If we want $int_0^⋅XdM$ to be a martingale, do we need to assume $E[int_0^tX_sd[M]_s]<∞$ for all $t$...
$begingroup$
Let
$(Omega,mathcal A,operatorname P)$ be a complete probability space
$(mathcal F_t)_{tge0}$ be a complete and right-continuous filtration on $(Omega,mathcal A,operatorname P)$
$(M_t)_{tge0}$ be a real-valued continous square-integrable $mathcal F$-martingale on $(Omega,mathcal A,operatorname P)$
$(X_t)_{tge0}$ be a real-valued $mathcal F$-predictable process on $(Omega,mathcal A,operatorname P)$
- $Tin[0,infty]$
In the construction of the Itō integral process $$Xcdot M=left(int_0^tX_s:{rm d}M_sright)_{tge0}$$ as a square-integrable martingale, do we need to assume $$operatorname Eleft[int_0^t|X_s|^2:{rm d}[M]_sright]<infty;;;text{for all }tin[0,T]setminusleft{inftyright}tag1$$ or even $$operatorname Eleft[int_0^T|X_s|^2:{rm d}[M]_sright]<inftytag2?$$
Clearly, the question is trivial if $T<infty$ and the whole construction is absolutely clear to me in that case. But what if $T=infty$? Is there any reason why we should even need $(2)$?
probability-theory stochastic-processes stochastic-calculus stochastic-integrals stochastic-analysis
asked Dec 19 '18 at 16:07
0xbadf00d0xbadf00d
1,97041531
$endgroup$
add a comment |
$begingroup$
Let
$(Omega,mathcal A,operatorname P)$ be a complete probability space
$(mathcal F_t)_{tge0}$ be a complete and right-continuous filtration on $(Omega,mathcal A,operatorname P)$
$(M_t)_{tge0}$ be a real-valued continous square-integrable $mathcal F$-martingale on $(Omega,mathcal A,operatorname P)$
$(X_t)_{tge0}$ be a real-valued $mathcal F$-predictable process on $(Omega,mathcal A,operatorname P)$
- $Tin[0,infty]$
In the construction of the Itō integral process $$Xcdot M=left(int_0^tX_s:{rm d}M_sright)_{tge0}$$ as a square-integrable martingale, do we need to assume $$operatorname Eleft[int_0^t|X_s|^2:{rm d}[M]_sright]<infty;;;text{for all }tin[0,T]setminusleft{inftyright}tag1$$ or even $$operatorname Eleft[int_0^T|X_s|^2:{rm d}[M]_sright]<inftytag2?$$
Clearly, the question is trivial if $T<infty$ and the whole construction is absolutely clear to me in that case. But what if $T=infty$? Is there any reason why we should even need $(2)$?
probability-theory stochastic-processes stochastic-calculus stochastic-integrals stochastic-analysis
asked Dec 19 '18 at 16:07
0xbadf00d0xbadf00d
1,97041531
$endgroup$
$begingroup$
(1) is enough to get a martingale. If you want e.g. an $L^2$-bounded martingale then you will need (2).
$endgroup$
– saz
Dec 19 '18 at 16:31
$begingroup$
@saz So, $(2)$ if we want, for example, define $(Xcdot M)_infty$.
$endgroup$
– 0xbadf00d
Dec 19 '18 at 18:07
$begingroup$
Yes, exactly...
$endgroup$
– saz
Dec 20 '18 at 6:03
$begingroup$
@saz Could you please take a look at this question: math.stackexchange.com/questions/3048457/…
$endgroup$
– 0xbadf00d
Dec 21 '18 at 18:05
add a comment |
$begingroup$
Let
$(Omega,mathcal A,operatorname P)$ be a complete probability space
$(mathcal F_t)_{tge0}$ be a complete and right-continuous filtration on $(Omega,mathcal A,operatorname P)$
$(M_t)_{tge0}$ be a real-valued continous square-integrable $mathcal F$-martingale on $(Omega,mathcal A,operatorname P)$
$(X_t)_{tge0}$ be a real-valued $mathcal F$-predictable process on $(Omega,mathcal A,operatorname P)$
- $Tin[0,infty]$
In the construction of the Itō integral process $$Xcdot M=left(int_0^tX_s:{rm d}M_sright)_{tge0}$$ as a square-integrable martingale, do we need to assume $$operatorname Eleft[int_0^t|X_s|^2:{rm d}[M]_sright]<infty;;;text{for all }tin[0,T]setminusleft{inftyright}tag1$$ or even $$operatorname Eleft[int_0^T|X_s|^2:{rm d}[M]_sright]<inftytag2?$$
Clearly, the question is trivial if $T<infty$ and the whole construction is absolutely clear to me in that case. But what if $T=infty$? Is there any reason why we should even need $(2)$?
probability-theory stochastic-processes stochastic-calculus stochastic-integrals stochastic-analysis
asked Dec 19 '18 at 16:07
0xbadf00d0xbadf00d
1,97041531
$endgroup$
Let
$(Omega,mathcal A,operatorname P)$ be a complete probability space
$(mathcal F_t)_{tge0}$ be a complete and right-continuous filtration on $(Omega,mathcal A,operatorname P)$
$(M_t)_{tge0}$ be a real-valued continous square-integrable $mathcal F$-martingale on $(Omega,mathcal A,operatorname P)$
$(X_t)_{tge0}$ be a real-valued $mathcal F$-predictable process on $(Omega,mathcal A,operatorname P)$
- $Tin[0,infty]$
In the construction of the Itō integral process $$Xcdot M=left(int_0^tX_s:{rm d}M_sright)_{tge0}$$ as a square-integrable martingale, do we need to assume $$operatorname Eleft[int_0^t|X_s|^2:{rm d}[M]_sright]<infty;;;text{for all }tin[0,T]setminusleft{inftyright}tag1$$ or even $$operatorname Eleft[int_0^T|X_s|^2:{rm d}[M]_sright]<inftytag2?$$
Clearly, the question is trivial if $T<infty$ and the whole construction is absolutely clear to me in that case. But what if $T=infty$? Is there any reason why we should even need $(2)$?
probability-theory stochastic-processes stochastic-calculus stochastic-integrals stochastic-analysis
probability-theory stochastic-processes stochastic-calculus stochastic-integrals stochastic-analysis
asked Dec 19 '18 at 16:07
0xbadf00d0xbadf00d
1,97041531
asked Dec 19 '18 at 16:07
0xbadf00d0xbadf00d
1,97041531
asked Dec 19 '18 at 16:07
0xbadf00d0xbadf00d
1,97041531
asked Dec 19 '18 at 16:07
0xbadf00d0xbadf00d
1,97041531
asked Dec 19 '18 at 16:07
0xbadf00d0xbadf00d
1,97041531
1,97041531
$begingroup$
(1) is enough to get a martingale. If you want e.g. an $L^2$-bounded martingale then you will need (2).
$endgroup$
– saz
Dec 19 '18 at 16:31
$begingroup$
@saz So, $(2)$ if we want, for example, define $(Xcdot M)_infty$.
$endgroup$
– 0xbadf00d
Dec 19 '18 at 18:07
$begingroup$
Yes, exactly...
$endgroup$
– saz
Dec 20 '18 at 6:03
$begingroup$
@saz Could you please take a look at this question: math.stackexchange.com/questions/3048457/…
$endgroup$
– 0xbadf00d
Dec 21 '18 at 18:05
add a comment |
$begingroup$
(1) is enough to get a martingale. If you want e.g. an $L^2$-bounded martingale then you will need (2).
$endgroup$
– saz
Dec 19 '18 at 16:31
$begingroup$
@saz So, $(2)$ if we want, for example, define $(Xcdot M)_infty$.
$endgroup$
– 0xbadf00d
Dec 19 '18 at 18:07
$begingroup$
Yes, exactly...
$endgroup$
– saz
Dec 20 '18 at 6:03
$begingroup$
@saz Could you please take a look at this question: math.stackexchange.com/questions/3048457/…
$endgroup$
– 0xbadf00d
Dec 21 '18 at 18:05
$begingroup$
(1) is enough to get a martingale. If you want e.g. an $L^2$-bounded martingale then you will need (2).
$endgroup$
– saz
Dec 19 '18 at 16:31
$begingroup$
(1) is enough to get a martingale. If you want e.g. an $L^2$-bounded martingale then you will need (2).
$endgroup$
– saz
Dec 19 '18 at 16:31
$begingroup$
@saz So, $(2)$ if we want, for example, define $(Xcdot M)_infty$.
$endgroup$
– 0xbadf00d
Dec 19 '18 at 18:07
$begingroup$
@saz So, $(2)$ if we want, for example, define $(Xcdot M)_infty$.
$endgroup$
– 0xbadf00d
Dec 19 '18 at 18:07
$begingroup$
Yes, exactly...
$endgroup$
– saz
Dec 20 '18 at 6:03
$begingroup$
Yes, exactly...
$endgroup$
– saz
Dec 20 '18 at 6:03
$begingroup$
@saz Could you please take a look at this question: math.stackexchange.com/questions/3048457/…
$endgroup$
– 0xbadf00d
Dec 21 '18 at 18:05
$begingroup$
@saz Could you please take a look at this question: math.stackexchange.com/questions/3048457/…
$endgroup$
– 0xbadf00d
Dec 21 '18 at 18:05
add a comment |
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$begingroup$
(1) is enough to get a martingale. If you want e.g. an $L^2$-bounded martingale then you will need (2).
$endgroup$
– saz
Dec 19 '18 at 16:31
$begingroup$
@saz So, $(2)$ if we want, for example, define $(Xcdot M)_infty$.
$endgroup$
– 0xbadf00d
Dec 19 '18 at 18:07
$begingroup$
Yes, exactly...
$endgroup$
– saz
Dec 20 '18 at 6:03
$begingroup$
@saz Could you please take a look at this question: math.stackexchange.com/questions/3048457/…
$endgroup$
– 0xbadf00d
Dec 21 '18 at 18:05