If $X$ is integrable then $mathbb{E}[sqrt{a_0 + a_1X^2}]$ is finite












1












$begingroup$


Let $X$ be an integrable random variable with possibly no moment of order 2.



Let $a_0$ and $a_1$ be two positive scalars. I want to prove that $mathbb{E}[sqrt{a_0 + a_1X^2}] < infty$



I tried to use Jensen or $sqrt{x} < 1 + x$, but to do that I have to assume that $mathbb{E}[X^2] < infty$ which is not the case.






probability-theory inequality







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$endgroup$








  • 2




    $begingroup$
    Hint: Prove the pointwise inequality $$sqrt{a_0+a_1X^2}leqslantsqrt{a_0}+sqrt{a_1},|X|$$ and integrate it.
    $endgroup$
    – Did
    Dec 13 '18 at 16:05










  • $begingroup$
    7 minutes. $ $ $ $
    $endgroup$
    – Did
    Dec 13 '18 at 16:05










  • $begingroup$
    Thanks, I proved $sqrt{a+x} leq sqrt{a} + sqrt{x}$ with $a geq 0$ and $x geq 0$ and it works
    $endgroup$
    – Dimitri Meunier
    Dec 13 '18 at 16:16










  • $begingroup$
    Exactly. $ $ $ $
    $endgroup$
    – Did
    Dec 13 '18 at 16:18
















1












$begingroup$


Let $X$ be an integrable random variable with possibly no moment of order 2.



Let $a_0$ and $a_1$ be two positive scalars. I want to prove that $mathbb{E}[sqrt{a_0 + a_1X^2}] < infty$



I tried to use Jensen or $sqrt{x} < 1 + x$, but to do that I have to assume that $mathbb{E}[X^2] < infty$ which is not the case.






probability-theory inequality







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Hint: Prove the pointwise inequality $$sqrt{a_0+a_1X^2}leqslantsqrt{a_0}+sqrt{a_1},|X|$$ and integrate it.
    $endgroup$
    – Did
    Dec 13 '18 at 16:05










  • $begingroup$
    7 minutes. $ $ $ $
    $endgroup$
    – Did
    Dec 13 '18 at 16:05










  • $begingroup$
    Thanks, I proved $sqrt{a+x} leq sqrt{a} + sqrt{x}$ with $a geq 0$ and $x geq 0$ and it works
    $endgroup$
    – Dimitri Meunier
    Dec 13 '18 at 16:16










  • $begingroup$
    Exactly. $ $ $ $
    $endgroup$
    – Did
    Dec 13 '18 at 16:18














1












1








1





$begingroup$


Let $X$ be an integrable random variable with possibly no moment of order 2.



Let $a_0$ and $a_1$ be two positive scalars. I want to prove that $mathbb{E}[sqrt{a_0 + a_1X^2}] < infty$



I tried to use Jensen or $sqrt{x} < 1 + x$, but to do that I have to assume that $mathbb{E}[X^2] < infty$ which is not the case.






probability-theory inequality







share|cite|improve this question











$endgroup$




Let $X$ be an integrable random variable with possibly no moment of order 2.



Let $a_0$ and $a_1$ be two positive scalars. I want to prove that $mathbb{E}[sqrt{a_0 + a_1X^2}] < infty$



I tried to use Jensen or $sqrt{x} < 1 + x$, but to do that I have to assume that $mathbb{E}[X^2] < infty$ which is not the case.






probability-theory inequality




probability-theory inequality






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 '18 at 16:06









Did

249k23227466




249k23227466










asked Dec 13 '18 at 15:45









Dimitri MeunierDimitri Meunier

1219




1219








  • 2




    $begingroup$
    Hint: Prove the pointwise inequality $$sqrt{a_0+a_1X^2}leqslantsqrt{a_0}+sqrt{a_1},|X|$$ and integrate it.
    $endgroup$
    – Did
    Dec 13 '18 at 16:05










  • $begingroup$
    7 minutes. $ $ $ $
    $endgroup$
    – Did
    Dec 13 '18 at 16:05










  • $begingroup$
    Thanks, I proved $sqrt{a+x} leq sqrt{a} + sqrt{x}$ with $a geq 0$ and $x geq 0$ and it works
    $endgroup$
    – Dimitri Meunier
    Dec 13 '18 at 16:16










  • $begingroup$
    Exactly. $ $ $ $
    $endgroup$
    – Did
    Dec 13 '18 at 16:18














  • 2




    $begingroup$
    Hint: Prove the pointwise inequality $$sqrt{a_0+a_1X^2}leqslantsqrt{a_0}+sqrt{a_1},|X|$$ and integrate it.
    $endgroup$
    – Did
    Dec 13 '18 at 16:05










  • $begingroup$
    7 minutes. $ $ $ $
    $endgroup$
    – Did
    Dec 13 '18 at 16:05










  • $begingroup$
    Thanks, I proved $sqrt{a+x} leq sqrt{a} + sqrt{x}$ with $a geq 0$ and $x geq 0$ and it works
    $endgroup$
    – Dimitri Meunier
    Dec 13 '18 at 16:16










  • $begingroup$
    Exactly. $ $ $ $
    $endgroup$
    – Did
    Dec 13 '18 at 16:18








2




2




$begingroup$
Hint: Prove the pointwise inequality $$sqrt{a_0+a_1X^2}leqslantsqrt{a_0}+sqrt{a_1},|X|$$ and integrate it.
$endgroup$
– Did
Dec 13 '18 at 16:05




$begingroup$
Hint: Prove the pointwise inequality $$sqrt{a_0+a_1X^2}leqslantsqrt{a_0}+sqrt{a_1},|X|$$ and integrate it.
$endgroup$
– Did
Dec 13 '18 at 16:05












$begingroup$
7 minutes. $ $ $ $
$endgroup$
– Did
Dec 13 '18 at 16:05




$begingroup$
7 minutes. $ $ $ $
$endgroup$
– Did
Dec 13 '18 at 16:05












$begingroup$
Thanks, I proved $sqrt{a+x} leq sqrt{a} + sqrt{x}$ with $a geq 0$ and $x geq 0$ and it works
$endgroup$
– Dimitri Meunier
Dec 13 '18 at 16:16




$begingroup$
Thanks, I proved $sqrt{a+x} leq sqrt{a} + sqrt{x}$ with $a geq 0$ and $x geq 0$ and it works
$endgroup$
– Dimitri Meunier
Dec 13 '18 at 16:16












$begingroup$
Exactly. $ $ $ $
$endgroup$
– Did
Dec 13 '18 at 16:18




$begingroup$
Exactly. $ $ $ $
$endgroup$
– Did
Dec 13 '18 at 16:18










1 Answer
1






active

oldest

votes


















1












$begingroup$

Note that $sqrt{a_1}|X|lesqrt{a_0+a_1X^2}$ implies $E|X|<infty$ is a necessary condition. That it is also sufficient follows from $0lesqrt{a_0+a_1X^2}lesqrt{a_0}+sqrt{a_1}|X|$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Interesting mistake: the fact that $Uleqslantmax{V,W}$ almost surely does not imply that $E(U)leqslantmax{E(V),E(W)}$. Thus, this step in your answer is wrong.
    $endgroup$
    – Did
    Dec 13 '18 at 16:02












  • $begingroup$
    And I fail to understand what you mean by: "...$E|X|$ is finite, which is necessary in the case $a_0=a_1$."
    $endgroup$
    – Did
    Dec 13 '18 at 16:03










  • $begingroup$
    And to be honest, I also fail to understand the two instant upvotes.
    $endgroup$
    – Did
    Dec 13 '18 at 16:06










  • $begingroup$
    "There's nothing wrong about these manipulations of these non-negative quantities" ?? Exercise: Find some almost surely nonnegative random variables $U$, $V$ and $W$ such that $Uleqslantmax{V,W}$ and $E(U)>max{E(V),E(W)}$.
    $endgroup$
    – Did
    Dec 13 '18 at 16:16










  • $begingroup$
    "final comment explains why a finite E|X| is also necessary" Except that nobody assumes $a_0=a_1$ here, so, what are you saying?
    $endgroup$
    – Did
    Dec 13 '18 at 16:17












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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Note that $sqrt{a_1}|X|lesqrt{a_0+a_1X^2}$ implies $E|X|<infty$ is a necessary condition. That it is also sufficient follows from $0lesqrt{a_0+a_1X^2}lesqrt{a_0}+sqrt{a_1}|X|$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Interesting mistake: the fact that $Uleqslantmax{V,W}$ almost surely does not imply that $E(U)leqslantmax{E(V),E(W)}$. Thus, this step in your answer is wrong.
    $endgroup$
    – Did
    Dec 13 '18 at 16:02












  • $begingroup$
    And I fail to understand what you mean by: "...$E|X|$ is finite, which is necessary in the case $a_0=a_1$."
    $endgroup$
    – Did
    Dec 13 '18 at 16:03










  • $begingroup$
    And to be honest, I also fail to understand the two instant upvotes.
    $endgroup$
    – Did
    Dec 13 '18 at 16:06










  • $begingroup$
    "There's nothing wrong about these manipulations of these non-negative quantities" ?? Exercise: Find some almost surely nonnegative random variables $U$, $V$ and $W$ such that $Uleqslantmax{V,W}$ and $E(U)>max{E(V),E(W)}$.
    $endgroup$
    – Did
    Dec 13 '18 at 16:16










  • $begingroup$
    "final comment explains why a finite E|X| is also necessary" Except that nobody assumes $a_0=a_1$ here, so, what are you saying?
    $endgroup$
    – Did
    Dec 13 '18 at 16:17
















1












$begingroup$

Note that $sqrt{a_1}|X|lesqrt{a_0+a_1X^2}$ implies $E|X|<infty$ is a necessary condition. That it is also sufficient follows from $0lesqrt{a_0+a_1X^2}lesqrt{a_0}+sqrt{a_1}|X|$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Interesting mistake: the fact that $Uleqslantmax{V,W}$ almost surely does not imply that $E(U)leqslantmax{E(V),E(W)}$. Thus, this step in your answer is wrong.
    $endgroup$
    – Did
    Dec 13 '18 at 16:02












  • $begingroup$
    And I fail to understand what you mean by: "...$E|X|$ is finite, which is necessary in the case $a_0=a_1$."
    $endgroup$
    – Did
    Dec 13 '18 at 16:03










  • $begingroup$
    And to be honest, I also fail to understand the two instant upvotes.
    $endgroup$
    – Did
    Dec 13 '18 at 16:06










  • $begingroup$
    "There's nothing wrong about these manipulations of these non-negative quantities" ?? Exercise: Find some almost surely nonnegative random variables $U$, $V$ and $W$ such that $Uleqslantmax{V,W}$ and $E(U)>max{E(V),E(W)}$.
    $endgroup$
    – Did
    Dec 13 '18 at 16:16










  • $begingroup$
    "final comment explains why a finite E|X| is also necessary" Except that nobody assumes $a_0=a_1$ here, so, what are you saying?
    $endgroup$
    – Did
    Dec 13 '18 at 16:17














1












1








1





$begingroup$

Note that $sqrt{a_1}|X|lesqrt{a_0+a_1X^2}$ implies $E|X|<infty$ is a necessary condition. That it is also sufficient follows from $0lesqrt{a_0+a_1X^2}lesqrt{a_0}+sqrt{a_1}|X|$.






share|cite|improve this answer











$endgroup$



Note that $sqrt{a_1}|X|lesqrt{a_0+a_1X^2}$ implies $E|X|<infty$ is a necessary condition. That it is also sufficient follows from $0lesqrt{a_0+a_1X^2}lesqrt{a_0}+sqrt{a_1}|X|$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 13 '18 at 16:42

























answered Dec 13 '18 at 15:53









J.G.J.G.

32.8k23250




32.8k23250












  • $begingroup$
    Interesting mistake: the fact that $Uleqslantmax{V,W}$ almost surely does not imply that $E(U)leqslantmax{E(V),E(W)}$. Thus, this step in your answer is wrong.
    $endgroup$
    – Did
    Dec 13 '18 at 16:02












  • $begingroup$
    And I fail to understand what you mean by: "...$E|X|$ is finite, which is necessary in the case $a_0=a_1$."
    $endgroup$
    – Did
    Dec 13 '18 at 16:03










  • $begingroup$
    And to be honest, I also fail to understand the two instant upvotes.
    $endgroup$
    – Did
    Dec 13 '18 at 16:06










  • $begingroup$
    "There's nothing wrong about these manipulations of these non-negative quantities" ?? Exercise: Find some almost surely nonnegative random variables $U$, $V$ and $W$ such that $Uleqslantmax{V,W}$ and $E(U)>max{E(V),E(W)}$.
    $endgroup$
    – Did
    Dec 13 '18 at 16:16










  • $begingroup$
    "final comment explains why a finite E|X| is also necessary" Except that nobody assumes $a_0=a_1$ here, so, what are you saying?
    $endgroup$
    – Did
    Dec 13 '18 at 16:17


















  • $begingroup$
    Interesting mistake: the fact that $Uleqslantmax{V,W}$ almost surely does not imply that $E(U)leqslantmax{E(V),E(W)}$. Thus, this step in your answer is wrong.
    $endgroup$
    – Did
    Dec 13 '18 at 16:02












  • $begingroup$
    And I fail to understand what you mean by: "...$E|X|$ is finite, which is necessary in the case $a_0=a_1$."
    $endgroup$
    – Did
    Dec 13 '18 at 16:03










  • $begingroup$
    And to be honest, I also fail to understand the two instant upvotes.
    $endgroup$
    – Did
    Dec 13 '18 at 16:06










  • $begingroup$
    "There's nothing wrong about these manipulations of these non-negative quantities" ?? Exercise: Find some almost surely nonnegative random variables $U$, $V$ and $W$ such that $Uleqslantmax{V,W}$ and $E(U)>max{E(V),E(W)}$.
    $endgroup$
    – Did
    Dec 13 '18 at 16:16










  • $begingroup$
    "final comment explains why a finite E|X| is also necessary" Except that nobody assumes $a_0=a_1$ here, so, what are you saying?
    $endgroup$
    – Did
    Dec 13 '18 at 16:17
















$begingroup$
Interesting mistake: the fact that $Uleqslantmax{V,W}$ almost surely does not imply that $E(U)leqslantmax{E(V),E(W)}$. Thus, this step in your answer is wrong.
$endgroup$
– Did
Dec 13 '18 at 16:02






$begingroup$
Interesting mistake: the fact that $Uleqslantmax{V,W}$ almost surely does not imply that $E(U)leqslantmax{E(V),E(W)}$. Thus, this step in your answer is wrong.
$endgroup$
– Did
Dec 13 '18 at 16:02














$begingroup$
And I fail to understand what you mean by: "...$E|X|$ is finite, which is necessary in the case $a_0=a_1$."
$endgroup$
– Did
Dec 13 '18 at 16:03




$begingroup$
And I fail to understand what you mean by: "...$E|X|$ is finite, which is necessary in the case $a_0=a_1$."
$endgroup$
– Did
Dec 13 '18 at 16:03












$begingroup$
And to be honest, I also fail to understand the two instant upvotes.
$endgroup$
– Did
Dec 13 '18 at 16:06




$begingroup$
And to be honest, I also fail to understand the two instant upvotes.
$endgroup$
– Did
Dec 13 '18 at 16:06












$begingroup$
"There's nothing wrong about these manipulations of these non-negative quantities" ?? Exercise: Find some almost surely nonnegative random variables $U$, $V$ and $W$ such that $Uleqslantmax{V,W}$ and $E(U)>max{E(V),E(W)}$.
$endgroup$
– Did
Dec 13 '18 at 16:16




$begingroup$
"There's nothing wrong about these manipulations of these non-negative quantities" ?? Exercise: Find some almost surely nonnegative random variables $U$, $V$ and $W$ such that $Uleqslantmax{V,W}$ and $E(U)>max{E(V),E(W)}$.
$endgroup$
– Did
Dec 13 '18 at 16:16












$begingroup$
"final comment explains why a finite E|X| is also necessary" Except that nobody assumes $a_0=a_1$ here, so, what are you saying?
$endgroup$
– Did
Dec 13 '18 at 16:17




$begingroup$
"final comment explains why a finite E|X| is also necessary" Except that nobody assumes $a_0=a_1$ here, so, what are you saying?
$endgroup$
– Did
Dec 13 '18 at 16:17


















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