Moment Estimate for Simple Linear SDE
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In a paper I am reading, it is claimed that the SDE
$$
dX_t = b X_t dt + (sigma X_t + beta_t) dB_t , quad t in [0,T],$$
satisfies
$$
Eleft[ sup_{t in [0,T]} |X_t|^p right] < infty, forall p in [2,infty),
$$
where $b, sigma in mathbb{R}$ are constants and $beta$ is a process satisfying
$$
E left[ int_0^T | beta_t |^2 dt right] < infty .
$$
Is this true?
sde
asked Jan 6 at 12:15
WhiteWhite
989
$endgroup$
add a comment |
$begingroup$
In a paper I am reading, it is claimed that the SDE
$$
dX_t = b X_t dt + (sigma X_t + beta_t) dB_t , quad t in [0,T],$$
satisfies
$$
Eleft[ sup_{t in [0,T]} |X_t|^p right] < infty, forall p in [2,infty),
$$
where $b, sigma in mathbb{R}$ are constants and $beta$ is a process satisfying
$$
E left[ int_0^T | beta_t |^2 dt right] < infty .
$$
Is this true?
sde
asked Jan 6 at 12:15
WhiteWhite
989
$endgroup$
$begingroup$
What is the definition of $X_t$ ?
$endgroup$
– user617446
Jan 6 at 12:16
1
$begingroup$
Without additional assumptions on $beta$ this is not true. Consider for instance $b=0$, $sigma=0$ and $beta_t := Z$ for some random variable $Z in L^2(mathbb{P})$ which is independent of $(B_t)_t$ and which fails to have a finite moment of order $p>2$ (i.e. $mathbb{E}(|Z|^p)=infty$)
$endgroup$
– saz
Jan 6 at 12:48
$begingroup$
that's what I thought, thank you.
$endgroup$
– White
Jan 6 at 12:53
add a comment |
$begingroup$
In a paper I am reading, it is claimed that the SDE
$$
dX_t = b X_t dt + (sigma X_t + beta_t) dB_t , quad t in [0,T],$$
satisfies
$$
Eleft[ sup_{t in [0,T]} |X_t|^p right] < infty, forall p in [2,infty),
$$
where $b, sigma in mathbb{R}$ are constants and $beta$ is a process satisfying
$$
E left[ int_0^T | beta_t |^2 dt right] < infty .
$$
Is this true?
sde
asked Jan 6 at 12:15
WhiteWhite
989
$endgroup$
In a paper I am reading, it is claimed that the SDE
$$
dX_t = b X_t dt + (sigma X_t + beta_t) dB_t , quad t in [0,T],$$
satisfies
$$
Eleft[ sup_{t in [0,T]} |X_t|^p right] < infty, forall p in [2,infty),
$$
where $b, sigma in mathbb{R}$ are constants and $beta$ is a process satisfying
$$
E left[ int_0^T | beta_t |^2 dt right] < infty .
$$
Is this true?
sde
sde
asked Jan 6 at 12:15
WhiteWhite
989
asked Jan 6 at 12:15
WhiteWhite
989
edited Jan 6 at 12:21
White
asked Jan 6 at 12:15
WhiteWhite
989
asked Jan 6 at 12:15
WhiteWhite
989
asked Jan 6 at 12:15
WhiteWhite
989
989
$begingroup$
What is the definition of $X_t$ ?
$endgroup$
– user617446
Jan 6 at 12:16
1
$begingroup$
Without additional assumptions on $beta$ this is not true. Consider for instance $b=0$, $sigma=0$ and $beta_t := Z$ for some random variable $Z in L^2(mathbb{P})$ which is independent of $(B_t)_t$ and which fails to have a finite moment of order $p>2$ (i.e. $mathbb{E}(|Z|^p)=infty$)
$endgroup$
– saz
Jan 6 at 12:48
$begingroup$
that's what I thought, thank you.
$endgroup$
– White
Jan 6 at 12:53
add a comment |
$begingroup$
What is the definition of $X_t$ ?
$endgroup$
– user617446
Jan 6 at 12:16
1
$begingroup$
Without additional assumptions on $beta$ this is not true. Consider for instance $b=0$, $sigma=0$ and $beta_t := Z$ for some random variable $Z in L^2(mathbb{P})$ which is independent of $(B_t)_t$ and which fails to have a finite moment of order $p>2$ (i.e. $mathbb{E}(|Z|^p)=infty$)
$endgroup$
– saz
Jan 6 at 12:48
$begingroup$
that's what I thought, thank you.
$endgroup$
– White
Jan 6 at 12:53
$begingroup$
What is the definition of $X_t$ ?
$endgroup$
– user617446
Jan 6 at 12:16
$begingroup$
What is the definition of $X_t$ ?
$endgroup$
– user617446
Jan 6 at 12:16
1
1
$begingroup$
Without additional assumptions on $beta$ this is not true. Consider for instance $b=0$, $sigma=0$ and $beta_t := Z$ for some random variable $Z in L^2(mathbb{P})$ which is independent of $(B_t)_t$ and which fails to have a finite moment of order $p>2$ (i.e. $mathbb{E}(|Z|^p)=infty$)
$endgroup$
– saz
Jan 6 at 12:48
$begingroup$
Without additional assumptions on $beta$ this is not true. Consider for instance $b=0$, $sigma=0$ and $beta_t := Z$ for some random variable $Z in L^2(mathbb{P})$ which is independent of $(B_t)_t$ and which fails to have a finite moment of order $p>2$ (i.e. $mathbb{E}(|Z|^p)=infty$)
$endgroup$
– saz
Jan 6 at 12:48
$begingroup$
that's what I thought, thank you.
$endgroup$
– White
Jan 6 at 12:53
$begingroup$
that's what I thought, thank you.
$endgroup$
– White
Jan 6 at 12:53
add a comment |
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$begingroup$
What is the definition of $X_t$ ?
$endgroup$
– user617446
Jan 6 at 12:16
1
$begingroup$
Without additional assumptions on $beta$ this is not true. Consider for instance $b=0$, $sigma=0$ and $beta_t := Z$ for some random variable $Z in L^2(mathbb{P})$ which is independent of $(B_t)_t$ and which fails to have a finite moment of order $p>2$ (i.e. $mathbb{E}(|Z|^p)=infty$)
$endgroup$
– saz
Jan 6 at 12:48
$begingroup$
that's what I thought, thank you.
$endgroup$
– White
Jan 6 at 12:53