Which inequality for the fourth central moment do we need to apply here?












6












$begingroup$


Let





  • $dinmathbb N$ with $d>1$


  • $lambda^d$ denote the Lebesuge measure on $mathcal Bleft(mathbb R^dright)$


  • $fin C^2(mathbb R)$ be positive and $$pi(x):=prod_{i=1}^df(x_i);;;text{for }xinmathbb R^d$$


  • $(Omega,mathcal A,operatorname P)$ be a probability space


  • $X:Omegatomathbb R^d$ with $X_astoperatorname P=pilambda^d$


Now, let $$g(x):=frac1{d-1}sum_{i=2}^dleft|frac{f'(x_i)}{f(x_i)}right|^2;;;text{for }xinmathbb R^d.$$



Assume $$M:=intfrac{|f'|^8}{f^7}:{rm d}lambda^1<inftytag1.$$



Note that $$operatorname Eleft[g(X)right]=intfrac{left|f'right|^2}f:{rm d}lambda^1=:I.tag2$$




I want to show that $$operatorname Eleft[left|g(X)-Iright|^4right]le d^{-frac12}(d-1)^{-frac32}3Mtag3.$$




Is there an easy estimate which yields $(3)$? Clearly, we can expand the left-hand side using the multinomial theorem, but then we deal with a complicated expression and annoying computations.



On the other hand, by applying the Cauchy-Schwarz inequality twice, we obtain $$operatorname Eleft[left|g(X)-Iright|^4right]lefrac1{d-1}sum_{i=2}^doperatorname Eleft[left|left|frac{f'(X)}{f(X)}right|^2-Iright|^4right],$$ but I don't know how we need to proceed from here.






probability-theory inequality expected-value







share|cite|improve this question











$endgroup$












  • $begingroup$
    What is the source of this problem?
    $endgroup$
    – LoveTooNap29
    Dec 9 '18 at 22:20






  • 1




    $begingroup$
    @LoveTooNap29 You can find it in the proof of Lemma 2.1 here: projecteuclid.org/euclid.aoap/1034625254
    $endgroup$
    – 0xbadf00d
    Dec 9 '18 at 22:27










  • $begingroup$
    @0xbadf00d Have you had a look at my answer? :)
    $endgroup$
    – Roberto Rastapopoulos
    Dec 15 '18 at 13:01
















6












$begingroup$


Let





  • $dinmathbb N$ with $d>1$


  • $lambda^d$ denote the Lebesuge measure on $mathcal Bleft(mathbb R^dright)$


  • $fin C^2(mathbb R)$ be positive and $$pi(x):=prod_{i=1}^df(x_i);;;text{for }xinmathbb R^d$$


  • $(Omega,mathcal A,operatorname P)$ be a probability space


  • $X:Omegatomathbb R^d$ with $X_astoperatorname P=pilambda^d$


Now, let $$g(x):=frac1{d-1}sum_{i=2}^dleft|frac{f'(x_i)}{f(x_i)}right|^2;;;text{for }xinmathbb R^d.$$



Assume $$M:=intfrac{|f'|^8}{f^7}:{rm d}lambda^1<inftytag1.$$



Note that $$operatorname Eleft[g(X)right]=intfrac{left|f'right|^2}f:{rm d}lambda^1=:I.tag2$$




I want to show that $$operatorname Eleft[left|g(X)-Iright|^4right]le d^{-frac12}(d-1)^{-frac32}3Mtag3.$$




Is there an easy estimate which yields $(3)$? Clearly, we can expand the left-hand side using the multinomial theorem, but then we deal with a complicated expression and annoying computations.



On the other hand, by applying the Cauchy-Schwarz inequality twice, we obtain $$operatorname Eleft[left|g(X)-Iright|^4right]lefrac1{d-1}sum_{i=2}^doperatorname Eleft[left|left|frac{f'(X)}{f(X)}right|^2-Iright|^4right],$$ but I don't know how we need to proceed from here.






probability-theory inequality expected-value







share|cite|improve this question











$endgroup$












  • $begingroup$
    What is the source of this problem?
    $endgroup$
    – LoveTooNap29
    Dec 9 '18 at 22:20






  • 1




    $begingroup$
    @LoveTooNap29 You can find it in the proof of Lemma 2.1 here: projecteuclid.org/euclid.aoap/1034625254
    $endgroup$
    – 0xbadf00d
    Dec 9 '18 at 22:27










  • $begingroup$
    @0xbadf00d Have you had a look at my answer? :)
    $endgroup$
    – Roberto Rastapopoulos
    Dec 15 '18 at 13:01














6












6








6


4



$begingroup$


Let





  • $dinmathbb N$ with $d>1$


  • $lambda^d$ denote the Lebesuge measure on $mathcal Bleft(mathbb R^dright)$


  • $fin C^2(mathbb R)$ be positive and $$pi(x):=prod_{i=1}^df(x_i);;;text{for }xinmathbb R^d$$


  • $(Omega,mathcal A,operatorname P)$ be a probability space


  • $X:Omegatomathbb R^d$ with $X_astoperatorname P=pilambda^d$


Now, let $$g(x):=frac1{d-1}sum_{i=2}^dleft|frac{f'(x_i)}{f(x_i)}right|^2;;;text{for }xinmathbb R^d.$$



Assume $$M:=intfrac{|f'|^8}{f^7}:{rm d}lambda^1<inftytag1.$$



Note that $$operatorname Eleft[g(X)right]=intfrac{left|f'right|^2}f:{rm d}lambda^1=:I.tag2$$




I want to show that $$operatorname Eleft[left|g(X)-Iright|^4right]le d^{-frac12}(d-1)^{-frac32}3Mtag3.$$




Is there an easy estimate which yields $(3)$? Clearly, we can expand the left-hand side using the multinomial theorem, but then we deal with a complicated expression and annoying computations.



On the other hand, by applying the Cauchy-Schwarz inequality twice, we obtain $$operatorname Eleft[left|g(X)-Iright|^4right]lefrac1{d-1}sum_{i=2}^doperatorname Eleft[left|left|frac{f'(X)}{f(X)}right|^2-Iright|^4right],$$ but I don't know how we need to proceed from here.






probability-theory inequality expected-value







share|cite|improve this question











$endgroup$




Let





  • $dinmathbb N$ with $d>1$


  • $lambda^d$ denote the Lebesuge measure on $mathcal Bleft(mathbb R^dright)$


  • $fin C^2(mathbb R)$ be positive and $$pi(x):=prod_{i=1}^df(x_i);;;text{for }xinmathbb R^d$$


  • $(Omega,mathcal A,operatorname P)$ be a probability space


  • $X:Omegatomathbb R^d$ with $X_astoperatorname P=pilambda^d$


Now, let $$g(x):=frac1{d-1}sum_{i=2}^dleft|frac{f'(x_i)}{f(x_i)}right|^2;;;text{for }xinmathbb R^d.$$



Assume $$M:=intfrac{|f'|^8}{f^7}:{rm d}lambda^1<inftytag1.$$



Note that $$operatorname Eleft[g(X)right]=intfrac{left|f'right|^2}f:{rm d}lambda^1=:I.tag2$$




I want to show that $$operatorname Eleft[left|g(X)-Iright|^4right]le d^{-frac12}(d-1)^{-frac32}3Mtag3.$$




Is there an easy estimate which yields $(3)$? Clearly, we can expand the left-hand side using the multinomial theorem, but then we deal with a complicated expression and annoying computations.



On the other hand, by applying the Cauchy-Schwarz inequality twice, we obtain $$operatorname Eleft[left|g(X)-Iright|^4right]lefrac1{d-1}sum_{i=2}^doperatorname Eleft[left|left|frac{f'(X)}{f(X)}right|^2-Iright|^4right],$$ but I don't know how we need to proceed from here.






probability-theory inequality expected-value




probability-theory inequality expected-value






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 11 '18 at 14:39







0xbadf00d

















asked Oct 31 '18 at 10:18









0xbadf00d0xbadf00d

1,91341531




1,91341531












  • $begingroup$
    What is the source of this problem?
    $endgroup$
    – LoveTooNap29
    Dec 9 '18 at 22:20






  • 1




    $begingroup$
    @LoveTooNap29 You can find it in the proof of Lemma 2.1 here: projecteuclid.org/euclid.aoap/1034625254
    $endgroup$
    – 0xbadf00d
    Dec 9 '18 at 22:27










  • $begingroup$
    @0xbadf00d Have you had a look at my answer? :)
    $endgroup$
    – Roberto Rastapopoulos
    Dec 15 '18 at 13:01


















  • $begingroup$
    What is the source of this problem?
    $endgroup$
    – LoveTooNap29
    Dec 9 '18 at 22:20






  • 1




    $begingroup$
    @LoveTooNap29 You can find it in the proof of Lemma 2.1 here: projecteuclid.org/euclid.aoap/1034625254
    $endgroup$
    – 0xbadf00d
    Dec 9 '18 at 22:27










  • $begingroup$
    @0xbadf00d Have you had a look at my answer? :)
    $endgroup$
    – Roberto Rastapopoulos
    Dec 15 '18 at 13:01
















$begingroup$
What is the source of this problem?
$endgroup$
– LoveTooNap29
Dec 9 '18 at 22:20




$begingroup$
What is the source of this problem?
$endgroup$
– LoveTooNap29
Dec 9 '18 at 22:20




1




1




$begingroup$
@LoveTooNap29 You can find it in the proof of Lemma 2.1 here: projecteuclid.org/euclid.aoap/1034625254
$endgroup$
– 0xbadf00d
Dec 9 '18 at 22:27




$begingroup$
@LoveTooNap29 You can find it in the proof of Lemma 2.1 here: projecteuclid.org/euclid.aoap/1034625254
$endgroup$
– 0xbadf00d
Dec 9 '18 at 22:27












$begingroup$
@0xbadf00d Have you had a look at my answer? :)
$endgroup$
– Roberto Rastapopoulos
Dec 15 '18 at 13:01




$begingroup$
@0xbadf00d Have you had a look at my answer? :)
$endgroup$
– Roberto Rastapopoulos
Dec 15 '18 at 13:01










1 Answer
1






active

oldest

votes


















2





+25







$begingroup$

Let us write $h = (f'/f)^2$.
By the definitions of $I$ and $M$, $E[h(X)] = I$ and $E[h^4(X)] = M$,
and we have
$$ g(x) = frac{1}{d-1} sum_{i=2}^{d} h(x_i).$$
Now we have
begin{align}
E left[ |g(X) - I|^4 right] &= E left[ left( frac{1}{d-1} sum_{i=2}^{d} (h(X_i) - I) right)^4 right] \
&= left( frac{1}{d-1} right)^4 E left[ left(sum_{i=2}^{d} (h(X_i) - I)^4 right) + 3 left( sum_{i=2}^{d} sum_{j=2, jneq i}^{d} (h(X_i) - I)^2(h(X_j) - I)^2 right) right],
end{align}
because the other terms of the product cancel out from the fact that $E(h(X_i) - I) = 0$.
Using the moment bound at our disposal and Holder's inequality,
we obtain
begin{align}
E left[ |g(X) - I|^4 right] &leq M , left( frac{1}{d-1} right)^4 left(d - 1 + 3 (d-1)(d-2) right), \
&= 3M , frac{d-5/3}{(d-1)^3} = 3 M , (d-1)^{-3/2} , frac{d - 5/3}{(d-1)^{3/2}}.
end{align}
and the last term is less than $d^{-1/2}$ when $d geq 2$.




EDIT: the fact that $E[(h(X_i) - I)^4] leq E[h(X_i)^4]$ follows from the non-negativity of $h$, see my answer here.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    The bound after "we obtain" seems to require $E[(h(X_i)-I)^4]leq E[h(X_i)^4],$ unless I've missed something, but this is not true in general: math.stackexchange.com/a/1591395.
    $endgroup$
    – Dap
    Dec 14 '18 at 14:06










  • $begingroup$
    @Dap I think I proved the moment bound. Let me know if you see any mistakes. :)
    $endgroup$
    – Roberto Rastapopoulos
    Dec 15 '18 at 13:00











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









2





+25







$begingroup$

Let us write $h = (f'/f)^2$.
By the definitions of $I$ and $M$, $E[h(X)] = I$ and $E[h^4(X)] = M$,
and we have
$$ g(x) = frac{1}{d-1} sum_{i=2}^{d} h(x_i).$$
Now we have
begin{align}
E left[ |g(X) - I|^4 right] &= E left[ left( frac{1}{d-1} sum_{i=2}^{d} (h(X_i) - I) right)^4 right] \
&= left( frac{1}{d-1} right)^4 E left[ left(sum_{i=2}^{d} (h(X_i) - I)^4 right) + 3 left( sum_{i=2}^{d} sum_{j=2, jneq i}^{d} (h(X_i) - I)^2(h(X_j) - I)^2 right) right],
end{align}
because the other terms of the product cancel out from the fact that $E(h(X_i) - I) = 0$.
Using the moment bound at our disposal and Holder's inequality,
we obtain
begin{align}
E left[ |g(X) - I|^4 right] &leq M , left( frac{1}{d-1} right)^4 left(d - 1 + 3 (d-1)(d-2) right), \
&= 3M , frac{d-5/3}{(d-1)^3} = 3 M , (d-1)^{-3/2} , frac{d - 5/3}{(d-1)^{3/2}}.
end{align}
and the last term is less than $d^{-1/2}$ when $d geq 2$.




EDIT: the fact that $E[(h(X_i) - I)^4] leq E[h(X_i)^4]$ follows from the non-negativity of $h$, see my answer here.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    The bound after "we obtain" seems to require $E[(h(X_i)-I)^4]leq E[h(X_i)^4],$ unless I've missed something, but this is not true in general: math.stackexchange.com/a/1591395.
    $endgroup$
    – Dap
    Dec 14 '18 at 14:06










  • $begingroup$
    @Dap I think I proved the moment bound. Let me know if you see any mistakes. :)
    $endgroup$
    – Roberto Rastapopoulos
    Dec 15 '18 at 13:00
















2





+25







$begingroup$

Let us write $h = (f'/f)^2$.
By the definitions of $I$ and $M$, $E[h(X)] = I$ and $E[h^4(X)] = M$,
and we have
$$ g(x) = frac{1}{d-1} sum_{i=2}^{d} h(x_i).$$
Now we have
begin{align}
E left[ |g(X) - I|^4 right] &= E left[ left( frac{1}{d-1} sum_{i=2}^{d} (h(X_i) - I) right)^4 right] \
&= left( frac{1}{d-1} right)^4 E left[ left(sum_{i=2}^{d} (h(X_i) - I)^4 right) + 3 left( sum_{i=2}^{d} sum_{j=2, jneq i}^{d} (h(X_i) - I)^2(h(X_j) - I)^2 right) right],
end{align}
because the other terms of the product cancel out from the fact that $E(h(X_i) - I) = 0$.
Using the moment bound at our disposal and Holder's inequality,
we obtain
begin{align}
E left[ |g(X) - I|^4 right] &leq M , left( frac{1}{d-1} right)^4 left(d - 1 + 3 (d-1)(d-2) right), \
&= 3M , frac{d-5/3}{(d-1)^3} = 3 M , (d-1)^{-3/2} , frac{d - 5/3}{(d-1)^{3/2}}.
end{align}
and the last term is less than $d^{-1/2}$ when $d geq 2$.




EDIT: the fact that $E[(h(X_i) - I)^4] leq E[h(X_i)^4]$ follows from the non-negativity of $h$, see my answer here.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    The bound after "we obtain" seems to require $E[(h(X_i)-I)^4]leq E[h(X_i)^4],$ unless I've missed something, but this is not true in general: math.stackexchange.com/a/1591395.
    $endgroup$
    – Dap
    Dec 14 '18 at 14:06










  • $begingroup$
    @Dap I think I proved the moment bound. Let me know if you see any mistakes. :)
    $endgroup$
    – Roberto Rastapopoulos
    Dec 15 '18 at 13:00














2





+25







2





+25



2




+25



$begingroup$

Let us write $h = (f'/f)^2$.
By the definitions of $I$ and $M$, $E[h(X)] = I$ and $E[h^4(X)] = M$,
and we have
$$ g(x) = frac{1}{d-1} sum_{i=2}^{d} h(x_i).$$
Now we have
begin{align}
E left[ |g(X) - I|^4 right] &= E left[ left( frac{1}{d-1} sum_{i=2}^{d} (h(X_i) - I) right)^4 right] \
&= left( frac{1}{d-1} right)^4 E left[ left(sum_{i=2}^{d} (h(X_i) - I)^4 right) + 3 left( sum_{i=2}^{d} sum_{j=2, jneq i}^{d} (h(X_i) - I)^2(h(X_j) - I)^2 right) right],
end{align}
because the other terms of the product cancel out from the fact that $E(h(X_i) - I) = 0$.
Using the moment bound at our disposal and Holder's inequality,
we obtain
begin{align}
E left[ |g(X) - I|^4 right] &leq M , left( frac{1}{d-1} right)^4 left(d - 1 + 3 (d-1)(d-2) right), \
&= 3M , frac{d-5/3}{(d-1)^3} = 3 M , (d-1)^{-3/2} , frac{d - 5/3}{(d-1)^{3/2}}.
end{align}
and the last term is less than $d^{-1/2}$ when $d geq 2$.




EDIT: the fact that $E[(h(X_i) - I)^4] leq E[h(X_i)^4]$ follows from the non-negativity of $h$, see my answer here.






share|cite|improve this answer











$endgroup$



Let us write $h = (f'/f)^2$.
By the definitions of $I$ and $M$, $E[h(X)] = I$ and $E[h^4(X)] = M$,
and we have
$$ g(x) = frac{1}{d-1} sum_{i=2}^{d} h(x_i).$$
Now we have
begin{align}
E left[ |g(X) - I|^4 right] &= E left[ left( frac{1}{d-1} sum_{i=2}^{d} (h(X_i) - I) right)^4 right] \
&= left( frac{1}{d-1} right)^4 E left[ left(sum_{i=2}^{d} (h(X_i) - I)^4 right) + 3 left( sum_{i=2}^{d} sum_{j=2, jneq i}^{d} (h(X_i) - I)^2(h(X_j) - I)^2 right) right],
end{align}
because the other terms of the product cancel out from the fact that $E(h(X_i) - I) = 0$.
Using the moment bound at our disposal and Holder's inequality,
we obtain
begin{align}
E left[ |g(X) - I|^4 right] &leq M , left( frac{1}{d-1} right)^4 left(d - 1 + 3 (d-1)(d-2) right), \
&= 3M , frac{d-5/3}{(d-1)^3} = 3 M , (d-1)^{-3/2} , frac{d - 5/3}{(d-1)^{3/2}}.
end{align}
and the last term is less than $d^{-1/2}$ when $d geq 2$.




EDIT: the fact that $E[(h(X_i) - I)^4] leq E[h(X_i)^4]$ follows from the non-negativity of $h$, see my answer here.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 15 '18 at 18:45

























answered Dec 13 '18 at 19:48









Roberto RastapopoulosRoberto Rastapopoulos

899424




899424












  • $begingroup$
    The bound after "we obtain" seems to require $E[(h(X_i)-I)^4]leq E[h(X_i)^4],$ unless I've missed something, but this is not true in general: math.stackexchange.com/a/1591395.
    $endgroup$
    – Dap
    Dec 14 '18 at 14:06










  • $begingroup$
    @Dap I think I proved the moment bound. Let me know if you see any mistakes. :)
    $endgroup$
    – Roberto Rastapopoulos
    Dec 15 '18 at 13:00


















  • $begingroup$
    The bound after "we obtain" seems to require $E[(h(X_i)-I)^4]leq E[h(X_i)^4],$ unless I've missed something, but this is not true in general: math.stackexchange.com/a/1591395.
    $endgroup$
    – Dap
    Dec 14 '18 at 14:06










  • $begingroup$
    @Dap I think I proved the moment bound. Let me know if you see any mistakes. :)
    $endgroup$
    – Roberto Rastapopoulos
    Dec 15 '18 at 13:00
















$begingroup$
The bound after "we obtain" seems to require $E[(h(X_i)-I)^4]leq E[h(X_i)^4],$ unless I've missed something, but this is not true in general: math.stackexchange.com/a/1591395.
$endgroup$
– Dap
Dec 14 '18 at 14:06




$begingroup$
The bound after "we obtain" seems to require $E[(h(X_i)-I)^4]leq E[h(X_i)^4],$ unless I've missed something, but this is not true in general: math.stackexchange.com/a/1591395.
$endgroup$
– Dap
Dec 14 '18 at 14:06












$begingroup$
@Dap I think I proved the moment bound. Let me know if you see any mistakes. :)
$endgroup$
– Roberto Rastapopoulos
Dec 15 '18 at 13:00




$begingroup$
@Dap I think I proved the moment bound. Let me know if you see any mistakes. :)
$endgroup$
– Roberto Rastapopoulos
Dec 15 '18 at 13:00


















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Probability when a professor distributes a quiz and homework assignment to a class of n students.

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0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
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Aardman Animations

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Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
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Are they similar matrix

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0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
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