Which inequality for the fourth central moment do we need to apply here?
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
asked Oct 31 at 10:18
0xbadf00d
1,84141430
add a comment |
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
asked Oct 31 at 10:18
0xbadf00d
1,84141430
What is the source of this problem?
– LoveTooNap29
Dec 9 at 22:20
1
@LoveTooNap29 You can find it in the proof of Lemma 2.1 here: projecteuclid.org/euclid.aoap/1034625254
– 0xbadf00d
Dec 9 at 22:27
@0xbadf00d Have you had a look at my answer? :)
– Roberto Rastapopoulos
Dec 15 at 13:01
add a comment |
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
asked Oct 31 at 10:18
0xbadf00d
1,84141430
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
asked Oct 31 at 10:18
0xbadf00d
1,84141430
asked Oct 31 at 10:18
0xbadf00d
1,84141430
edited Dec 11 at 14:39
asked Oct 31 at 10:18
0xbadf00d
1,84141430
asked Oct 31 at 10:18
0xbadf00d
1,84141430
asked Oct 31 at 10:18
0xbadf00d
1,84141430
1,84141430
What is the source of this problem?
– LoveTooNap29
Dec 9 at 22:20
1
@LoveTooNap29 You can find it in the proof of Lemma 2.1 here: projecteuclid.org/euclid.aoap/1034625254
– 0xbadf00d
Dec 9 at 22:27
@0xbadf00d Have you had a look at my answer? :)
– Roberto Rastapopoulos
Dec 15 at 13:01
add a comment |
What is the source of this problem?
– LoveTooNap29
Dec 9 at 22:20
1
@LoveTooNap29 You can find it in the proof of Lemma 2.1 here: projecteuclid.org/euclid.aoap/1034625254
– 0xbadf00d
Dec 9 at 22:27
@0xbadf00d Have you had a look at my answer? :)
– Roberto Rastapopoulos
Dec 15 at 13:01
What is the source of this problem?
– LoveTooNap29
Dec 9 at 22:20
What is the source of this problem?
– LoveTooNap29
Dec 9 at 22:20
1
1
@LoveTooNap29 You can find it in the proof of Lemma 2.1 here: projecteuclid.org/euclid.aoap/1034625254
– 0xbadf00d
Dec 9 at 22:27
@LoveTooNap29 You can find it in the proof of Lemma 2.1 here: projecteuclid.org/euclid.aoap/1034625254
– 0xbadf00d
Dec 9 at 22:27
@0xbadf00d Have you had a look at my answer? :)
– Roberto Rastapopoulos
Dec 15 at 13:01
@0xbadf00d Have you had a look at my answer? :)
– Roberto Rastapopoulos
Dec 15 at 13:01
add a comment |
1 Answer
1
active
oldest
votes
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.
answered Dec 13 at 19:48
Roberto Rastapopoulos
869424
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.
– Dap
Dec 14 at 14:06
@Dap I think I proved the moment bound. Let me know if you see any mistakes. :)
– Roberto Rastapopoulos
Dec 15 at 13:00
add a comment |
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1 Answer
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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.
answered Dec 13 at 19:48
Roberto Rastapopoulos
869424
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.
– Dap
Dec 14 at 14:06
@Dap I think I proved the moment bound. Let me know if you see any mistakes. :)
– Roberto Rastapopoulos
Dec 15 at 13:00
add a comment |
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.
answered Dec 13 at 19:48
Roberto Rastapopoulos
869424
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.
– Dap
Dec 14 at 14:06
@Dap I think I proved the moment bound. Let me know if you see any mistakes. :)
– Roberto Rastapopoulos
Dec 15 at 13:00
add a comment |
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.
answered Dec 13 at 19:48
Roberto Rastapopoulos
869424
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.
answered Dec 13 at 19:48
Roberto Rastapopoulos
869424
edited Dec 15 at 18:45
answered Dec 13 at 19:48
Roberto Rastapopoulos
869424
answered Dec 13 at 19:48
Roberto Rastapopoulos
869424
answered Dec 13 at 19:48
Roberto Rastapopoulos
869424
869424
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.
– Dap
Dec 14 at 14:06
@Dap I think I proved the moment bound. Let me know if you see any mistakes. :)
– Roberto Rastapopoulos
Dec 15 at 13:00
add a comment |
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.
– Dap
Dec 14 at 14:06
@Dap I think I proved the moment bound. Let me know if you see any mistakes. :)
– Roberto Rastapopoulos
Dec 15 at 13:00
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.
– Dap
Dec 14 at 14:06
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.
– Dap
Dec 14 at 14:06
@Dap I think I proved the moment bound. Let me know if you see any mistakes. :)
– Roberto Rastapopoulos
Dec 15 at 13:00
@Dap I think I proved the moment bound. Let me know if you see any mistakes. :)
– Roberto Rastapopoulos
Dec 15 at 13:00
add a comment |
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What is the source of this problem?
– LoveTooNap29
Dec 9 at 22:20
1
@LoveTooNap29 You can find it in the proof of Lemma 2.1 here: projecteuclid.org/euclid.aoap/1034625254
– 0xbadf00d
Dec 9 at 22:27
@0xbadf00d Have you had a look at my answer? :)
– Roberto Rastapopoulos
Dec 15 at 13:01