How to check that a random variable is in $L^p(Omega,F,mathbb{P})$
$begingroup$
How to check that a random variable is in $L^p(Omega,F,mathbb{P})$?
Maybe I should provide an example to get my point across.
Let $X_alpha$ be a random variable with CFD $F_alpha(x) = (1 - frac{1}{x^alpha})mathbb{1}_{[1,infty]}(x)$
I want to find for which values $alpha$ $X_alpha$ belongs to $L^p$.
Should I look at integrating the density function to the power p in absolute value or rather the CFD? I am a bit confused.
probability lebesgue-integral
asked Jan 15 at 20:22
qcc101qcc101
629213
$endgroup$
add a comment |
$begingroup$
How to check that a random variable is in $L^p(Omega,F,mathbb{P})$?
Maybe I should provide an example to get my point across.
Let $X_alpha$ be a random variable with CFD $F_alpha(x) = (1 - frac{1}{x^alpha})mathbb{1}_{[1,infty]}(x)$
I want to find for which values $alpha$ $X_alpha$ belongs to $L^p$.
Should I look at integrating the density function to the power p in absolute value or rather the CFD? I am a bit confused.
probability lebesgue-integral
asked Jan 15 at 20:22
qcc101qcc101
629213
$endgroup$
add a comment |
$begingroup$
How to check that a random variable is in $L^p(Omega,F,mathbb{P})$?
Maybe I should provide an example to get my point across.
Let $X_alpha$ be a random variable with CFD $F_alpha(x) = (1 - frac{1}{x^alpha})mathbb{1}_{[1,infty]}(x)$
I want to find for which values $alpha$ $X_alpha$ belongs to $L^p$.
Should I look at integrating the density function to the power p in absolute value or rather the CFD? I am a bit confused.
probability lebesgue-integral
asked Jan 15 at 20:22
qcc101qcc101
629213
$endgroup$
How to check that a random variable is in $L^p(Omega,F,mathbb{P})$?
Maybe I should provide an example to get my point across.
Let $X_alpha$ be a random variable with CFD $F_alpha(x) = (1 - frac{1}{x^alpha})mathbb{1}_{[1,infty]}(x)$
I want to find for which values $alpha$ $X_alpha$ belongs to $L^p$.
Should I look at integrating the density function to the power p in absolute value or rather the CFD? I am a bit confused.
probability lebesgue-integral
probability lebesgue-integral
asked Jan 15 at 20:22
qcc101qcc101
629213
asked Jan 15 at 20:22
qcc101qcc101
629213
asked Jan 15 at 20:22
qcc101qcc101
629213
asked Jan 15 at 20:22
qcc101qcc101
629213
asked Jan 15 at 20:22
qcc101qcc101
629213
629213
add a comment |
add a comment |
1 Answer
1
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$begingroup$
$X_alpha$ is nonnegative, so it suffices to check $E X_alpha^p < infty$.
Since you have the CDF, it is easiest to use the tail sum probability for this expectation.
$$E X_alpha^p = int_0^infty P(X_alpha^p > t) , dt = int_0^infty (1 - F_alpha(t^{1/p})) , dt.$$
Hopefully this helps you finish the question.
answered Jan 15 at 20:28
angryavianangryavian
42.5k23481
$endgroup$
$begingroup$
What happened to $p$ in the last expression?
$endgroup$
– Robert Israel
Jan 15 at 20:32
$begingroup$
@RobertIsrael Thanks for catching my mistake.
$endgroup$
– angryavian
Jan 15 at 20:33
$begingroup$
What if the sign $X_alpha$ is not known?
$endgroup$
– qcc101
Jan 15 at 20:40
$begingroup$
@qcc101 The CDF you gave implies $X_alpha ge 0$ almost surely.
$endgroup$
– angryavian
Jan 15 at 20:47
$begingroup$
Yes, I mean: what if in another case I do not know the sign of my random variable?
$endgroup$
– qcc101
Jan 15 at 20:57
|
show 1 more comment
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$X_alpha$ is nonnegative, so it suffices to check $E X_alpha^p < infty$.
Since you have the CDF, it is easiest to use the tail sum probability for this expectation.
$$E X_alpha^p = int_0^infty P(X_alpha^p > t) , dt = int_0^infty (1 - F_alpha(t^{1/p})) , dt.$$
Hopefully this helps you finish the question.
answered Jan 15 at 20:28
angryavianangryavian
42.5k23481
$endgroup$
$begingroup$
What happened to $p$ in the last expression?
$endgroup$
– Robert Israel
Jan 15 at 20:32
$begingroup$
@RobertIsrael Thanks for catching my mistake.
$endgroup$
– angryavian
Jan 15 at 20:33
$begingroup$
What if the sign $X_alpha$ is not known?
$endgroup$
– qcc101
Jan 15 at 20:40
$begingroup$
@qcc101 The CDF you gave implies $X_alpha ge 0$ almost surely.
$endgroup$
– angryavian
Jan 15 at 20:47
$begingroup$
Yes, I mean: what if in another case I do not know the sign of my random variable?
$endgroup$
– qcc101
Jan 15 at 20:57
|
show 1 more comment
$begingroup$
$X_alpha$ is nonnegative, so it suffices to check $E X_alpha^p < infty$.
Since you have the CDF, it is easiest to use the tail sum probability for this expectation.
$$E X_alpha^p = int_0^infty P(X_alpha^p > t) , dt = int_0^infty (1 - F_alpha(t^{1/p})) , dt.$$
Hopefully this helps you finish the question.
answered Jan 15 at 20:28
angryavianangryavian
42.5k23481
$endgroup$
$begingroup$
What happened to $p$ in the last expression?
$endgroup$
– Robert Israel
Jan 15 at 20:32
$begingroup$
@RobertIsrael Thanks for catching my mistake.
$endgroup$
– angryavian
Jan 15 at 20:33
$begingroup$
What if the sign $X_alpha$ is not known?
$endgroup$
– qcc101
Jan 15 at 20:40
$begingroup$
@qcc101 The CDF you gave implies $X_alpha ge 0$ almost surely.
$endgroup$
– angryavian
Jan 15 at 20:47
$begingroup$
Yes, I mean: what if in another case I do not know the sign of my random variable?
$endgroup$
– qcc101
Jan 15 at 20:57
|
show 1 more comment
$begingroup$
$X_alpha$ is nonnegative, so it suffices to check $E X_alpha^p < infty$.
Since you have the CDF, it is easiest to use the tail sum probability for this expectation.
$$E X_alpha^p = int_0^infty P(X_alpha^p > t) , dt = int_0^infty (1 - F_alpha(t^{1/p})) , dt.$$
Hopefully this helps you finish the question.
answered Jan 15 at 20:28
angryavianangryavian
42.5k23481
$endgroup$
$X_alpha$ is nonnegative, so it suffices to check $E X_alpha^p < infty$.
Since you have the CDF, it is easiest to use the tail sum probability for this expectation.
$$E X_alpha^p = int_0^infty P(X_alpha^p > t) , dt = int_0^infty (1 - F_alpha(t^{1/p})) , dt.$$
Hopefully this helps you finish the question.
answered Jan 15 at 20:28
angryavianangryavian
42.5k23481
edited Jan 15 at 20:32
answered Jan 15 at 20:28
angryavianangryavian
42.5k23481
answered Jan 15 at 20:28
angryavianangryavian
42.5k23481
answered Jan 15 at 20:28
angryavianangryavian
42.5k23481
42.5k23481
$begingroup$
What happened to $p$ in the last expression?
$endgroup$
– Robert Israel
Jan 15 at 20:32
$begingroup$
@RobertIsrael Thanks for catching my mistake.
$endgroup$
– angryavian
Jan 15 at 20:33
$begingroup$
What if the sign $X_alpha$ is not known?
$endgroup$
– qcc101
Jan 15 at 20:40
$begingroup$
@qcc101 The CDF you gave implies $X_alpha ge 0$ almost surely.
$endgroup$
– angryavian
Jan 15 at 20:47
$begingroup$
Yes, I mean: what if in another case I do not know the sign of my random variable?
$endgroup$
– qcc101
Jan 15 at 20:57
|
show 1 more comment
$begingroup$
What happened to $p$ in the last expression?
$endgroup$
– Robert Israel
Jan 15 at 20:32
$begingroup$
@RobertIsrael Thanks for catching my mistake.
$endgroup$
– angryavian
Jan 15 at 20:33
$begingroup$
What if the sign $X_alpha$ is not known?
$endgroup$
– qcc101
Jan 15 at 20:40
$begingroup$
@qcc101 The CDF you gave implies $X_alpha ge 0$ almost surely.
$endgroup$
– angryavian
Jan 15 at 20:47
$begingroup$
Yes, I mean: what if in another case I do not know the sign of my random variable?
$endgroup$
– qcc101
Jan 15 at 20:57
$begingroup$
What happened to $p$ in the last expression?
$endgroup$
– Robert Israel
Jan 15 at 20:32
$begingroup$
What happened to $p$ in the last expression?
$endgroup$
– Robert Israel
Jan 15 at 20:32
$begingroup$
@RobertIsrael Thanks for catching my mistake.
$endgroup$
– angryavian
Jan 15 at 20:33
$begingroup$
@RobertIsrael Thanks for catching my mistake.
$endgroup$
– angryavian
Jan 15 at 20:33
$begingroup$
What if the sign $X_alpha$ is not known?
$endgroup$
– qcc101
Jan 15 at 20:40
$begingroup$
What if the sign $X_alpha$ is not known?
$endgroup$
– qcc101
Jan 15 at 20:40
$begingroup$
@qcc101 The CDF you gave implies $X_alpha ge 0$ almost surely.
$endgroup$
– angryavian
Jan 15 at 20:47
$begingroup$
@qcc101 The CDF you gave implies $X_alpha ge 0$ almost surely.
$endgroup$
– angryavian
Jan 15 at 20:47
$begingroup$
Yes, I mean: what if in another case I do not know the sign of my random variable?
$endgroup$
– qcc101
Jan 15 at 20:57
$begingroup$
Yes, I mean: what if in another case I do not know the sign of my random variable?
$endgroup$
– qcc101
Jan 15 at 20:57
|
show 1 more comment
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