Finiteness of the number of big jumps of a Lévy process on a finite interval
$begingroup$
Let $Y(t)$, $0 leq t leq 1$, be a Lévy process. Denote by ${ Delta Y_i }$ the jumps of the process. I would like to show that the set ${i: |Delta Y_i| > r}$ is finite almost surely. However, I do not entirely understand what should be shown and how to approach this problem. I would be grateful for any help.
Thank you.
Regards,
Ivan
probability-theory levy-processes
asked Nov 25 '14 at 13:38
IvanIvan
548516
$endgroup$
add a comment |
$begingroup$
Let $Y(t)$, $0 leq t leq 1$, be a Lévy process. Denote by ${ Delta Y_i }$ the jumps of the process. I would like to show that the set ${i: |Delta Y_i| > r}$ is finite almost surely. However, I do not entirely understand what should be shown and how to approach this problem. I would be grateful for any help.
Thank you.
Regards,
Ivan
probability-theory levy-processes
asked Nov 25 '14 at 13:38
IvanIvan
548516
$endgroup$
add a comment |
$begingroup$
Let $Y(t)$, $0 leq t leq 1$, be a Lévy process. Denote by ${ Delta Y_i }$ the jumps of the process. I would like to show that the set ${i: |Delta Y_i| > r}$ is finite almost surely. However, I do not entirely understand what should be shown and how to approach this problem. I would be grateful for any help.
Thank you.
Regards,
Ivan
probability-theory levy-processes
asked Nov 25 '14 at 13:38
IvanIvan
548516
$endgroup$
Let $Y(t)$, $0 leq t leq 1$, be a Lévy process. Denote by ${ Delta Y_i }$ the jumps of the process. I would like to show that the set ${i: |Delta Y_i| > r}$ is finite almost surely. However, I do not entirely understand what should be shown and how to approach this problem. I would be grateful for any help.
Thank you.
Regards,
Ivan
probability-theory levy-processes
probability-theory levy-processes
asked Nov 25 '14 at 13:38
IvanIvan
548516
asked Nov 25 '14 at 13:38
IvanIvan
548516
edited Nov 25 '14 at 15:26
Ivan
asked Nov 25 '14 at 13:38
IvanIvan
548516
asked Nov 25 '14 at 13:38
IvanIvan
548516
asked Nov 25 '14 at 13:38
IvanIvan
548516
548516
add a comment |
add a comment |
1 Answer
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$begingroup$
Let $(Y_t)_{t in [0,1]}$ be a Lévy process and denote by $Delta Y_t := Y_t-Y_{t-}$ the jump height at time $t$. The aim is to show that $(Y_t)_{t in [0,1]}$ has only finitely many jumps with jump size larger than $r$, i.e. that
$${t in [0,1]; |Delta Y_t(omega)|>r}$$
is a finite set for (almost) all $omega in Omega$. This follows from the following lemma (applied to the sample paths $[0,1] ni t mapsto Y_t(omega)$ for fixed $omega in Omega$):
Lemma Let $f:[0,1] to mathbb{R}$ be a càdlàg function (i.e. right-continuous with left-hand limits). Then the set $${t in [0,1]; |Delta f(t)|>r}$$ is finite for each $r>0$.
Proof:
Suppose that the set is not finite. Then we can choose a sequence $(t_n)_{n in mathbb{N}} subseteq [0,1]$ such that $|Delta f(t_n)|>r$ for each $n in mathbb{N}$. Since $[0,1]$ is (sequentially) compact, we can pick a subsequence $(t_{n(k)})_{k in mathbb{N}}$ such that $t_{n(k)} to s$ for some $s in [0,1]$.
Since $f$ is right-continuous at $s$, there exists $delta_1>0$ such that $$|f(s)-f(t)| < frac{r}{4}$$ for any $t in [s,s+delta_1]$. This means in particular that there cannot exist $t in [s,s+delta_1]$ such that $|Delta f(t)|>r$. Similarly, one can show that the existence of the left-limit implies the existence of $delta_2>0$ such that $|Delta f(t)| leq r$ for all $t in [s-delta_2,s)$. Obviously, this contradicts the first step of our proof where we constructed a sequence converging to $s$ with jumps heights $>r$.
answered Nov 25 '14 at 15:10
sazsaz
78.8k858123
$endgroup$
$begingroup$
Thank you for the answer. What is left to do is to mention that every Lévy process has a version whose paths are càdlàg, right?
$endgroup$
– Ivan
Nov 25 '14 at 19:26
$begingroup$
@Ivan Usually, a Lévy process has by definition (almost surely) càdlàg sample paths, but obviously this depends on the definition you are using.
$endgroup$
– saz
Nov 25 '14 at 19:27
$begingroup$
Sorry for necroing this old answer, but in a first step, we only obtain the existence of $delta_2>0$ such that $|f(t)-f(s-)|le r$ for all $tin[s-delta_2,s]$ by the existence of the left-limit at $t$. How do you proceed?
$endgroup$
– 0xbadf00d
Dec 1 '18 at 17:20
1
$begingroup$
@0xbadf00d If $|f(t)-f(s-)| leq r/4$ for $t in [s-delta_2,s)$, then $|f(t)-f(u)| leq r/2$ for any $u,v in [s-delta_2,s)$, and this clearly implies $|Delta f(t)| leq r$ for any $r in [s-delta_2,s)$.
$endgroup$
– saz
Dec 1 '18 at 17:29
add a comment |
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$begingroup$
Let $(Y_t)_{t in [0,1]}$ be a Lévy process and denote by $Delta Y_t := Y_t-Y_{t-}$ the jump height at time $t$. The aim is to show that $(Y_t)_{t in [0,1]}$ has only finitely many jumps with jump size larger than $r$, i.e. that
$${t in [0,1]; |Delta Y_t(omega)|>r}$$
is a finite set for (almost) all $omega in Omega$. This follows from the following lemma (applied to the sample paths $[0,1] ni t mapsto Y_t(omega)$ for fixed $omega in Omega$):
Lemma Let $f:[0,1] to mathbb{R}$ be a càdlàg function (i.e. right-continuous with left-hand limits). Then the set $${t in [0,1]; |Delta f(t)|>r}$$ is finite for each $r>0$.
Proof:
Suppose that the set is not finite. Then we can choose a sequence $(t_n)_{n in mathbb{N}} subseteq [0,1]$ such that $|Delta f(t_n)|>r$ for each $n in mathbb{N}$. Since $[0,1]$ is (sequentially) compact, we can pick a subsequence $(t_{n(k)})_{k in mathbb{N}}$ such that $t_{n(k)} to s$ for some $s in [0,1]$.
Since $f$ is right-continuous at $s$, there exists $delta_1>0$ such that $$|f(s)-f(t)| < frac{r}{4}$$ for any $t in [s,s+delta_1]$. This means in particular that there cannot exist $t in [s,s+delta_1]$ such that $|Delta f(t)|>r$. Similarly, one can show that the existence of the left-limit implies the existence of $delta_2>0$ such that $|Delta f(t)| leq r$ for all $t in [s-delta_2,s)$. Obviously, this contradicts the first step of our proof where we constructed a sequence converging to $s$ with jumps heights $>r$.
answered Nov 25 '14 at 15:10
sazsaz
78.8k858123
$endgroup$
$begingroup$
Thank you for the answer. What is left to do is to mention that every Lévy process has a version whose paths are càdlàg, right?
$endgroup$
– Ivan
Nov 25 '14 at 19:26
$begingroup$
@Ivan Usually, a Lévy process has by definition (almost surely) càdlàg sample paths, but obviously this depends on the definition you are using.
$endgroup$
– saz
Nov 25 '14 at 19:27
$begingroup$
Sorry for necroing this old answer, but in a first step, we only obtain the existence of $delta_2>0$ such that $|f(t)-f(s-)|le r$ for all $tin[s-delta_2,s]$ by the existence of the left-limit at $t$. How do you proceed?
$endgroup$
– 0xbadf00d
Dec 1 '18 at 17:20
1
$begingroup$
@0xbadf00d If $|f(t)-f(s-)| leq r/4$ for $t in [s-delta_2,s)$, then $|f(t)-f(u)| leq r/2$ for any $u,v in [s-delta_2,s)$, and this clearly implies $|Delta f(t)| leq r$ for any $r in [s-delta_2,s)$.
$endgroup$
– saz
Dec 1 '18 at 17:29
add a comment |
$begingroup$
Let $(Y_t)_{t in [0,1]}$ be a Lévy process and denote by $Delta Y_t := Y_t-Y_{t-}$ the jump height at time $t$. The aim is to show that $(Y_t)_{t in [0,1]}$ has only finitely many jumps with jump size larger than $r$, i.e. that
$${t in [0,1]; |Delta Y_t(omega)|>r}$$
is a finite set for (almost) all $omega in Omega$. This follows from the following lemma (applied to the sample paths $[0,1] ni t mapsto Y_t(omega)$ for fixed $omega in Omega$):
Lemma Let $f:[0,1] to mathbb{R}$ be a càdlàg function (i.e. right-continuous with left-hand limits). Then the set $${t in [0,1]; |Delta f(t)|>r}$$ is finite for each $r>0$.
Proof:
Suppose that the set is not finite. Then we can choose a sequence $(t_n)_{n in mathbb{N}} subseteq [0,1]$ such that $|Delta f(t_n)|>r$ for each $n in mathbb{N}$. Since $[0,1]$ is (sequentially) compact, we can pick a subsequence $(t_{n(k)})_{k in mathbb{N}}$ such that $t_{n(k)} to s$ for some $s in [0,1]$.
Since $f$ is right-continuous at $s$, there exists $delta_1>0$ such that $$|f(s)-f(t)| < frac{r}{4}$$ for any $t in [s,s+delta_1]$. This means in particular that there cannot exist $t in [s,s+delta_1]$ such that $|Delta f(t)|>r$. Similarly, one can show that the existence of the left-limit implies the existence of $delta_2>0$ such that $|Delta f(t)| leq r$ for all $t in [s-delta_2,s)$. Obviously, this contradicts the first step of our proof where we constructed a sequence converging to $s$ with jumps heights $>r$.
answered Nov 25 '14 at 15:10
sazsaz
78.8k858123
$endgroup$
$begingroup$
Thank you for the answer. What is left to do is to mention that every Lévy process has a version whose paths are càdlàg, right?
$endgroup$
– Ivan
Nov 25 '14 at 19:26
$begingroup$
@Ivan Usually, a Lévy process has by definition (almost surely) càdlàg sample paths, but obviously this depends on the definition you are using.
$endgroup$
– saz
Nov 25 '14 at 19:27
$begingroup$
Sorry for necroing this old answer, but in a first step, we only obtain the existence of $delta_2>0$ such that $|f(t)-f(s-)|le r$ for all $tin[s-delta_2,s]$ by the existence of the left-limit at $t$. How do you proceed?
$endgroup$
– 0xbadf00d
Dec 1 '18 at 17:20
1
$begingroup$
@0xbadf00d If $|f(t)-f(s-)| leq r/4$ for $t in [s-delta_2,s)$, then $|f(t)-f(u)| leq r/2$ for any $u,v in [s-delta_2,s)$, and this clearly implies $|Delta f(t)| leq r$ for any $r in [s-delta_2,s)$.
$endgroup$
– saz
Dec 1 '18 at 17:29
add a comment |
$begingroup$
Let $(Y_t)_{t in [0,1]}$ be a Lévy process and denote by $Delta Y_t := Y_t-Y_{t-}$ the jump height at time $t$. The aim is to show that $(Y_t)_{t in [0,1]}$ has only finitely many jumps with jump size larger than $r$, i.e. that
$${t in [0,1]; |Delta Y_t(omega)|>r}$$
is a finite set for (almost) all $omega in Omega$. This follows from the following lemma (applied to the sample paths $[0,1] ni t mapsto Y_t(omega)$ for fixed $omega in Omega$):
Lemma Let $f:[0,1] to mathbb{R}$ be a càdlàg function (i.e. right-continuous with left-hand limits). Then the set $${t in [0,1]; |Delta f(t)|>r}$$ is finite for each $r>0$.
Proof:
Suppose that the set is not finite. Then we can choose a sequence $(t_n)_{n in mathbb{N}} subseteq [0,1]$ such that $|Delta f(t_n)|>r$ for each $n in mathbb{N}$. Since $[0,1]$ is (sequentially) compact, we can pick a subsequence $(t_{n(k)})_{k in mathbb{N}}$ such that $t_{n(k)} to s$ for some $s in [0,1]$.
Since $f$ is right-continuous at $s$, there exists $delta_1>0$ such that $$|f(s)-f(t)| < frac{r}{4}$$ for any $t in [s,s+delta_1]$. This means in particular that there cannot exist $t in [s,s+delta_1]$ such that $|Delta f(t)|>r$. Similarly, one can show that the existence of the left-limit implies the existence of $delta_2>0$ such that $|Delta f(t)| leq r$ for all $t in [s-delta_2,s)$. Obviously, this contradicts the first step of our proof where we constructed a sequence converging to $s$ with jumps heights $>r$.
answered Nov 25 '14 at 15:10
sazsaz
78.8k858123
$endgroup$
Let $(Y_t)_{t in [0,1]}$ be a Lévy process and denote by $Delta Y_t := Y_t-Y_{t-}$ the jump height at time $t$. The aim is to show that $(Y_t)_{t in [0,1]}$ has only finitely many jumps with jump size larger than $r$, i.e. that
$${t in [0,1]; |Delta Y_t(omega)|>r}$$
is a finite set for (almost) all $omega in Omega$. This follows from the following lemma (applied to the sample paths $[0,1] ni t mapsto Y_t(omega)$ for fixed $omega in Omega$):
Lemma Let $f:[0,1] to mathbb{R}$ be a càdlàg function (i.e. right-continuous with left-hand limits). Then the set $${t in [0,1]; |Delta f(t)|>r}$$ is finite for each $r>0$.
Proof:
Suppose that the set is not finite. Then we can choose a sequence $(t_n)_{n in mathbb{N}} subseteq [0,1]$ such that $|Delta f(t_n)|>r$ for each $n in mathbb{N}$. Since $[0,1]$ is (sequentially) compact, we can pick a subsequence $(t_{n(k)})_{k in mathbb{N}}$ such that $t_{n(k)} to s$ for some $s in [0,1]$.
Since $f$ is right-continuous at $s$, there exists $delta_1>0$ such that $$|f(s)-f(t)| < frac{r}{4}$$ for any $t in [s,s+delta_1]$. This means in particular that there cannot exist $t in [s,s+delta_1]$ such that $|Delta f(t)|>r$. Similarly, one can show that the existence of the left-limit implies the existence of $delta_2>0$ such that $|Delta f(t)| leq r$ for all $t in [s-delta_2,s)$. Obviously, this contradicts the first step of our proof where we constructed a sequence converging to $s$ with jumps heights $>r$.
answered Nov 25 '14 at 15:10
sazsaz
78.8k858123
edited Dec 1 '18 at 17:27
answered Nov 25 '14 at 15:10
sazsaz
78.8k858123
answered Nov 25 '14 at 15:10
sazsaz
78.8k858123
answered Nov 25 '14 at 15:10
sazsaz
78.8k858123
78.8k858123
$begingroup$
Thank you for the answer. What is left to do is to mention that every Lévy process has a version whose paths are càdlàg, right?
$endgroup$
– Ivan
Nov 25 '14 at 19:26
$begingroup$
@Ivan Usually, a Lévy process has by definition (almost surely) càdlàg sample paths, but obviously this depends on the definition you are using.
$endgroup$
– saz
Nov 25 '14 at 19:27
$begingroup$
Sorry for necroing this old answer, but in a first step, we only obtain the existence of $delta_2>0$ such that $|f(t)-f(s-)|le r$ for all $tin[s-delta_2,s]$ by the existence of the left-limit at $t$. How do you proceed?
$endgroup$
– 0xbadf00d
Dec 1 '18 at 17:20
1
$begingroup$
@0xbadf00d If $|f(t)-f(s-)| leq r/4$ for $t in [s-delta_2,s)$, then $|f(t)-f(u)| leq r/2$ for any $u,v in [s-delta_2,s)$, and this clearly implies $|Delta f(t)| leq r$ for any $r in [s-delta_2,s)$.
$endgroup$
– saz
Dec 1 '18 at 17:29
add a comment |
$begingroup$
Thank you for the answer. What is left to do is to mention that every Lévy process has a version whose paths are càdlàg, right?
$endgroup$
– Ivan
Nov 25 '14 at 19:26
$begingroup$
@Ivan Usually, a Lévy process has by definition (almost surely) càdlàg sample paths, but obviously this depends on the definition you are using.
$endgroup$
– saz
Nov 25 '14 at 19:27
$begingroup$
Sorry for necroing this old answer, but in a first step, we only obtain the existence of $delta_2>0$ such that $|f(t)-f(s-)|le r$ for all $tin[s-delta_2,s]$ by the existence of the left-limit at $t$. How do you proceed?
$endgroup$
– 0xbadf00d
Dec 1 '18 at 17:20
1
$begingroup$
@0xbadf00d If $|f(t)-f(s-)| leq r/4$ for $t in [s-delta_2,s)$, then $|f(t)-f(u)| leq r/2$ for any $u,v in [s-delta_2,s)$, and this clearly implies $|Delta f(t)| leq r$ for any $r in [s-delta_2,s)$.
$endgroup$
– saz
Dec 1 '18 at 17:29
$begingroup$
Thank you for the answer. What is left to do is to mention that every Lévy process has a version whose paths are càdlàg, right?
$endgroup$
– Ivan
Nov 25 '14 at 19:26
$begingroup$
Thank you for the answer. What is left to do is to mention that every Lévy process has a version whose paths are càdlàg, right?
$endgroup$
– Ivan
Nov 25 '14 at 19:26
$begingroup$
@Ivan Usually, a Lévy process has by definition (almost surely) càdlàg sample paths, but obviously this depends on the definition you are using.
$endgroup$
– saz
Nov 25 '14 at 19:27
$begingroup$
@Ivan Usually, a Lévy process has by definition (almost surely) càdlàg sample paths, but obviously this depends on the definition you are using.
$endgroup$
– saz
Nov 25 '14 at 19:27
$begingroup$
Sorry for necroing this old answer, but in a first step, we only obtain the existence of $delta_2>0$ such that $|f(t)-f(s-)|le r$ for all $tin[s-delta_2,s]$ by the existence of the left-limit at $t$. How do you proceed?
$endgroup$
– 0xbadf00d
Dec 1 '18 at 17:20
$begingroup$
Sorry for necroing this old answer, but in a first step, we only obtain the existence of $delta_2>0$ such that $|f(t)-f(s-)|le r$ for all $tin[s-delta_2,s]$ by the existence of the left-limit at $t$. How do you proceed?
$endgroup$
– 0xbadf00d
Dec 1 '18 at 17:20
1
1
$begingroup$
@0xbadf00d If $|f(t)-f(s-)| leq r/4$ for $t in [s-delta_2,s)$, then $|f(t)-f(u)| leq r/2$ for any $u,v in [s-delta_2,s)$, and this clearly implies $|Delta f(t)| leq r$ for any $r in [s-delta_2,s)$.
$endgroup$
– saz
Dec 1 '18 at 17:29
$begingroup$
@0xbadf00d If $|f(t)-f(s-)| leq r/4$ for $t in [s-delta_2,s)$, then $|f(t)-f(u)| leq r/2$ for any $u,v in [s-delta_2,s)$, and this clearly implies $|Delta f(t)| leq r$ for any $r in [s-delta_2,s)$.
$endgroup$
– saz
Dec 1 '18 at 17:29
add a comment |
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