Is a continuous function of (integrable) Brownian motion always integrable?
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Let $Z_{t}=e^{-W(t)}$ with ${W(t):tgeq0}$ a Brownian motion. Is $Z_{t}$ integrable, since it is a continuous function of Brownian motion? Furthermore, are all continuous functions of Brownian motion integrable, since Bownian motion itself is integrable?
measure-theory brownian-motion uniform-integrability
asked Jan 22 at 9:25
rs4rs35rs4rs35
237
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add a comment |
$begingroup$
Let $Z_{t}=e^{-W(t)}$ with ${W(t):tgeq0}$ a Brownian motion. Is $Z_{t}$ integrable, since it is a continuous function of Brownian motion? Furthermore, are all continuous functions of Brownian motion integrable, since Bownian motion itself is integrable?
measure-theory brownian-motion uniform-integrability
asked Jan 22 at 9:25
rs4rs35rs4rs35
237
$endgroup$
$begingroup$
Do you meant the integrability in $t$? Then yes, any continuous function is (locally) integrable.
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– zhoraster
Jan 22 at 9:28
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Yes, for a fixed $t$. So, my question: is the expectation of a continuous function of Brownian motion always finite implying that this function is integrable?
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– rs4rs35
Jan 22 at 9:41
$begingroup$
"For a fixed $t$" means integrability in $omega$, not in $t$. See the response of Kavi Rama Murthy.
$endgroup$
– zhoraster
Jan 23 at 11:27
add a comment |
$begingroup$
Let $Z_{t}=e^{-W(t)}$ with ${W(t):tgeq0}$ a Brownian motion. Is $Z_{t}$ integrable, since it is a continuous function of Brownian motion? Furthermore, are all continuous functions of Brownian motion integrable, since Bownian motion itself is integrable?
measure-theory brownian-motion uniform-integrability
asked Jan 22 at 9:25
rs4rs35rs4rs35
237
$endgroup$
Let $Z_{t}=e^{-W(t)}$ with ${W(t):tgeq0}$ a Brownian motion. Is $Z_{t}$ integrable, since it is a continuous function of Brownian motion? Furthermore, are all continuous functions of Brownian motion integrable, since Bownian motion itself is integrable?
measure-theory brownian-motion uniform-integrability
measure-theory brownian-motion uniform-integrability
asked Jan 22 at 9:25
rs4rs35rs4rs35
237
asked Jan 22 at 9:25
rs4rs35rs4rs35
237
asked Jan 22 at 9:25
rs4rs35rs4rs35
237
asked Jan 22 at 9:25
rs4rs35rs4rs35
237
asked Jan 22 at 9:25
rs4rs35rs4rs35
237
237
$begingroup$
Do you meant the integrability in $t$? Then yes, any continuous function is (locally) integrable.
$endgroup$
– zhoraster
Jan 22 at 9:28
$begingroup$
Yes, for a fixed $t$. So, my question: is the expectation of a continuous function of Brownian motion always finite implying that this function is integrable?
$endgroup$
– rs4rs35
Jan 22 at 9:41
$begingroup$
"For a fixed $t$" means integrability in $omega$, not in $t$. See the response of Kavi Rama Murthy.
$endgroup$
– zhoraster
Jan 23 at 11:27
add a comment |
$begingroup$
Do you meant the integrability in $t$? Then yes, any continuous function is (locally) integrable.
$endgroup$
– zhoraster
Jan 22 at 9:28
$begingroup$
Yes, for a fixed $t$. So, my question: is the expectation of a continuous function of Brownian motion always finite implying that this function is integrable?
$endgroup$
– rs4rs35
Jan 22 at 9:41
$begingroup$
"For a fixed $t$" means integrability in $omega$, not in $t$. See the response of Kavi Rama Murthy.
$endgroup$
– zhoraster
Jan 23 at 11:27
$begingroup$
Do you meant the integrability in $t$? Then yes, any continuous function is (locally) integrable.
$endgroup$
– zhoraster
Jan 22 at 9:28
$begingroup$
Do you meant the integrability in $t$? Then yes, any continuous function is (locally) integrable.
$endgroup$
– zhoraster
Jan 22 at 9:28
$begingroup$
Yes, for a fixed $t$. So, my question: is the expectation of a continuous function of Brownian motion always finite implying that this function is integrable?
$endgroup$
– rs4rs35
Jan 22 at 9:41
$begingroup$
Yes, for a fixed $t$. So, my question: is the expectation of a continuous function of Brownian motion always finite implying that this function is integrable?
$endgroup$
– rs4rs35
Jan 22 at 9:41
$begingroup$
"For a fixed $t$" means integrability in $omega$, not in $t$. See the response of Kavi Rama Murthy.
$endgroup$
– zhoraster
Jan 23 at 11:27
$begingroup$
"For a fixed $t$" means integrability in $omega$, not in $t$. See the response of Kavi Rama Murthy.
$endgroup$
– zhoraster
Jan 23 at 11:27
add a comment |
2 Answers
2
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oldest
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Continuous functions of integrable functions need not be intergable. In this case we can use the fact that if $X$ is normally distributed then $Ee^{cX}$ is integrable for any constant $c$. In fact $Ee^{cX}=e^{cmu} e^{sigma^{2}c^{2}/2}$ where $mu$ is the mean and $sigma^{2}$ the variance.
answered Jan 22 at 9:28
Kavi Rama MurthyKavi Rama Murthy
63k42362
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add a comment |
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$e^{-W_t}$ is integrable: $Ee^{-W_t}=e^{frac{1}{2}t}$.
Other continuous functions of $W_t$ need not be integrable. For example, $e^{W_1^2}$.
answered Jan 22 at 9:36
AddSupAddSup
4731316
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2 Answers
2
active
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votes
2 Answers
2
active
oldest
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active
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$begingroup$
Continuous functions of integrable functions need not be intergable. In this case we can use the fact that if $X$ is normally distributed then $Ee^{cX}$ is integrable for any constant $c$. In fact $Ee^{cX}=e^{cmu} e^{sigma^{2}c^{2}/2}$ where $mu$ is the mean and $sigma^{2}$ the variance.
answered Jan 22 at 9:28
Kavi Rama MurthyKavi Rama Murthy
63k42362
$endgroup$
add a comment |
$begingroup$
Continuous functions of integrable functions need not be intergable. In this case we can use the fact that if $X$ is normally distributed then $Ee^{cX}$ is integrable for any constant $c$. In fact $Ee^{cX}=e^{cmu} e^{sigma^{2}c^{2}/2}$ where $mu$ is the mean and $sigma^{2}$ the variance.
answered Jan 22 at 9:28
Kavi Rama MurthyKavi Rama Murthy
63k42362
$endgroup$
add a comment |
$begingroup$
Continuous functions of integrable functions need not be intergable. In this case we can use the fact that if $X$ is normally distributed then $Ee^{cX}$ is integrable for any constant $c$. In fact $Ee^{cX}=e^{cmu} e^{sigma^{2}c^{2}/2}$ where $mu$ is the mean and $sigma^{2}$ the variance.
answered Jan 22 at 9:28
Kavi Rama MurthyKavi Rama Murthy
63k42362
$endgroup$
Continuous functions of integrable functions need not be intergable. In this case we can use the fact that if $X$ is normally distributed then $Ee^{cX}$ is integrable for any constant $c$. In fact $Ee^{cX}=e^{cmu} e^{sigma^{2}c^{2}/2}$ where $mu$ is the mean and $sigma^{2}$ the variance.
answered Jan 22 at 9:28
Kavi Rama MurthyKavi Rama Murthy
63k42362
answered Jan 22 at 9:28
Kavi Rama MurthyKavi Rama Murthy
63k42362
answered Jan 22 at 9:28
Kavi Rama MurthyKavi Rama Murthy
63k42362
answered Jan 22 at 9:28
Kavi Rama MurthyKavi Rama Murthy
63k42362
63k42362
add a comment |
add a comment |
$begingroup$
$e^{-W_t}$ is integrable: $Ee^{-W_t}=e^{frac{1}{2}t}$.
Other continuous functions of $W_t$ need not be integrable. For example, $e^{W_1^2}$.
answered Jan 22 at 9:36
AddSupAddSup
4731316
$endgroup$
add a comment |
$begingroup$
$e^{-W_t}$ is integrable: $Ee^{-W_t}=e^{frac{1}{2}t}$.
Other continuous functions of $W_t$ need not be integrable. For example, $e^{W_1^2}$.
answered Jan 22 at 9:36
AddSupAddSup
4731316
$endgroup$
add a comment |
$begingroup$
$e^{-W_t}$ is integrable: $Ee^{-W_t}=e^{frac{1}{2}t}$.
Other continuous functions of $W_t$ need not be integrable. For example, $e^{W_1^2}$.
answered Jan 22 at 9:36
AddSupAddSup
4731316
$endgroup$
$e^{-W_t}$ is integrable: $Ee^{-W_t}=e^{frac{1}{2}t}$.
Other continuous functions of $W_t$ need not be integrable. For example, $e^{W_1^2}$.
answered Jan 22 at 9:36
AddSupAddSup
4731316
answered Jan 22 at 9:36
AddSupAddSup
4731316
answered Jan 22 at 9:36
AddSupAddSup
4731316
answered Jan 22 at 9:36
AddSupAddSup
4731316
4731316
add a comment |
add a comment |
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$begingroup$
Do you meant the integrability in $t$? Then yes, any continuous function is (locally) integrable.
$endgroup$
– zhoraster
Jan 22 at 9:28
$begingroup$
Yes, for a fixed $t$. So, my question: is the expectation of a continuous function of Brownian motion always finite implying that this function is integrable?
$endgroup$
– rs4rs35
Jan 22 at 9:41
$begingroup$
"For a fixed $t$" means integrability in $omega$, not in $t$. See the response of Kavi Rama Murthy.
$endgroup$
– zhoraster
Jan 23 at 11:27