Is the supremum of an almost surely continuous random function random variable?
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Let {$X_t, tin[0,1]$} on {$R, mathfrak B(R) $} be random, almost surely continuous, function. How to show that $X^+=sup_{t in[0,1]} X_t$ is random variable ?
Perhaps here I can say that $X_t$ it will be a random variable $forall t$ аnd prove the statement like for random variables ?
continuity proof-writing stochastic-processes supremum-and-infimum
asked Nov 18 at 11:07
Emerald
358
add a comment |
up vote
0
down vote
favorite
Let {$X_t, tin[0,1]$} on {$R, mathfrak B(R) $} be random, almost surely continuous, function. How to show that $X^+=sup_{t in[0,1]} X_t$ is random variable ?
Perhaps here I can say that $X_t$ it will be a random variable $forall t$ аnd prove the statement like for random variables ?
continuity proof-writing stochastic-processes supremum-and-infimum
asked Nov 18 at 11:07
Emerald
358
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let {$X_t, tin[0,1]$} on {$R, mathfrak B(R) $} be random, almost surely continuous, function. How to show that $X^+=sup_{t in[0,1]} X_t$ is random variable ?
Perhaps here I can say that $X_t$ it will be a random variable $forall t$ аnd prove the statement like for random variables ?
continuity proof-writing stochastic-processes supremum-and-infimum
asked Nov 18 at 11:07
Emerald
358
Let {$X_t, tin[0,1]$} on {$R, mathfrak B(R) $} be random, almost surely continuous, function. How to show that $X^+=sup_{t in[0,1]} X_t$ is random variable ?
Perhaps here I can say that $X_t$ it will be a random variable $forall t$ аnd prove the statement like for random variables ?
continuity proof-writing stochastic-processes supremum-and-infimum
continuity proof-writing stochastic-processes supremum-and-infimum
asked Nov 18 at 11:07
Emerald
358
asked Nov 18 at 11:07
Emerald
358
asked Nov 18 at 11:07
Emerald
358
asked Nov 18 at 11:07
Emerald
358
asked Nov 18 at 11:07
Emerald
358
358
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
$X^{+}=sup{X_t: 0leq t leq 1,tin mathbb Q}$ outside a null set, so $X^{+}$ is almost everywhere equal to a random variable. So $X^{+}$ is Lebesgue measurable. It need not be measurable w.r.t the Borel sigma field.
answered Nov 18 at 11:46
Kavi Rama Murthy
43.9k31852
Thanks for your answer. But I don't understand, why can we say, that $X^+$ is random variable? I would be grateful for any tips.
– Emerald
Nov 18 at 11:55
I think you have to look at the source for definitions. A random function is usually defined as a collection of random variables. so it is assumed that each $X_t$ is a random variable. Otherwise there is no hope whatsoever of proving that $X^{+}$ is a random variable.
– Kavi Rama Murthy
Nov 18 at 11:59
Thank you very much. Now it became clearer.
– Emerald
Nov 18 at 12:05
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
$X^{+}=sup{X_t: 0leq t leq 1,tin mathbb Q}$ outside a null set, so $X^{+}$ is almost everywhere equal to a random variable. So $X^{+}$ is Lebesgue measurable. It need not be measurable w.r.t the Borel sigma field.
answered Nov 18 at 11:46
Kavi Rama Murthy
43.9k31852
Thanks for your answer. But I don't understand, why can we say, that $X^+$ is random variable? I would be grateful for any tips.
– Emerald
Nov 18 at 11:55
I think you have to look at the source for definitions. A random function is usually defined as a collection of random variables. so it is assumed that each $X_t$ is a random variable. Otherwise there is no hope whatsoever of proving that $X^{+}$ is a random variable.
– Kavi Rama Murthy
Nov 18 at 11:59
Thank you very much. Now it became clearer.
– Emerald
Nov 18 at 12:05
add a comment |
up vote
0
down vote
accepted
$X^{+}=sup{X_t: 0leq t leq 1,tin mathbb Q}$ outside a null set, so $X^{+}$ is almost everywhere equal to a random variable. So $X^{+}$ is Lebesgue measurable. It need not be measurable w.r.t the Borel sigma field.
answered Nov 18 at 11:46
Kavi Rama Murthy
43.9k31852
Thanks for your answer. But I don't understand, why can we say, that $X^+$ is random variable? I would be grateful for any tips.
– Emerald
Nov 18 at 11:55
I think you have to look at the source for definitions. A random function is usually defined as a collection of random variables. so it is assumed that each $X_t$ is a random variable. Otherwise there is no hope whatsoever of proving that $X^{+}$ is a random variable.
– Kavi Rama Murthy
Nov 18 at 11:59
Thank you very much. Now it became clearer.
– Emerald
Nov 18 at 12:05
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
$X^{+}=sup{X_t: 0leq t leq 1,tin mathbb Q}$ outside a null set, so $X^{+}$ is almost everywhere equal to a random variable. So $X^{+}$ is Lebesgue measurable. It need not be measurable w.r.t the Borel sigma field.
answered Nov 18 at 11:46
Kavi Rama Murthy
43.9k31852
$X^{+}=sup{X_t: 0leq t leq 1,tin mathbb Q}$ outside a null set, so $X^{+}$ is almost everywhere equal to a random variable. So $X^{+}$ is Lebesgue measurable. It need not be measurable w.r.t the Borel sigma field.
answered Nov 18 at 11:46
Kavi Rama Murthy
43.9k31852
answered Nov 18 at 11:46
Kavi Rama Murthy
43.9k31852
answered Nov 18 at 11:46
Kavi Rama Murthy
43.9k31852
answered Nov 18 at 11:46
Kavi Rama Murthy
43.9k31852
43.9k31852
Thanks for your answer. But I don't understand, why can we say, that $X^+$ is random variable? I would be grateful for any tips.
– Emerald
Nov 18 at 11:55
I think you have to look at the source for definitions. A random function is usually defined as a collection of random variables. so it is assumed that each $X_t$ is a random variable. Otherwise there is no hope whatsoever of proving that $X^{+}$ is a random variable.
– Kavi Rama Murthy
Nov 18 at 11:59
Thank you very much. Now it became clearer.
– Emerald
Nov 18 at 12:05
add a comment |
Thanks for your answer. But I don't understand, why can we say, that $X^+$ is random variable? I would be grateful for any tips.
– Emerald
Nov 18 at 11:55
I think you have to look at the source for definitions. A random function is usually defined as a collection of random variables. so it is assumed that each $X_t$ is a random variable. Otherwise there is no hope whatsoever of proving that $X^{+}$ is a random variable.
– Kavi Rama Murthy
Nov 18 at 11:59
Thank you very much. Now it became clearer.
– Emerald
Nov 18 at 12:05
Thanks for your answer. But I don't understand, why can we say, that $X^+$ is random variable? I would be grateful for any tips.
– Emerald
Nov 18 at 11:55
Thanks for your answer. But I don't understand, why can we say, that $X^+$ is random variable? I would be grateful for any tips.
– Emerald
Nov 18 at 11:55
I think you have to look at the source for definitions. A random function is usually defined as a collection of random variables. so it is assumed that each $X_t$ is a random variable. Otherwise there is no hope whatsoever of proving that $X^{+}$ is a random variable.
– Kavi Rama Murthy
Nov 18 at 11:59
I think you have to look at the source for definitions. A random function is usually defined as a collection of random variables. so it is assumed that each $X_t$ is a random variable. Otherwise there is no hope whatsoever of proving that $X^{+}$ is a random variable.
– Kavi Rama Murthy
Nov 18 at 11:59
Thank you very much. Now it became clearer.
– Emerald
Nov 18 at 12:05
Thank you very much. Now it became clearer.
– Emerald
Nov 18 at 12:05
add a comment |
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