Mixing
Local martingale up to a stopping time vs local martingale
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Let $M$ be a stochastic process, and let $T$ be a stopping time. We call $M$ a local martingale up to $T$ if there exists a sequence $(T_n)$ of stopping times such that $T = text{sup} T_n$ and $M^{T_n}$ is a martingale for all $n$. Moreover, a martingale is a local martingale if it is a local martingale up to $infty$.
Now if $M$ is a continuous local martingale up to $T$, how can I show that $M^T$ is a local martingale? Is continuity necessary for this statement? Thanks for any idea :-)
probability-theory stochastic-processes stochastic-calculus
asked Jan 18 at 18:31
RenRen
317
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add a comment |
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Let $M$ be a stochastic process, and let $T$ be a stopping time. We call $M$ a local martingale up to $T$ if there exists a sequence $(T_n)$ of stopping times such that $T = text{sup} T_n$ and $M^{T_n}$ is a martingale for all $n$. Moreover, a martingale is a local martingale if it is a local martingale up to $infty$.
Now if $M$ is a continuous local martingale up to $T$, how can I show that $M^T$ is a local martingale? Is continuity necessary for this statement? Thanks for any idea :-)
probability-theory stochastic-processes stochastic-calculus
asked Jan 18 at 18:31
RenRen
317
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$begingroup$
math.stackexchange.com/questions/2329717/…
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– d.k.o.
Jan 18 at 19:09
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@d.k.o the $M$ discussed there is already a local martingale up to time $infty$
$endgroup$
– Ren
Jan 18 at 19:51
add a comment |
$begingroup$
Let $M$ be a stochastic process, and let $T$ be a stopping time. We call $M$ a local martingale up to $T$ if there exists a sequence $(T_n)$ of stopping times such that $T = text{sup} T_n$ and $M^{T_n}$ is a martingale for all $n$. Moreover, a martingale is a local martingale if it is a local martingale up to $infty$.
Now if $M$ is a continuous local martingale up to $T$, how can I show that $M^T$ is a local martingale? Is continuity necessary for this statement? Thanks for any idea :-)
probability-theory stochastic-processes stochastic-calculus
asked Jan 18 at 18:31
RenRen
317
$endgroup$
Let $M$ be a stochastic process, and let $T$ be a stopping time. We call $M$ a local martingale up to $T$ if there exists a sequence $(T_n)$ of stopping times such that $T = text{sup} T_n$ and $M^{T_n}$ is a martingale for all $n$. Moreover, a martingale is a local martingale if it is a local martingale up to $infty$.
Now if $M$ is a continuous local martingale up to $T$, how can I show that $M^T$ is a local martingale? Is continuity necessary for this statement? Thanks for any idea :-)
probability-theory stochastic-processes stochastic-calculus
probability-theory stochastic-processes stochastic-calculus
asked Jan 18 at 18:31
RenRen
317
asked Jan 18 at 18:31
RenRen
317
asked Jan 18 at 18:31
RenRen
317
asked Jan 18 at 18:31
RenRen
317
asked Jan 18 at 18:31
RenRen
317
317
$begingroup$
math.stackexchange.com/questions/2329717/…
$endgroup$
– d.k.o.
Jan 18 at 19:09
$begingroup$
@d.k.o the $M$ discussed there is already a local martingale up to time $infty$
$endgroup$
– Ren
Jan 18 at 19:51
add a comment |
$begingroup$
math.stackexchange.com/questions/2329717/…
$endgroup$
– d.k.o.
Jan 18 at 19:09
$begingroup$
@d.k.o the $M$ discussed there is already a local martingale up to time $infty$
$endgroup$
– Ren
Jan 18 at 19:51
$begingroup$
math.stackexchange.com/questions/2329717/…
$endgroup$
– d.k.o.
Jan 18 at 19:09
$begingroup$
math.stackexchange.com/questions/2329717/…
$endgroup$
– d.k.o.
Jan 18 at 19:09
$begingroup$
@d.k.o the $M$ discussed there is already a local martingale up to time $infty$
$endgroup$
– Ren
Jan 18 at 19:51
$begingroup$
@d.k.o the $M$ discussed there is already a local martingale up to time $infty$
$endgroup$
– Ren
Jan 18 at 19:51
add a comment |
1 Answer
1
active
oldest
votes
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Let $S_n:=inf{tge 0:|M_t|ge n}$. Since $M$ is continuous $S_nnearrowinfty$ a.s. It tsuffices to show that ${S_n}$ reduces $M^{T}$, i.e. we nts that $(M_{twedge Twedge S_n})$ is a martingale. Let $Xequiv M^T$. Then for each $kge 1$, $X^{S_nwedge T_k}$ is a bounded martingale and $T_knearrow T$ a.s. Therefore, for all $sle t$,
$$
mathsf{E}[X_{twedge S_n}mid mathcal{F}_s]=lim_{ktoinfty}mathsf{E}[X_{twedge S_nwedge T_k}mid mathcal{F}_s]=lim_{ktoinfty}X_{swedge S_nwedge T_k}=X_{swedge S_n} quadtext{a.s.}
$$
answered Jan 18 at 21:42
d.k.o.d.k.o.
9,930629
$endgroup$
$begingroup$
for the first equality, the dominant convergence theorem is used. Why is its condition satisfied?
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– Ren
Jan 19 at 9:00
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$lvert (X^{S_nwedge T_k})_trvertle n$.
$endgroup$
– d.k.o.
Jan 19 at 9:23
$begingroup$
oh thanks a lot!
$endgroup$
– Ren
Jan 19 at 13:10
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $S_n:=inf{tge 0:|M_t|ge n}$. Since $M$ is continuous $S_nnearrowinfty$ a.s. It tsuffices to show that ${S_n}$ reduces $M^{T}$, i.e. we nts that $(M_{twedge Twedge S_n})$ is a martingale. Let $Xequiv M^T$. Then for each $kge 1$, $X^{S_nwedge T_k}$ is a bounded martingale and $T_knearrow T$ a.s. Therefore, for all $sle t$,
$$
mathsf{E}[X_{twedge S_n}mid mathcal{F}_s]=lim_{ktoinfty}mathsf{E}[X_{twedge S_nwedge T_k}mid mathcal{F}_s]=lim_{ktoinfty}X_{swedge S_nwedge T_k}=X_{swedge S_n} quadtext{a.s.}
$$
answered Jan 18 at 21:42
d.k.o.d.k.o.
9,930629
$endgroup$
$begingroup$
for the first equality, the dominant convergence theorem is used. Why is its condition satisfied?
$endgroup$
– Ren
Jan 19 at 9:00
$begingroup$
$lvert (X^{S_nwedge T_k})_trvertle n$.
$endgroup$
– d.k.o.
Jan 19 at 9:23
$begingroup$
oh thanks a lot!
$endgroup$
– Ren
Jan 19 at 13:10
add a comment |
$begingroup$
Let $S_n:=inf{tge 0:|M_t|ge n}$. Since $M$ is continuous $S_nnearrowinfty$ a.s. It tsuffices to show that ${S_n}$ reduces $M^{T}$, i.e. we nts that $(M_{twedge Twedge S_n})$ is a martingale. Let $Xequiv M^T$. Then for each $kge 1$, $X^{S_nwedge T_k}$ is a bounded martingale and $T_knearrow T$ a.s. Therefore, for all $sle t$,
$$
mathsf{E}[X_{twedge S_n}mid mathcal{F}_s]=lim_{ktoinfty}mathsf{E}[X_{twedge S_nwedge T_k}mid mathcal{F}_s]=lim_{ktoinfty}X_{swedge S_nwedge T_k}=X_{swedge S_n} quadtext{a.s.}
$$
answered Jan 18 at 21:42
d.k.o.d.k.o.
9,930629
$endgroup$
$begingroup$
for the first equality, the dominant convergence theorem is used. Why is its condition satisfied?
$endgroup$
– Ren
Jan 19 at 9:00
$begingroup$
$lvert (X^{S_nwedge T_k})_trvertle n$.
$endgroup$
– d.k.o.
Jan 19 at 9:23
$begingroup$
oh thanks a lot!
$endgroup$
– Ren
Jan 19 at 13:10
add a comment |
$begingroup$
Let $S_n:=inf{tge 0:|M_t|ge n}$. Since $M$ is continuous $S_nnearrowinfty$ a.s. It tsuffices to show that ${S_n}$ reduces $M^{T}$, i.e. we nts that $(M_{twedge Twedge S_n})$ is a martingale. Let $Xequiv M^T$. Then for each $kge 1$, $X^{S_nwedge T_k}$ is a bounded martingale and $T_knearrow T$ a.s. Therefore, for all $sle t$,
$$
mathsf{E}[X_{twedge S_n}mid mathcal{F}_s]=lim_{ktoinfty}mathsf{E}[X_{twedge S_nwedge T_k}mid mathcal{F}_s]=lim_{ktoinfty}X_{swedge S_nwedge T_k}=X_{swedge S_n} quadtext{a.s.}
$$
answered Jan 18 at 21:42
d.k.o.d.k.o.
9,930629
$endgroup$
Let $S_n:=inf{tge 0:|M_t|ge n}$. Since $M$ is continuous $S_nnearrowinfty$ a.s. It tsuffices to show that ${S_n}$ reduces $M^{T}$, i.e. we nts that $(M_{twedge Twedge S_n})$ is a martingale. Let $Xequiv M^T$. Then for each $kge 1$, $X^{S_nwedge T_k}$ is a bounded martingale and $T_knearrow T$ a.s. Therefore, for all $sle t$,
$$
mathsf{E}[X_{twedge S_n}mid mathcal{F}_s]=lim_{ktoinfty}mathsf{E}[X_{twedge S_nwedge T_k}mid mathcal{F}_s]=lim_{ktoinfty}X_{swedge S_nwedge T_k}=X_{swedge S_n} quadtext{a.s.}
$$
answered Jan 18 at 21:42
d.k.o.d.k.o.
9,930629
answered Jan 18 at 21:42
d.k.o.d.k.o.
9,930629
answered Jan 18 at 21:42
d.k.o.d.k.o.
9,930629
answered Jan 18 at 21:42
d.k.o.d.k.o.
9,930629
9,930629
$begingroup$
for the first equality, the dominant convergence theorem is used. Why is its condition satisfied?
$endgroup$
– Ren
Jan 19 at 9:00
$begingroup$
$lvert (X^{S_nwedge T_k})_trvertle n$.
$endgroup$
– d.k.o.
Jan 19 at 9:23
$begingroup$
oh thanks a lot!
$endgroup$
– Ren
Jan 19 at 13:10
add a comment |
$begingroup$
for the first equality, the dominant convergence theorem is used. Why is its condition satisfied?
$endgroup$
– Ren
Jan 19 at 9:00
$begingroup$
$lvert (X^{S_nwedge T_k})_trvertle n$.
$endgroup$
– d.k.o.
Jan 19 at 9:23
$begingroup$
oh thanks a lot!
$endgroup$
– Ren
Jan 19 at 13:10
$begingroup$
for the first equality, the dominant convergence theorem is used. Why is its condition satisfied?
$endgroup$
– Ren
Jan 19 at 9:00
$begingroup$
for the first equality, the dominant convergence theorem is used. Why is its condition satisfied?
$endgroup$
– Ren
Jan 19 at 9:00
$begingroup$
$lvert (X^{S_nwedge T_k})_trvertle n$.
$endgroup$
– d.k.o.
Jan 19 at 9:23
$begingroup$
$lvert (X^{S_nwedge T_k})_trvertle n$.
$endgroup$
– d.k.o.
Jan 19 at 9:23
$begingroup$
oh thanks a lot!
$endgroup$
– Ren
Jan 19 at 13:10
$begingroup$
oh thanks a lot!
$endgroup$
– Ren
Jan 19 at 13:10
add a comment |
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$begingroup$
math.stackexchange.com/questions/2329717/…
$endgroup$
– d.k.o.
Jan 18 at 19:09
$begingroup$
@d.k.o the $M$ discussed there is already a local martingale up to time $infty$
$endgroup$
– Ren
Jan 18 at 19:51