obviously $ E[X_{t} | mathcal{F_{s}} ] = X_{t}$ for every $t > s$ which breaks our martingale property. So adaptability and integrability for ${ X_{t},tgeq 0}$ and $ E[X_{t} | mathcal{F_{s}} ] = X_{s}$ is not enough to be a martingale? Do we need some extra assumtions concerning ${mathcal{F_{t}},tgeq 0}$?
probability-theory stochastic-processes martingales stochastic-analysis filtrations
edited Jan 22 at 13:33
Davide Giraudo
127k16152266
asked Jan 20 at 14:18
Est MayhemEst Mayhem
534
$endgroup$
add a comment |
$begingroup$
Suppose we have a filtration ${mathcal{F_{t}},tgeq 0}$ and a stochastic process ${ X_{t},tgeq 0}$ which is adapted to this filtration and also integrable. All we need for this process to be a martingale is to hold $ E[X_{t} | mathcal{F_{s}} ] = X_{s}$ for every $t > s$. But suppose we take a trivial filtration i.e. $mathcal{F_{t}}=mathcal{F}$ for every $tgeq 0$
where $mathcal{F}$ is some sigma-algebra in our probability space. Then obviously $ E[X_{t} | mathcal{F_{s}} ] = X_{t}$ for every $t > s$ which breaks our martingale property. So adaptability and integrability for ${ X_{t},tgeq 0}$ and $ E[X_{t} | mathcal{F_{s}} ] = X_{s}$ is not enough to be a martingale? Do we need some extra assumtions concerning ${mathcal{F_{t}},tgeq 0}$?
probability-theory stochastic-processes martingales stochastic-analysis filtrations
edited Jan 22 at 13:33
Davide Giraudo
127k16152266
asked Jan 20 at 14:18
Est MayhemEst Mayhem
534
$endgroup$
$begingroup$
A martingale is a martingale always with respect to a filtration. A process that is a martingale with respect to one filtration may not be a martingale with respect to another filtration.
$endgroup$
– AddSup
Jan 20 at 14:22
$begingroup$
@AddSup I mean adaptability to a trivial filtration which I described in the question, the fact that X is adaptable to such filtration makes this process somehow trivial as well or no?
$endgroup$
– Est Mayhem
Jan 20 at 14:24
add a comment |
$begingroup$
Suppose we have a filtration ${mathcal{F_{t}},tgeq 0}$ and a stochastic process ${ X_{t},tgeq 0}$ which is adapted to this filtration and also integrable. All we need for this process to be a martingale is to hold $ E[X_{t} | mathcal{F_{s}} ] = X_{s}$ for every $t > s$. But suppose we take a trivial filtration i.e. $mathcal{F_{t}}=mathcal{F}$ for every $tgeq 0$
where $mathcal{F}$ is some sigma-algebra in our probability space. Then obviously $ E[X_{t} | mathcal{F_{s}} ] = X_{t}$ for every $t > s$ which breaks our martingale property. So adaptability and integrability for ${ X_{t},tgeq 0}$ and $ E[X_{t} | mathcal{F_{s}} ] = X_{s}$ is not enough to be a martingale? Do we need some extra assumtions concerning ${mathcal{F_{t}},tgeq 0}$?
probability-theory stochastic-processes martingales stochastic-analysis filtrations
edited Jan 22 at 13:33
Davide Giraudo
127k16152266
asked Jan 20 at 14:18
Est MayhemEst Mayhem
534
$endgroup$
Suppose we have a filtration ${mathcal{F_{t}},tgeq 0}$ and a stochastic process ${ X_{t},tgeq 0}$ which is adapted to this filtration and also integrable. All we need for this process to be a martingale is to hold $ E[X_{t} | mathcal{F_{s}} ] = X_{s}$ for every $t > s$. But suppose we take a trivial filtration i.e. $mathcal{F_{t}}=mathcal{F}$ for every $tgeq 0$
where $mathcal{F}$ is some sigma-algebra in our probability space. Then obviously $ E[X_{t} | mathcal{F_{s}} ] = X_{t}$ for every $t > s$ which breaks our martingale property. So adaptability and integrability for ${ X_{t},tgeq 0}$ and $ E[X_{t} | mathcal{F_{s}} ] = X_{s}$ is not enough to be a martingale? Do we need some extra assumtions concerning ${mathcal{F_{t}},tgeq 0}$?
probability-theory stochastic-processes martingales stochastic-analysis filtrations
probability-theory stochastic-processes martingales stochastic-analysis filtrations
edited Jan 22 at 13:33
Davide Giraudo
127k16152266
asked Jan 20 at 14:18
Est MayhemEst Mayhem
534
edited Jan 22 at 13:33
Davide Giraudo
127k16152266
asked Jan 20 at 14:18
Est MayhemEst Mayhem
534
edited Jan 22 at 13:33
Davide Giraudo
127k16152266
edited Jan 22 at 13:33
Davide Giraudo
127k16152266
edited Jan 22 at 13:33
Davide Giraudo
127k16152266
127k16152266
asked Jan 20 at 14:18
Est MayhemEst Mayhem
534
asked Jan 20 at 14:18
Est MayhemEst Mayhem
534
asked Jan 20 at 14:18
Est MayhemEst Mayhem
534
534
$begingroup$
A martingale is a martingale always with respect to a filtration. A process that is a martingale with respect to one filtration may not be a martingale with respect to another filtration.
$endgroup$
– AddSup
Jan 20 at 14:22
$begingroup$
@AddSup I mean adaptability to a trivial filtration which I described in the question, the fact that X is adaptable to such filtration makes this process somehow trivial as well or no?
$endgroup$
– Est Mayhem
Jan 20 at 14:24
add a comment |
$begingroup$
A martingale is a martingale always with respect to a filtration. A process that is a martingale with respect to one filtration may not be a martingale with respect to another filtration.
$endgroup$
– AddSup
Jan 20 at 14:22
$begingroup$
@AddSup I mean adaptability to a trivial filtration which I described in the question, the fact that X is adaptable to such filtration makes this process somehow trivial as well or no?
$endgroup$
– Est Mayhem
Jan 20 at 14:24
$begingroup$
A martingale is a martingale always with respect to a filtration. A process that is a martingale with respect to one filtration may not be a martingale with respect to another filtration.
$endgroup$
– AddSup
Jan 20 at 14:22
$begingroup$
A martingale is a martingale always with respect to a filtration. A process that is a martingale with respect to one filtration may not be a martingale with respect to another filtration.
$endgroup$
– AddSup
Jan 20 at 14:22
$begingroup$
@AddSup I mean adaptability to a trivial filtration which I described in the question, the fact that X is adaptable to such filtration makes this process somehow trivial as well or no?
$endgroup$
– Est Mayhem
Jan 20 at 14:24
$begingroup$
@AddSup I mean adaptability to a trivial filtration which I described in the question, the fact that X is adaptable to such filtration makes this process somehow trivial as well or no?
$endgroup$
– Est Mayhem
Jan 20 at 14:24
add a comment |
1 Answer
1
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$begingroup$
Suppose that $left(X_tright)_{tgeqslant 0}$ is a martingale with respect to the filtration $left(mathcal F_tright)_{tgeqslant 0}$ where $mathcal F_t=mathcal F$ for all $t$.
The condition of adaptedness implies that for all $t$, $X_t$ is $mathcal F$-measurable.
The condition $mathbb Eleft[X_tmidmathcal F_sright]=X_s$ for $0leqslant slt t$ reads therefore $X_s=mathbb Eleft[X_tmidmathcal Fright]=X_t$, since $X_t$ is $mathcal F$-measurable hence $X_t=X_0$ for all $t$.
answered Jan 20 at 14:30
Davide GiraudoDavide Giraudo
127k16152266
$endgroup$
$begingroup$
So such a process would be stationary?
$endgroup$
– Est Mayhem
Jan 20 at 14:30
$begingroup$
It would be even "constant", in sense that $X_t$ does not depend on $t$.
$endgroup$
– Davide Giraudo
Jan 20 at 14:32
$begingroup$
So if we take some fancy combination of Wiener process (e.g. $frac{1}{3}W_{t}^3-tW_{t}$ as $X_{t}$) and say that it is adapted to some arbitrary filtration then it's not gonna be a martingale in general because of this possibility of taking "trivial filtration"? Even though martingale property seems to work here.
$endgroup$
– Est Mayhem
Jan 20 at 14:40
$begingroup$
This fancy combination could be a martingale, but for an other filtration.
$endgroup$
– Davide Giraudo
Jan 20 at 21:25
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Suppose that $left(X_tright)_{tgeqslant 0}$ is a martingale with respect to the filtration $left(mathcal F_tright)_{tgeqslant 0}$ where $mathcal F_t=mathcal F$ for all $t$.
The condition of adaptedness implies that for all $t$, $X_t$ is $mathcal F$-measurable.
The condition $mathbb Eleft[X_tmidmathcal F_sright]=X_s$ for $0leqslant slt t$ reads therefore $X_s=mathbb Eleft[X_tmidmathcal Fright]=X_t$, since $X_t$ is $mathcal F$-measurable hence $X_t=X_0$ for all $t$.
answered Jan 20 at 14:30
Davide GiraudoDavide Giraudo
127k16152266
$endgroup$
$begingroup$
So such a process would be stationary?
$endgroup$
– Est Mayhem
Jan 20 at 14:30
$begingroup$
It would be even "constant", in sense that $X_t$ does not depend on $t$.
$endgroup$
– Davide Giraudo
Jan 20 at 14:32
$begingroup$
So if we take some fancy combination of Wiener process (e.g. $frac{1}{3}W_{t}^3-tW_{t}$ as $X_{t}$) and say that it is adapted to some arbitrary filtration then it's not gonna be a martingale in general because of this possibility of taking "trivial filtration"? Even though martingale property seems to work here.
$endgroup$
– Est Mayhem
Jan 20 at 14:40
$begingroup$
This fancy combination could be a martingale, but for an other filtration.
$endgroup$
– Davide Giraudo
Jan 20 at 21:25
add a comment |
$begingroup$
Suppose that $left(X_tright)_{tgeqslant 0}$ is a martingale with respect to the filtration $left(mathcal F_tright)_{tgeqslant 0}$ where $mathcal F_t=mathcal F$ for all $t$.
The condition of adaptedness implies that for all $t$, $X_t$ is $mathcal F$-measurable.
The condition $mathbb Eleft[X_tmidmathcal F_sright]=X_s$ for $0leqslant slt t$ reads therefore $X_s=mathbb Eleft[X_tmidmathcal Fright]=X_t$, since $X_t$ is $mathcal F$-measurable hence $X_t=X_0$ for all $t$.
answered Jan 20 at 14:30
Davide GiraudoDavide Giraudo
127k16152266
$endgroup$
$begingroup$
So such a process would be stationary?
$endgroup$
– Est Mayhem
Jan 20 at 14:30
$begingroup$
It would be even "constant", in sense that $X_t$ does not depend on $t$.
$endgroup$
– Davide Giraudo
Jan 20 at 14:32
$begingroup$
So if we take some fancy combination of Wiener process (e.g. $frac{1}{3}W_{t}^3-tW_{t}$ as $X_{t}$) and say that it is adapted to some arbitrary filtration then it's not gonna be a martingale in general because of this possibility of taking "trivial filtration"? Even though martingale property seems to work here.
$endgroup$
– Est Mayhem
Jan 20 at 14:40
$begingroup$
This fancy combination could be a martingale, but for an other filtration.
$endgroup$
– Davide Giraudo
Jan 20 at 21:25
add a comment |
$begingroup$
Suppose that $left(X_tright)_{tgeqslant 0}$ is a martingale with respect to the filtration $left(mathcal F_tright)_{tgeqslant 0}$ where $mathcal F_t=mathcal F$ for all $t$.
The condition of adaptedness implies that for all $t$, $X_t$ is $mathcal F$-measurable.
The condition $mathbb Eleft[X_tmidmathcal F_sright]=X_s$ for $0leqslant slt t$ reads therefore $X_s=mathbb Eleft[X_tmidmathcal Fright]=X_t$, since $X_t$ is $mathcal F$-measurable hence $X_t=X_0$ for all $t$.
answered Jan 20 at 14:30
Davide GiraudoDavide Giraudo
127k16152266
$endgroup$
Suppose that $left(X_tright)_{tgeqslant 0}$ is a martingale with respect to the filtration $left(mathcal F_tright)_{tgeqslant 0}$ where $mathcal F_t=mathcal F$ for all $t$.
The condition of adaptedness implies that for all $t$, $X_t$ is $mathcal F$-measurable.
The condition $mathbb Eleft[X_tmidmathcal F_sright]=X_s$ for $0leqslant slt t$ reads therefore $X_s=mathbb Eleft[X_tmidmathcal Fright]=X_t$, since $X_t$ is $mathcal F$-measurable hence $X_t=X_0$ for all $t$.
answered Jan 20 at 14:30
Davide GiraudoDavide Giraudo
127k16152266
answered Jan 20 at 14:30
Davide GiraudoDavide Giraudo
127k16152266
answered Jan 20 at 14:30
Davide GiraudoDavide Giraudo
127k16152266
answered Jan 20 at 14:30
Davide GiraudoDavide Giraudo
127k16152266
127k16152266
$begingroup$
So such a process would be stationary?
$endgroup$
– Est Mayhem
Jan 20 at 14:30
$begingroup$
It would be even "constant", in sense that $X_t$ does not depend on $t$.
$endgroup$
– Davide Giraudo
Jan 20 at 14:32
$begingroup$
So if we take some fancy combination of Wiener process (e.g. $frac{1}{3}W_{t}^3-tW_{t}$ as $X_{t}$) and say that it is adapted to some arbitrary filtration then it's not gonna be a martingale in general because of this possibility of taking "trivial filtration"? Even though martingale property seems to work here.
$endgroup$
– Est Mayhem
Jan 20 at 14:40
$begingroup$
This fancy combination could be a martingale, but for an other filtration.
$endgroup$
– Davide Giraudo
Jan 20 at 21:25
add a comment |
$begingroup$
So such a process would be stationary?
$endgroup$
– Est Mayhem
Jan 20 at 14:30
$begingroup$
It would be even "constant", in sense that $X_t$ does not depend on $t$.
$endgroup$
– Davide Giraudo
Jan 20 at 14:32
$begingroup$
So if we take some fancy combination of Wiener process (e.g. $frac{1}{3}W_{t}^3-tW_{t}$ as $X_{t}$) and say that it is adapted to some arbitrary filtration then it's not gonna be a martingale in general because of this possibility of taking "trivial filtration"? Even though martingale property seems to work here.
$endgroup$
– Est Mayhem
Jan 20 at 14:40
$begingroup$
This fancy combination could be a martingale, but for an other filtration.
$endgroup$
– Davide Giraudo
Jan 20 at 21:25
$begingroup$
So such a process would be stationary?
$endgroup$
– Est Mayhem
Jan 20 at 14:30
$begingroup$
So such a process would be stationary?
$endgroup$
– Est Mayhem
Jan 20 at 14:30
$begingroup$
It would be even "constant", in sense that $X_t$ does not depend on $t$.
$endgroup$
– Davide Giraudo
Jan 20 at 14:32
$begingroup$
It would be even "constant", in sense that $X_t$ does not depend on $t$.
$endgroup$
– Davide Giraudo
Jan 20 at 14:32
$begingroup$
So if we take some fancy combination of Wiener process (e.g. $frac{1}{3}W_{t}^3-tW_{t}$ as $X_{t}$) and say that it is adapted to some arbitrary filtration then it's not gonna be a martingale in general because of this possibility of taking "trivial filtration"? Even though martingale property seems to work here.
$endgroup$
– Est Mayhem
Jan 20 at 14:40
$begingroup$
So if we take some fancy combination of Wiener process (e.g. $frac{1}{3}W_{t}^3-tW_{t}$ as $X_{t}$) and say that it is adapted to some arbitrary filtration then it's not gonna be a martingale in general because of this possibility of taking "trivial filtration"? Even though martingale property seems to work here.
$endgroup$
– Est Mayhem
Jan 20 at 14:40
$begingroup$
This fancy combination could be a martingale, but for an other filtration.
$endgroup$
– Davide Giraudo
Jan 20 at 21:25
$begingroup$
This fancy combination could be a martingale, but for an other filtration.
$endgroup$
– Davide Giraudo
Jan 20 at 21:25
add a comment |
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$begingroup$
A martingale is a martingale always with respect to a filtration. A process that is a martingale with respect to one filtration may not be a martingale with respect to another filtration.
$endgroup$
– AddSup
Jan 20 at 14:22
$begingroup$
@AddSup I mean adaptability to a trivial filtration which I described in the question, the fact that X is adaptable to such filtration makes this process somehow trivial as well or no?
$endgroup$
– Est Mayhem
Jan 20 at 14:24