Boundedness of Moments of Random Variables
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I'm sure this must be well-known so hints and/or a reference are fine. Let $epsilon>0$. Must there exist a random variable $X$ such that:
(i) $X geq 0$ a.s.
(ii) $E[X] leq epsilon$
(iii) $E[X^2] geq 1$
(iv) $E[X^3] leq 10^{100}$?
If we just want the first 3 conditions we can construct a solution along the lines of $X=epsilon n$ with probability $frac{1}{nsqrt{n}}$. But this won't satisfy the fourth property.
probability-theory random-variables
asked Jan 24 at 11:19
user3131035user3131035
51128
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add a comment |
$begingroup$
I'm sure this must be well-known so hints and/or a reference are fine. Let $epsilon>0$. Must there exist a random variable $X$ such that:
(i) $X geq 0$ a.s.
(ii) $E[X] leq epsilon$
(iii) $E[X^2] geq 1$
(iv) $E[X^3] leq 10^{100}$?
If we just want the first 3 conditions we can construct a solution along the lines of $X=epsilon n$ with probability $frac{1}{nsqrt{n}}$. But this won't satisfy the fourth property.
probability-theory random-variables
asked Jan 24 at 11:19
user3131035user3131035
51128
$endgroup$
add a comment |
$begingroup$
I'm sure this must be well-known so hints and/or a reference are fine. Let $epsilon>0$. Must there exist a random variable $X$ such that:
(i) $X geq 0$ a.s.
(ii) $E[X] leq epsilon$
(iii) $E[X^2] geq 1$
(iv) $E[X^3] leq 10^{100}$?
If we just want the first 3 conditions we can construct a solution along the lines of $X=epsilon n$ with probability $frac{1}{nsqrt{n}}$. But this won't satisfy the fourth property.
probability-theory random-variables
asked Jan 24 at 11:19
user3131035user3131035
51128
$endgroup$
I'm sure this must be well-known so hints and/or a reference are fine. Let $epsilon>0$. Must there exist a random variable $X$ such that:
(i) $X geq 0$ a.s.
(ii) $E[X] leq epsilon$
(iii) $E[X^2] geq 1$
(iv) $E[X^3] leq 10^{100}$?
If we just want the first 3 conditions we can construct a solution along the lines of $X=epsilon n$ with probability $frac{1}{nsqrt{n}}$. But this won't satisfy the fourth property.
probability-theory random-variables
probability-theory random-variables
asked Jan 24 at 11:19
user3131035user3131035
51128
asked Jan 24 at 11:19
user3131035user3131035
51128
asked Jan 24 at 11:19
user3131035user3131035
51128
asked Jan 24 at 11:19
user3131035user3131035
51128
asked Jan 24 at 11:19
user3131035user3131035
51128
51128
add a comment |
add a comment |
1 Answer
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If $epsilon>0$ is very small, then we cannot find such $X$. It is because, informally speaking, if we have a large $E[X^2]/E[X]$, then we should also have a large $E[X^3]/E[X^2]$. As $epsilon to 0^+$, the former diverges to $infty$, while the latter is bounded by $10^{100}$. Hence it cannot happen.
To see this in a formal argument note first that
$$
X(X-1/epsilon)^2=X^3-(2/epsilon)X^2+X/epsilon^2ge 0.
$$ By taking expectation, we have
$$
10^{100}+1/epsilon ge E[X^3]+E[X]/epsilon^2ge 2E[X^2]/epsilonge 2/epsilon.
$$ This gives
$$
epsilon ge 10^{-100},
$$ hence for $epsilon in (0,10^{-100})$ $X$ satisfying the given conditions does not exist.
If $epsilon in [ 10^{-100},1]$, we may define
$$
P(X=1/epsilon)=epsilon^2,quad P(X=0)=1-epsilon^2.
$$ By this, we get $E[X]=epsilon, $ $E[X^2]=1$ and
$E[X^3]=1/epsilon le 10^{100}$. For $epsilonge 1$, we can choose $Xequiv 1$.
To conclude, a random variable satisfying all the given conditions exists if and only if $epsilonge 10^{-100}$.
answered Jan 24 at 11:40
SongSong
16.7k11044
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1 Answer
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1 Answer
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active
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$begingroup$
If $epsilon>0$ is very small, then we cannot find such $X$. It is because, informally speaking, if we have a large $E[X^2]/E[X]$, then we should also have a large $E[X^3]/E[X^2]$. As $epsilon to 0^+$, the former diverges to $infty$, while the latter is bounded by $10^{100}$. Hence it cannot happen.
To see this in a formal argument note first that
$$
X(X-1/epsilon)^2=X^3-(2/epsilon)X^2+X/epsilon^2ge 0.
$$ By taking expectation, we have
$$
10^{100}+1/epsilon ge E[X^3]+E[X]/epsilon^2ge 2E[X^2]/epsilonge 2/epsilon.
$$ This gives
$$
epsilon ge 10^{-100},
$$ hence for $epsilon in (0,10^{-100})$ $X$ satisfying the given conditions does not exist.
If $epsilon in [ 10^{-100},1]$, we may define
$$
P(X=1/epsilon)=epsilon^2,quad P(X=0)=1-epsilon^2.
$$ By this, we get $E[X]=epsilon, $ $E[X^2]=1$ and
$E[X^3]=1/epsilon le 10^{100}$. For $epsilonge 1$, we can choose $Xequiv 1$.
To conclude, a random variable satisfying all the given conditions exists if and only if $epsilonge 10^{-100}$.
answered Jan 24 at 11:40
SongSong
16.7k11044
$endgroup$
add a comment |
$begingroup$
If $epsilon>0$ is very small, then we cannot find such $X$. It is because, informally speaking, if we have a large $E[X^2]/E[X]$, then we should also have a large $E[X^3]/E[X^2]$. As $epsilon to 0^+$, the former diverges to $infty$, while the latter is bounded by $10^{100}$. Hence it cannot happen.
To see this in a formal argument note first that
$$
X(X-1/epsilon)^2=X^3-(2/epsilon)X^2+X/epsilon^2ge 0.
$$ By taking expectation, we have
$$
10^{100}+1/epsilon ge E[X^3]+E[X]/epsilon^2ge 2E[X^2]/epsilonge 2/epsilon.
$$ This gives
$$
epsilon ge 10^{-100},
$$ hence for $epsilon in (0,10^{-100})$ $X$ satisfying the given conditions does not exist.
If $epsilon in [ 10^{-100},1]$, we may define
$$
P(X=1/epsilon)=epsilon^2,quad P(X=0)=1-epsilon^2.
$$ By this, we get $E[X]=epsilon, $ $E[X^2]=1$ and
$E[X^3]=1/epsilon le 10^{100}$. For $epsilonge 1$, we can choose $Xequiv 1$.
To conclude, a random variable satisfying all the given conditions exists if and only if $epsilonge 10^{-100}$.
answered Jan 24 at 11:40
SongSong
16.7k11044
$endgroup$
add a comment |
$begingroup$
If $epsilon>0$ is very small, then we cannot find such $X$. It is because, informally speaking, if we have a large $E[X^2]/E[X]$, then we should also have a large $E[X^3]/E[X^2]$. As $epsilon to 0^+$, the former diverges to $infty$, while the latter is bounded by $10^{100}$. Hence it cannot happen.
To see this in a formal argument note first that
$$
X(X-1/epsilon)^2=X^3-(2/epsilon)X^2+X/epsilon^2ge 0.
$$ By taking expectation, we have
$$
10^{100}+1/epsilon ge E[X^3]+E[X]/epsilon^2ge 2E[X^2]/epsilonge 2/epsilon.
$$ This gives
$$
epsilon ge 10^{-100},
$$ hence for $epsilon in (0,10^{-100})$ $X$ satisfying the given conditions does not exist.
If $epsilon in [ 10^{-100},1]$, we may define
$$
P(X=1/epsilon)=epsilon^2,quad P(X=0)=1-epsilon^2.
$$ By this, we get $E[X]=epsilon, $ $E[X^2]=1$ and
$E[X^3]=1/epsilon le 10^{100}$. For $epsilonge 1$, we can choose $Xequiv 1$.
To conclude, a random variable satisfying all the given conditions exists if and only if $epsilonge 10^{-100}$.
answered Jan 24 at 11:40
SongSong
16.7k11044
$endgroup$
If $epsilon>0$ is very small, then we cannot find such $X$. It is because, informally speaking, if we have a large $E[X^2]/E[X]$, then we should also have a large $E[X^3]/E[X^2]$. As $epsilon to 0^+$, the former diverges to $infty$, while the latter is bounded by $10^{100}$. Hence it cannot happen.
To see this in a formal argument note first that
$$
X(X-1/epsilon)^2=X^3-(2/epsilon)X^2+X/epsilon^2ge 0.
$$ By taking expectation, we have
$$
10^{100}+1/epsilon ge E[X^3]+E[X]/epsilon^2ge 2E[X^2]/epsilonge 2/epsilon.
$$ This gives
$$
epsilon ge 10^{-100},
$$ hence for $epsilon in (0,10^{-100})$ $X$ satisfying the given conditions does not exist.
If $epsilon in [ 10^{-100},1]$, we may define
$$
P(X=1/epsilon)=epsilon^2,quad P(X=0)=1-epsilon^2.
$$ By this, we get $E[X]=epsilon, $ $E[X^2]=1$ and
$E[X^3]=1/epsilon le 10^{100}$. For $epsilonge 1$, we can choose $Xequiv 1$.
To conclude, a random variable satisfying all the given conditions exists if and only if $epsilonge 10^{-100}$.
answered Jan 24 at 11:40
SongSong
16.7k11044
edited Jan 24 at 12:07
answered Jan 24 at 11:40
SongSong
16.7k11044
answered Jan 24 at 11:40
SongSong
16.7k11044
answered Jan 24 at 11:40
SongSong
16.7k11044
16.7k11044
add a comment |
add a comment |
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