A nice inequality about characteristic function [closed]
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I need help with this please.
$psi$ is characteristic function of a probability measure, then:
$ |psi(t)-psi(s)|^2leq 2|1-psi(t-s)|$
probability inequality characteristic-functions
edited Dec 20 '18 at 15:14
saz
80.7k860125
asked Dec 19 '18 at 17:17
Giancarlos Giancarlos
42
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closed as off-topic by Davide Giraudo, Adrian Keister, Lord_Farin, amWhy, Dando18 Dec 21 '18 at 17:00
This question appears to be off-topic. The users who voted to close gave this specific reason:
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$begingroup$
I need help with this please.
$psi$ is characteristic function of a probability measure, then:
$ |psi(t)-psi(s)|^2leq 2|1-psi(t-s)|$
probability inequality characteristic-functions
edited Dec 20 '18 at 15:14
saz
80.7k860125
asked Dec 19 '18 at 17:17
Giancarlos Giancarlos
42
$endgroup$
closed as off-topic by Davide Giraudo, Adrian Keister, Lord_Farin, amWhy, Dando18 Dec 21 '18 at 17:00
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Davide Giraudo, Adrian Keister, Lord_Farin, amWhy, Dando18
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I need help with this please.
$psi$ is characteristic function of a probability measure, then:
$ |psi(t)-psi(s)|^2leq 2|1-psi(t-s)|$
probability inequality characteristic-functions
edited Dec 20 '18 at 15:14
saz
80.7k860125
asked Dec 19 '18 at 17:17
Giancarlos Giancarlos
42
$endgroup$
I need help with this please.
$psi$ is characteristic function of a probability measure, then:
$ |psi(t)-psi(s)|^2leq 2|1-psi(t-s)|$
probability inequality characteristic-functions
probability inequality characteristic-functions
edited Dec 20 '18 at 15:14
saz
80.7k860125
asked Dec 19 '18 at 17:17
Giancarlos Giancarlos
42
edited Dec 20 '18 at 15:14
saz
80.7k860125
asked Dec 19 '18 at 17:17
Giancarlos Giancarlos
42
edited Dec 20 '18 at 15:14
saz
80.7k860125
edited Dec 20 '18 at 15:14
saz
80.7k860125
edited Dec 20 '18 at 15:14
saz
80.7k860125
80.7k860125
asked Dec 19 '18 at 17:17
Giancarlos Giancarlos
42
asked Dec 19 '18 at 17:17
Giancarlos Giancarlos
42
asked Dec 19 '18 at 17:17
Giancarlos Giancarlos
42
42
closed as off-topic by Davide Giraudo, Adrian Keister, Lord_Farin, amWhy, Dando18 Dec 21 '18 at 17:00
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Davide Giraudo, Adrian Keister, Lord_Farin, amWhy, Dando18
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Davide Giraudo, Adrian Keister, Lord_Farin, amWhy, Dando18 Dec 21 '18 at 17:00
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Davide Giraudo, Adrian Keister, Lord_Farin, amWhy, Dando18
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
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1 Answer
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$begingroup$
Denote by $mu$ the associated probability measure, i.e. $$psi(t) = int e^{ixt} , mu(dx).$$
Since
$$|psi(t)-psi(s)|^2 = int (sin(tx)-sin(sx))^2 , mu(dx) + int (cos(tx)-cos(sx))^2 , mu(dx)$$
it follows from
$$sin u- sin v = 2 cos frac{u+v}{2} sin frac{u-v}{2} qquad cos u - cos v = - 2 sin frac{u+v}{2} sin frac{u-v}{2}$$
that
$$|psi(t)-psi(s)|^2 leq 4 int sin^2 frac{(t-s)x}{2} , mu(dx).$$
As
$$sin^2(u) = frac{1}{2} (1-cos(2u))$$
this gives
$$|psi(t)-psi(s)|^2 leq 2 int (1-cos((t-s)x)) , mu(dx) = 2 |1-text{Re} , psi(t-s)|;$$
in particular,
$$|psi(t)-psi(s)|^2 leq 2 |1-psi(t-s)|.$$
answered Dec 20 '18 at 9:05
sazsaz
80.7k860125
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Denote by $mu$ the associated probability measure, i.e. $$psi(t) = int e^{ixt} , mu(dx).$$
Since
$$|psi(t)-psi(s)|^2 = int (sin(tx)-sin(sx))^2 , mu(dx) + int (cos(tx)-cos(sx))^2 , mu(dx)$$
it follows from
$$sin u- sin v = 2 cos frac{u+v}{2} sin frac{u-v}{2} qquad cos u - cos v = - 2 sin frac{u+v}{2} sin frac{u-v}{2}$$
that
$$|psi(t)-psi(s)|^2 leq 4 int sin^2 frac{(t-s)x}{2} , mu(dx).$$
As
$$sin^2(u) = frac{1}{2} (1-cos(2u))$$
this gives
$$|psi(t)-psi(s)|^2 leq 2 int (1-cos((t-s)x)) , mu(dx) = 2 |1-text{Re} , psi(t-s)|;$$
in particular,
$$|psi(t)-psi(s)|^2 leq 2 |1-psi(t-s)|.$$
answered Dec 20 '18 at 9:05
sazsaz
80.7k860125
$endgroup$
add a comment |
$begingroup$
Denote by $mu$ the associated probability measure, i.e. $$psi(t) = int e^{ixt} , mu(dx).$$
Since
$$|psi(t)-psi(s)|^2 = int (sin(tx)-sin(sx))^2 , mu(dx) + int (cos(tx)-cos(sx))^2 , mu(dx)$$
it follows from
$$sin u- sin v = 2 cos frac{u+v}{2} sin frac{u-v}{2} qquad cos u - cos v = - 2 sin frac{u+v}{2} sin frac{u-v}{2}$$
that
$$|psi(t)-psi(s)|^2 leq 4 int sin^2 frac{(t-s)x}{2} , mu(dx).$$
As
$$sin^2(u) = frac{1}{2} (1-cos(2u))$$
this gives
$$|psi(t)-psi(s)|^2 leq 2 int (1-cos((t-s)x)) , mu(dx) = 2 |1-text{Re} , psi(t-s)|;$$
in particular,
$$|psi(t)-psi(s)|^2 leq 2 |1-psi(t-s)|.$$
answered Dec 20 '18 at 9:05
sazsaz
80.7k860125
$endgroup$
add a comment |
$begingroup$
Denote by $mu$ the associated probability measure, i.e. $$psi(t) = int e^{ixt} , mu(dx).$$
Since
$$|psi(t)-psi(s)|^2 = int (sin(tx)-sin(sx))^2 , mu(dx) + int (cos(tx)-cos(sx))^2 , mu(dx)$$
it follows from
$$sin u- sin v = 2 cos frac{u+v}{2} sin frac{u-v}{2} qquad cos u - cos v = - 2 sin frac{u+v}{2} sin frac{u-v}{2}$$
that
$$|psi(t)-psi(s)|^2 leq 4 int sin^2 frac{(t-s)x}{2} , mu(dx).$$
As
$$sin^2(u) = frac{1}{2} (1-cos(2u))$$
this gives
$$|psi(t)-psi(s)|^2 leq 2 int (1-cos((t-s)x)) , mu(dx) = 2 |1-text{Re} , psi(t-s)|;$$
in particular,
$$|psi(t)-psi(s)|^2 leq 2 |1-psi(t-s)|.$$
answered Dec 20 '18 at 9:05
sazsaz
80.7k860125
$endgroup$
Denote by $mu$ the associated probability measure, i.e. $$psi(t) = int e^{ixt} , mu(dx).$$
Since
$$|psi(t)-psi(s)|^2 = int (sin(tx)-sin(sx))^2 , mu(dx) + int (cos(tx)-cos(sx))^2 , mu(dx)$$
it follows from
$$sin u- sin v = 2 cos frac{u+v}{2} sin frac{u-v}{2} qquad cos u - cos v = - 2 sin frac{u+v}{2} sin frac{u-v}{2}$$
that
$$|psi(t)-psi(s)|^2 leq 4 int sin^2 frac{(t-s)x}{2} , mu(dx).$$
As
$$sin^2(u) = frac{1}{2} (1-cos(2u))$$
this gives
$$|psi(t)-psi(s)|^2 leq 2 int (1-cos((t-s)x)) , mu(dx) = 2 |1-text{Re} , psi(t-s)|;$$
in particular,
$$|psi(t)-psi(s)|^2 leq 2 |1-psi(t-s)|.$$
answered Dec 20 '18 at 9:05
sazsaz
80.7k860125
answered Dec 20 '18 at 9:05
sazsaz
80.7k860125
answered Dec 20 '18 at 9:05
sazsaz
80.7k860125
answered Dec 20 '18 at 9:05
sazsaz
80.7k860125
80.7k860125
add a comment |
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