Finiteness of exponential moments of a stochastic integral implies finite moments
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Background: Hello everybody. I'm reading the paper Optimal Portfolio under Fractional Stochastic Environment by Fouque and Hu. The proof of Proposition 2.2 contains an estimation (Abschätzung) of an expected value I cannot deduce myself. Thanks a lot for your help!
Setup: Let $[0,T]$ be a time-index set and $(Omega,mathcal{F},mathbb{P})$ a probability space supporting an $mathbb{R}$-valued Brownian motion $W$. Let $mathbb{F}$ be the augmented filtration generated by $W$. Let $xiin L^{2}(W)$. Let $cinmathbb{R}_{>0}$ and $kinmathbb{R}_{>1}$. We assume that $$mathbb{E}Big[e^{cint_{0}^{T}xi_{s}^{2} ds}Big] < infty.tag{1}$$
Question: Can we deduce that $$mathbb{E}bigg[int_{0}^{T}xi_{s}^{2k}dsbigg] < infty? tag{2}$$
My thoughts: (1) implies that the moment generating function of the random variable $int_{0}^{T}xi_{s}^{2}ds$ is well-defined in a neighborhood of $0$. Hence, $$forall linmathbb{N}:, mathbb{E}Bigg[bigg(int_{0}^{T}xi_{s}^{2}dsbigg)^{l}Bigg] < infty. tag{3}$$ However, I cannot prove that $$exists linmathbb{N}:, mathbb{E}bigg[int_{0}^{T}xi_{s}^{2k}dsbigg] leq mathbb{E}Bigg[bigg(int_{0}^{T}xi_{s}^{2}dsbigg)^{l}Bigg],$$ so (3) is not of much help.
probability-theory stochastic-processes stochastic-calculus stochastic-integrals
asked Dec 12 '18 at 21:58
HarryTuttleHarryTuttle
154
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add a comment |
$begingroup$
Background: Hello everybody. I'm reading the paper Optimal Portfolio under Fractional Stochastic Environment by Fouque and Hu. The proof of Proposition 2.2 contains an estimation (Abschätzung) of an expected value I cannot deduce myself. Thanks a lot for your help!
Setup: Let $[0,T]$ be a time-index set and $(Omega,mathcal{F},mathbb{P})$ a probability space supporting an $mathbb{R}$-valued Brownian motion $W$. Let $mathbb{F}$ be the augmented filtration generated by $W$. Let $xiin L^{2}(W)$. Let $cinmathbb{R}_{>0}$ and $kinmathbb{R}_{>1}$. We assume that $$mathbb{E}Big[e^{cint_{0}^{T}xi_{s}^{2} ds}Big] < infty.tag{1}$$
Question: Can we deduce that $$mathbb{E}bigg[int_{0}^{T}xi_{s}^{2k}dsbigg] < infty? tag{2}$$
My thoughts: (1) implies that the moment generating function of the random variable $int_{0}^{T}xi_{s}^{2}ds$ is well-defined in a neighborhood of $0$. Hence, $$forall linmathbb{N}:, mathbb{E}Bigg[bigg(int_{0}^{T}xi_{s}^{2}dsbigg)^{l}Bigg] < infty. tag{3}$$ However, I cannot prove that $$exists linmathbb{N}:, mathbb{E}bigg[int_{0}^{T}xi_{s}^{2k}dsbigg] leq mathbb{E}Bigg[bigg(int_{0}^{T}xi_{s}^{2}dsbigg)^{l}Bigg],$$ so (3) is not of much help.
probability-theory stochastic-processes stochastic-calculus stochastic-integrals
asked Dec 12 '18 at 21:58
HarryTuttleHarryTuttle
154
$endgroup$
add a comment |
$begingroup$
Background: Hello everybody. I'm reading the paper Optimal Portfolio under Fractional Stochastic Environment by Fouque and Hu. The proof of Proposition 2.2 contains an estimation (Abschätzung) of an expected value I cannot deduce myself. Thanks a lot for your help!
Setup: Let $[0,T]$ be a time-index set and $(Omega,mathcal{F},mathbb{P})$ a probability space supporting an $mathbb{R}$-valued Brownian motion $W$. Let $mathbb{F}$ be the augmented filtration generated by $W$. Let $xiin L^{2}(W)$. Let $cinmathbb{R}_{>0}$ and $kinmathbb{R}_{>1}$. We assume that $$mathbb{E}Big[e^{cint_{0}^{T}xi_{s}^{2} ds}Big] < infty.tag{1}$$
Question: Can we deduce that $$mathbb{E}bigg[int_{0}^{T}xi_{s}^{2k}dsbigg] < infty? tag{2}$$
My thoughts: (1) implies that the moment generating function of the random variable $int_{0}^{T}xi_{s}^{2}ds$ is well-defined in a neighborhood of $0$. Hence, $$forall linmathbb{N}:, mathbb{E}Bigg[bigg(int_{0}^{T}xi_{s}^{2}dsbigg)^{l}Bigg] < infty. tag{3}$$ However, I cannot prove that $$exists linmathbb{N}:, mathbb{E}bigg[int_{0}^{T}xi_{s}^{2k}dsbigg] leq mathbb{E}Bigg[bigg(int_{0}^{T}xi_{s}^{2}dsbigg)^{l}Bigg],$$ so (3) is not of much help.
probability-theory stochastic-processes stochastic-calculus stochastic-integrals
asked Dec 12 '18 at 21:58
HarryTuttleHarryTuttle
154
$endgroup$
Background: Hello everybody. I'm reading the paper Optimal Portfolio under Fractional Stochastic Environment by Fouque and Hu. The proof of Proposition 2.2 contains an estimation (Abschätzung) of an expected value I cannot deduce myself. Thanks a lot for your help!
Setup: Let $[0,T]$ be a time-index set and $(Omega,mathcal{F},mathbb{P})$ a probability space supporting an $mathbb{R}$-valued Brownian motion $W$. Let $mathbb{F}$ be the augmented filtration generated by $W$. Let $xiin L^{2}(W)$. Let $cinmathbb{R}_{>0}$ and $kinmathbb{R}_{>1}$. We assume that $$mathbb{E}Big[e^{cint_{0}^{T}xi_{s}^{2} ds}Big] < infty.tag{1}$$
Question: Can we deduce that $$mathbb{E}bigg[int_{0}^{T}xi_{s}^{2k}dsbigg] < infty? tag{2}$$
My thoughts: (1) implies that the moment generating function of the random variable $int_{0}^{T}xi_{s}^{2}ds$ is well-defined in a neighborhood of $0$. Hence, $$forall linmathbb{N}:, mathbb{E}Bigg[bigg(int_{0}^{T}xi_{s}^{2}dsbigg)^{l}Bigg] < infty. tag{3}$$ However, I cannot prove that $$exists linmathbb{N}:, mathbb{E}bigg[int_{0}^{T}xi_{s}^{2k}dsbigg] leq mathbb{E}Bigg[bigg(int_{0}^{T}xi_{s}^{2}dsbigg)^{l}Bigg],$$ so (3) is not of much help.
probability-theory stochastic-processes stochastic-calculus stochastic-integrals
probability-theory stochastic-processes stochastic-calculus stochastic-integrals
asked Dec 12 '18 at 21:58
HarryTuttleHarryTuttle
154
asked Dec 12 '18 at 21:58
HarryTuttleHarryTuttle
154
asked Dec 12 '18 at 21:58
HarryTuttleHarryTuttle
154
asked Dec 12 '18 at 21:58
HarryTuttleHarryTuttle
154
asked Dec 12 '18 at 21:58
HarryTuttleHarryTuttle
154
154
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1 Answer
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No, $(1)$ does, in general, not imply $(2)$. Define
$$xi(s,omega) := frac{1}{s^{1/4}} 1_{(0,infty)}(s) .$$
Clearly,
$$mathbb{E} exp left( int_0^T xi_s^2 , ds right) = exp left( int_0^T s^{-1/2} , ds right)<infty$$
for any $T>0$ but
$$mathbb{E} left( int_0^T xi_s^{2k} , ds right) = int_0^T s^{-k/2} , ds = infty$$
for all $k geq 2$.
answered Dec 13 '18 at 8:19
sazsaz
82.1k862131
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1 Answer
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$begingroup$
No, $(1)$ does, in general, not imply $(2)$. Define
$$xi(s,omega) := frac{1}{s^{1/4}} 1_{(0,infty)}(s) .$$
Clearly,
$$mathbb{E} exp left( int_0^T xi_s^2 , ds right) = exp left( int_0^T s^{-1/2} , ds right)<infty$$
for any $T>0$ but
$$mathbb{E} left( int_0^T xi_s^{2k} , ds right) = int_0^T s^{-k/2} , ds = infty$$
for all $k geq 2$.
answered Dec 13 '18 at 8:19
sazsaz
82.1k862131
$endgroup$
add a comment |
$begingroup$
No, $(1)$ does, in general, not imply $(2)$. Define
$$xi(s,omega) := frac{1}{s^{1/4}} 1_{(0,infty)}(s) .$$
Clearly,
$$mathbb{E} exp left( int_0^T xi_s^2 , ds right) = exp left( int_0^T s^{-1/2} , ds right)<infty$$
for any $T>0$ but
$$mathbb{E} left( int_0^T xi_s^{2k} , ds right) = int_0^T s^{-k/2} , ds = infty$$
for all $k geq 2$.
answered Dec 13 '18 at 8:19
sazsaz
82.1k862131
$endgroup$
add a comment |
$begingroup$
No, $(1)$ does, in general, not imply $(2)$. Define
$$xi(s,omega) := frac{1}{s^{1/4}} 1_{(0,infty)}(s) .$$
Clearly,
$$mathbb{E} exp left( int_0^T xi_s^2 , ds right) = exp left( int_0^T s^{-1/2} , ds right)<infty$$
for any $T>0$ but
$$mathbb{E} left( int_0^T xi_s^{2k} , ds right) = int_0^T s^{-k/2} , ds = infty$$
for all $k geq 2$.
answered Dec 13 '18 at 8:19
sazsaz
82.1k862131
$endgroup$
No, $(1)$ does, in general, not imply $(2)$. Define
$$xi(s,omega) := frac{1}{s^{1/4}} 1_{(0,infty)}(s) .$$
Clearly,
$$mathbb{E} exp left( int_0^T xi_s^2 , ds right) = exp left( int_0^T s^{-1/2} , ds right)<infty$$
for any $T>0$ but
$$mathbb{E} left( int_0^T xi_s^{2k} , ds right) = int_0^T s^{-k/2} , ds = infty$$
for all $k geq 2$.
answered Dec 13 '18 at 8:19
sazsaz
82.1k862131
edited Dec 13 '18 at 19:14
answered Dec 13 '18 at 8:19
sazsaz
82.1k862131
answered Dec 13 '18 at 8:19
sazsaz
82.1k862131
answered Dec 13 '18 at 8:19
sazsaz
82.1k862131
82.1k862131
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