Finiteness of the number of big jumps of a Lévy process on a finite interval












0












$begingroup$


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







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    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







    share|cite|improve this question











    $endgroup$















      0












      0








      0


      3



      $begingroup$


      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







      share|cite|improve this question











      $endgroup$




      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 25 '14 at 15:26







      Ivan

















      asked Nov 25 '14 at 13:38









      IvanIvan

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      553516






















          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 $>r$.







          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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











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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 $>r$.







          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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
















          2












          $begingroup$

          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 $>r$.







          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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














          2












          2








          2





          $begingroup$

          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 $>r$.







          share|cite|improve this answer











          $endgroup$



          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 $>r$.








          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 1 '18 at 17:27

























          answered Nov 25 '14 at 15:10









          sazsaz

          79k858123




          79k858123












          • $begingroup$
            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










          • $begingroup$
            @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


















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          Ida-Boy-Ed-Garten

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          Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
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