When is this stochastic integral a martingale
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Let $B = (B_t)_{t in [0,T} $ a Brownian motion.
Let $H = (H_t)_{t in [0,T}, G = (G_t)_{t in [0,T} $ be progressively measurable.
What are the conditions on $H,G$ such that $(int_0^t H_s G_s dB_s)_{t in [0,T]} $ is a true martingale?
From what I know we should have
$$
E left[ int_0^T mid H_s G_s mid^2 ; ds right] < infty.
$$
This in turn would be guaranteed if
$$
E left[ int_0^T |H_s|^4 + |G_s|^4 ; ds right] < infty.
$$
Is it possible to have a weaker condition to get a martingale?
Edit: Is it enough to ask for
$$
int_0^T | H_s G_s |^2 ; ds < infty, quad P-a.s. ?
$$
Edit: I was missing some information. $G$ is $dtotimes P$ square integrable and $E[ sup_{t in [0,T] } |H_s|^2 ] < infty. $ Then
$$
Eleft[ left(int_0^T | H_s G_s |^2 ; ds right)^{frac{1}{2}} right] < infty.
$$
Thus, $ int_0^cdot H_s G_s ds $ is a true martingale by BDG-inequality.
stochastic-calculus stochastic-integrals
asked Nov 14 at 18:51
White
789
add a comment |
up vote
0
down vote
favorite
Let $B = (B_t)_{t in [0,T} $ a Brownian motion.
Let $H = (H_t)_{t in [0,T}, G = (G_t)_{t in [0,T} $ be progressively measurable.
What are the conditions on $H,G$ such that $(int_0^t H_s G_s dB_s)_{t in [0,T]} $ is a true martingale?
From what I know we should have
$$
E left[ int_0^T mid H_s G_s mid^2 ; ds right] < infty.
$$
This in turn would be guaranteed if
$$
E left[ int_0^T |H_s|^4 + |G_s|^4 ; ds right] < infty.
$$
Is it possible to have a weaker condition to get a martingale?
Edit: Is it enough to ask for
$$
int_0^T | H_s G_s |^2 ; ds < infty, quad P-a.s. ?
$$
Edit: I was missing some information. $G$ is $dtotimes P$ square integrable and $E[ sup_{t in [0,T] } |H_s|^2 ] < infty. $ Then
$$
Eleft[ left(int_0^T | H_s G_s |^2 ; ds right)^{frac{1}{2}} right] < infty.
$$
Thus, $ int_0^cdot H_s G_s ds $ is a true martingale by BDG-inequality.
stochastic-calculus stochastic-integrals
asked Nov 14 at 18:51
White
789
Do you have a particular application in mind or are you just asking out of curiousity?
– saz
Nov 14 at 19:57
Im reading a paper. They claim that this particular stochastic integral is a martingale.
– White
Nov 15 at 11:25
1
I see; I would say it's not enough to get a (true) martingale... it's only going to be a local martingale.
– saz
Nov 15 at 11:41
Thank you, I missed out on some information, see the edit. I think it works now. Have a nice day.
– White
Nov 15 at 12:02
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $B = (B_t)_{t in [0,T} $ a Brownian motion.
Let $H = (H_t)_{t in [0,T}, G = (G_t)_{t in [0,T} $ be progressively measurable.
What are the conditions on $H,G$ such that $(int_0^t H_s G_s dB_s)_{t in [0,T]} $ is a true martingale?
From what I know we should have
$$
E left[ int_0^T mid H_s G_s mid^2 ; ds right] < infty.
$$
This in turn would be guaranteed if
$$
E left[ int_0^T |H_s|^4 + |G_s|^4 ; ds right] < infty.
$$
Is it possible to have a weaker condition to get a martingale?
Edit: Is it enough to ask for
$$
int_0^T | H_s G_s |^2 ; ds < infty, quad P-a.s. ?
$$
Edit: I was missing some information. $G$ is $dtotimes P$ square integrable and $E[ sup_{t in [0,T] } |H_s|^2 ] < infty. $ Then
$$
Eleft[ left(int_0^T | H_s G_s |^2 ; ds right)^{frac{1}{2}} right] < infty.
$$
Thus, $ int_0^cdot H_s G_s ds $ is a true martingale by BDG-inequality.
stochastic-calculus stochastic-integrals
asked Nov 14 at 18:51
White
789
Let $B = (B_t)_{t in [0,T} $ a Brownian motion.
Let $H = (H_t)_{t in [0,T}, G = (G_t)_{t in [0,T} $ be progressively measurable.
What are the conditions on $H,G$ such that $(int_0^t H_s G_s dB_s)_{t in [0,T]} $ is a true martingale?
From what I know we should have
$$
E left[ int_0^T mid H_s G_s mid^2 ; ds right] < infty.
$$
This in turn would be guaranteed if
$$
E left[ int_0^T |H_s|^4 + |G_s|^4 ; ds right] < infty.
$$
Is it possible to have a weaker condition to get a martingale?
Edit: Is it enough to ask for
$$
int_0^T | H_s G_s |^2 ; ds < infty, quad P-a.s. ?
$$
Edit: I was missing some information. $G$ is $dtotimes P$ square integrable and $E[ sup_{t in [0,T] } |H_s|^2 ] < infty. $ Then
$$
Eleft[ left(int_0^T | H_s G_s |^2 ; ds right)^{frac{1}{2}} right] < infty.
$$
Thus, $ int_0^cdot H_s G_s ds $ is a true martingale by BDG-inequality.
stochastic-calculus stochastic-integrals
stochastic-calculus stochastic-integrals
asked Nov 14 at 18:51
White
789
asked Nov 14 at 18:51
White
789
edited Nov 15 at 12:01
asked Nov 14 at 18:51
White
789
asked Nov 14 at 18:51
White
789
asked Nov 14 at 18:51
White
789
789
Do you have a particular application in mind or are you just asking out of curiousity?
– saz
Nov 14 at 19:57
Im reading a paper. They claim that this particular stochastic integral is a martingale.
– White
Nov 15 at 11:25
1
I see; I would say it's not enough to get a (true) martingale... it's only going to be a local martingale.
– saz
Nov 15 at 11:41
Thank you, I missed out on some information, see the edit. I think it works now. Have a nice day.
– White
Nov 15 at 12:02
add a comment |
Do you have a particular application in mind or are you just asking out of curiousity?
– saz
Nov 14 at 19:57
Im reading a paper. They claim that this particular stochastic integral is a martingale.
– White
Nov 15 at 11:25
1
I see; I would say it's not enough to get a (true) martingale... it's only going to be a local martingale.
– saz
Nov 15 at 11:41
Thank you, I missed out on some information, see the edit. I think it works now. Have a nice day.
– White
Nov 15 at 12:02
Do you have a particular application in mind or are you just asking out of curiousity?
– saz
Nov 14 at 19:57
Do you have a particular application in mind or are you just asking out of curiousity?
– saz
Nov 14 at 19:57
Im reading a paper. They claim that this particular stochastic integral is a martingale.
– White
Nov 15 at 11:25
Im reading a paper. They claim that this particular stochastic integral is a martingale.
– White
Nov 15 at 11:25
1
1
I see; I would say it's not enough to get a (true) martingale... it's only going to be a local martingale.
– saz
Nov 15 at 11:41
I see; I would say it's not enough to get a (true) martingale... it's only going to be a local martingale.
– saz
Nov 15 at 11:41
Thank you, I missed out on some information, see the edit. I think it works now. Have a nice day.
– White
Nov 15 at 12:02
Thank you, I missed out on some information, see the edit. I think it works now. Have a nice day.
– White
Nov 15 at 12:02
add a comment |
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Do you have a particular application in mind or are you just asking out of curiousity?
– saz
Nov 14 at 19:57
Im reading a paper. They claim that this particular stochastic integral is a martingale.
– White
Nov 15 at 11:25
1
I see; I would say it's not enough to get a (true) martingale... it's only going to be a local martingale.
– saz
Nov 15 at 11:41
Thank you, I missed out on some information, see the edit. I think it works now. Have a nice day.
– White
Nov 15 at 12:02