adapted, increasing, (locally) integrable variation process is a (local) submartingale
$begingroup$
I read a theorem that an adapted, increasing, (locally) integrable integrable, variation process is a (local) submartingale. (here increasing includes right continuity).
Definition: A process is $textbf{increasing}$ if each sample path is increasing and right continuous
Definition: A process has $textbf{finite variation}$ if it is the difference of two increasing processes
Definition: Let $X$ be a process with finite variation. We define the variation process $V_X$ by setting $V_X(t,omega)$ to be the variation of $X(cdot, omega)$ on $[0,t]$
Definition: A process is said to be $textbf{integrable}$ if $sup_{tgeq 0 } mathbb{E}|X_t|<infty$
Note: For an increasing, nonnegative process $X$ is integrable is equivalent to $X_{infty}$ existing a.s. and being an integrable random variable.
Definition: A finite variation process $X$ is said to have $textbf{integrable variation}$ if $V_X$ is integrable.
Definiton: A finite variation process $X$ is said to have $textbf{locally integrable variation}$ if there exists a sequence of stopping times $(T_n)_{n in mathbb{N}}$ such that $T_n nearrow infty$ a.s. and for each $n in mathbb{N}$, $X^{T_n}$ has integrable variation
Put $A$ to be our increasing process with locally integrable variation. So I know the adapted condition is satisfied, as is the increasing conditional expectation condition. But I cannot see how locally integrable variation implies for each $t geq 0$, $mathbb{E}|A_t| < infty$. The proof in the book (Klebner) is short but mentions (regardless of integrable or locally integrable variation) that it has to do with the localizing sequence. Since I'm new to localized properties, could someone explain?
From assumption, we can choose our localizing sequence $(T_n)_{n in mathbb{N}}$ such that $T_n nearrow infty$ a.s. and for each $n in mathbb{N}$, $sup_{t geq 0} mathbb{E} [{V_A}_{t wedge T_n}] < infty$
probability-theory stochastic-processes stochastic-calculus conditional-probability stochastic-analysis
asked Jan 16 at 7:34
CeeersonCeeerson
766
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|
show 2 more comments
$begingroup$
I read a theorem that an adapted, increasing, (locally) integrable integrable, variation process is a (local) submartingale. (here increasing includes right continuity).
Definition: A process is $textbf{increasing}$ if each sample path is increasing and right continuous
Definition: A process has $textbf{finite variation}$ if it is the difference of two increasing processes
Definition: Let $X$ be a process with finite variation. We define the variation process $V_X$ by setting $V_X(t,omega)$ to be the variation of $X(cdot, omega)$ on $[0,t]$
Definition: A process is said to be $textbf{integrable}$ if $sup_{tgeq 0 } mathbb{E}|X_t|<infty$
Note: For an increasing, nonnegative process $X$ is integrable is equivalent to $X_{infty}$ existing a.s. and being an integrable random variable.
Definition: A finite variation process $X$ is said to have $textbf{integrable variation}$ if $V_X$ is integrable.
Definiton: A finite variation process $X$ is said to have $textbf{locally integrable variation}$ if there exists a sequence of stopping times $(T_n)_{n in mathbb{N}}$ such that $T_n nearrow infty$ a.s. and for each $n in mathbb{N}$, $X^{T_n}$ has integrable variation
Put $A$ to be our increasing process with locally integrable variation. So I know the adapted condition is satisfied, as is the increasing conditional expectation condition. But I cannot see how locally integrable variation implies for each $t geq 0$, $mathbb{E}|A_t| < infty$. The proof in the book (Klebner) is short but mentions (regardless of integrable or locally integrable variation) that it has to do with the localizing sequence. Since I'm new to localized properties, could someone explain?
From assumption, we can choose our localizing sequence $(T_n)_{n in mathbb{N}}$ such that $T_n nearrow infty$ a.s. and for each $n in mathbb{N}$, $sup_{t geq 0} mathbb{E} [{V_A}_{t wedge T_n}] < infty$
probability-theory stochastic-processes stochastic-calculus conditional-probability stochastic-analysis
asked Jan 16 at 7:34
CeeersonCeeerson
766
$endgroup$
$begingroup$
What exactly do you mean by locally integrable variation? Could you add the definition?
$endgroup$
– saz
Jan 16 at 7:56
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Oh yeah, I'll edit that in
$endgroup$
– Ceeerson
Jan 16 at 8:24
$begingroup$
You added a lot of definitions, but as far as I can see you didn't define "locally integrable variation"? Do you define "locally integrable" via a localizing sequence of stopping times?
$endgroup$
– saz
Jan 16 at 8:49
$begingroup$
exactly, I'll add that in too
$endgroup$
– Ceeerson
Jan 16 at 8:53
1
$begingroup$
Well, if the variation process is integrable, then the assertion is a direct consequence of the fact that $$|X_t-X_0| leq V_X(t).$$ (Actually, they are equal because of the monotonicity of the sample paths.) However, I currently don't see why the assertion should be true if the variation process is only locally integrable.
$endgroup$
– saz
Jan 16 at 16:07
|
show 2 more comments
$begingroup$
I read a theorem that an adapted, increasing, (locally) integrable integrable, variation process is a (local) submartingale. (here increasing includes right continuity).
Definition: A process is $textbf{increasing}$ if each sample path is increasing and right continuous
Definition: A process has $textbf{finite variation}$ if it is the difference of two increasing processes
Definition: Let $X$ be a process with finite variation. We define the variation process $V_X$ by setting $V_X(t,omega)$ to be the variation of $X(cdot, omega)$ on $[0,t]$
Definition: A process is said to be $textbf{integrable}$ if $sup_{tgeq 0 } mathbb{E}|X_t|<infty$
Note: For an increasing, nonnegative process $X$ is integrable is equivalent to $X_{infty}$ existing a.s. and being an integrable random variable.
Definition: A finite variation process $X$ is said to have $textbf{integrable variation}$ if $V_X$ is integrable.
Definiton: A finite variation process $X$ is said to have $textbf{locally integrable variation}$ if there exists a sequence of stopping times $(T_n)_{n in mathbb{N}}$ such that $T_n nearrow infty$ a.s. and for each $n in mathbb{N}$, $X^{T_n}$ has integrable variation
Put $A$ to be our increasing process with locally integrable variation. So I know the adapted condition is satisfied, as is the increasing conditional expectation condition. But I cannot see how locally integrable variation implies for each $t geq 0$, $mathbb{E}|A_t| < infty$. The proof in the book (Klebner) is short but mentions (regardless of integrable or locally integrable variation) that it has to do with the localizing sequence. Since I'm new to localized properties, could someone explain?
From assumption, we can choose our localizing sequence $(T_n)_{n in mathbb{N}}$ such that $T_n nearrow infty$ a.s. and for each $n in mathbb{N}$, $sup_{t geq 0} mathbb{E} [{V_A}_{t wedge T_n}] < infty$
probability-theory stochastic-processes stochastic-calculus conditional-probability stochastic-analysis
asked Jan 16 at 7:34
CeeersonCeeerson
766
$endgroup$
I read a theorem that an adapted, increasing, (locally) integrable integrable, variation process is a (local) submartingale. (here increasing includes right continuity).
Definition: A process is $textbf{increasing}$ if each sample path is increasing and right continuous
Definition: A process has $textbf{finite variation}$ if it is the difference of two increasing processes
Definition: Let $X$ be a process with finite variation. We define the variation process $V_X$ by setting $V_X(t,omega)$ to be the variation of $X(cdot, omega)$ on $[0,t]$
Definition: A process is said to be $textbf{integrable}$ if $sup_{tgeq 0 } mathbb{E}|X_t|<infty$
Note: For an increasing, nonnegative process $X$ is integrable is equivalent to $X_{infty}$ existing a.s. and being an integrable random variable.
Definition: A finite variation process $X$ is said to have $textbf{integrable variation}$ if $V_X$ is integrable.
Definiton: A finite variation process $X$ is said to have $textbf{locally integrable variation}$ if there exists a sequence of stopping times $(T_n)_{n in mathbb{N}}$ such that $T_n nearrow infty$ a.s. and for each $n in mathbb{N}$, $X^{T_n}$ has integrable variation
Put $A$ to be our increasing process with locally integrable variation. So I know the adapted condition is satisfied, as is the increasing conditional expectation condition. But I cannot see how locally integrable variation implies for each $t geq 0$, $mathbb{E}|A_t| < infty$. The proof in the book (Klebner) is short but mentions (regardless of integrable or locally integrable variation) that it has to do with the localizing sequence. Since I'm new to localized properties, could someone explain?
From assumption, we can choose our localizing sequence $(T_n)_{n in mathbb{N}}$ such that $T_n nearrow infty$ a.s. and for each $n in mathbb{N}$, $sup_{t geq 0} mathbb{E} [{V_A}_{t wedge T_n}] < infty$
probability-theory stochastic-processes stochastic-calculus conditional-probability stochastic-analysis
probability-theory stochastic-processes stochastic-calculus conditional-probability stochastic-analysis
asked Jan 16 at 7:34
CeeersonCeeerson
766
asked Jan 16 at 7:34
CeeersonCeeerson
766
edited Jan 16 at 19:49
Ceeerson
asked Jan 16 at 7:34
CeeersonCeeerson
766
asked Jan 16 at 7:34
CeeersonCeeerson
766
asked Jan 16 at 7:34
CeeersonCeeerson
766
766
$begingroup$
What exactly do you mean by locally integrable variation? Could you add the definition?
$endgroup$
– saz
Jan 16 at 7:56
$begingroup$
Oh yeah, I'll edit that in
$endgroup$
– Ceeerson
Jan 16 at 8:24
$begingroup$
You added a lot of definitions, but as far as I can see you didn't define "locally integrable variation"? Do you define "locally integrable" via a localizing sequence of stopping times?
$endgroup$
– saz
Jan 16 at 8:49
$begingroup$
exactly, I'll add that in too
$endgroup$
– Ceeerson
Jan 16 at 8:53
1
$begingroup$
Well, if the variation process is integrable, then the assertion is a direct consequence of the fact that $$|X_t-X_0| leq V_X(t).$$ (Actually, they are equal because of the monotonicity of the sample paths.) However, I currently don't see why the assertion should be true if the variation process is only locally integrable.
$endgroup$
– saz
Jan 16 at 16:07
|
show 2 more comments
$begingroup$
What exactly do you mean by locally integrable variation? Could you add the definition?
$endgroup$
– saz
Jan 16 at 7:56
$begingroup$
Oh yeah, I'll edit that in
$endgroup$
– Ceeerson
Jan 16 at 8:24
$begingroup$
You added a lot of definitions, but as far as I can see you didn't define "locally integrable variation"? Do you define "locally integrable" via a localizing sequence of stopping times?
$endgroup$
– saz
Jan 16 at 8:49
$begingroup$
exactly, I'll add that in too
$endgroup$
– Ceeerson
Jan 16 at 8:53
1
$begingroup$
Well, if the variation process is integrable, then the assertion is a direct consequence of the fact that $$|X_t-X_0| leq V_X(t).$$ (Actually, they are equal because of the monotonicity of the sample paths.) However, I currently don't see why the assertion should be true if the variation process is only locally integrable.
$endgroup$
– saz
Jan 16 at 16:07
$begingroup$
What exactly do you mean by locally integrable variation? Could you add the definition?
$endgroup$
– saz
Jan 16 at 7:56
$begingroup$
What exactly do you mean by locally integrable variation? Could you add the definition?
$endgroup$
– saz
Jan 16 at 7:56
$begingroup$
Oh yeah, I'll edit that in
$endgroup$
– Ceeerson
Jan 16 at 8:24
$begingroup$
Oh yeah, I'll edit that in
$endgroup$
– Ceeerson
Jan 16 at 8:24
$begingroup$
You added a lot of definitions, but as far as I can see you didn't define "locally integrable variation"? Do you define "locally integrable" via a localizing sequence of stopping times?
$endgroup$
– saz
Jan 16 at 8:49
$begingroup$
You added a lot of definitions, but as far as I can see you didn't define "locally integrable variation"? Do you define "locally integrable" via a localizing sequence of stopping times?
$endgroup$
– saz
Jan 16 at 8:49
$begingroup$
exactly, I'll add that in too
$endgroup$
– Ceeerson
Jan 16 at 8:53
$begingroup$
exactly, I'll add that in too
$endgroup$
– Ceeerson
Jan 16 at 8:53
1
1
$begingroup$
Well, if the variation process is integrable, then the assertion is a direct consequence of the fact that $$|X_t-X_0| leq V_X(t).$$ (Actually, they are equal because of the monotonicity of the sample paths.) However, I currently don't see why the assertion should be true if the variation process is only locally integrable.
$endgroup$
– saz
Jan 16 at 16:07
$begingroup$
Well, if the variation process is integrable, then the assertion is a direct consequence of the fact that $$|X_t-X_0| leq V_X(t).$$ (Actually, they are equal because of the monotonicity of the sample paths.) However, I currently don't see why the assertion should be true if the variation process is only locally integrable.
$endgroup$
– saz
Jan 16 at 16:07
|
show 2 more comments
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$begingroup$
What exactly do you mean by locally integrable variation? Could you add the definition?
$endgroup$
– saz
Jan 16 at 7:56
$begingroup$
Oh yeah, I'll edit that in
$endgroup$
– Ceeerson
Jan 16 at 8:24
$begingroup$
You added a lot of definitions, but as far as I can see you didn't define "locally integrable variation"? Do you define "locally integrable" via a localizing sequence of stopping times?
$endgroup$
– saz
Jan 16 at 8:49
$begingroup$
exactly, I'll add that in too
$endgroup$
– Ceeerson
Jan 16 at 8:53
1
$begingroup$
Well, if the variation process is integrable, then the assertion is a direct consequence of the fact that $$|X_t-X_0| leq V_X(t).$$ (Actually, they are equal because of the monotonicity of the sample paths.) However, I currently don't see why the assertion should be true if the variation process is only locally integrable.
$endgroup$
– saz
Jan 16 at 16:07