Integration limits of central moments

Clash Royale CLAN TAG#URR8PPP

up vote
0
down vote

favorite

Could anyone clarify as to why central moments have integration limits from 0 to infinity rather than minus infinity to positive infinity? I am asking with particular reference to moment analysis chromatographic peaks (a distribution of a compound's concentration as a function of time). The chromatographic peaks are slightly asymmetric Gaussians which appear at a certain time "t" (>0) as an output of a detector. This characteristic time is the first moment M1. The area under the peak is taken as the zeroth moment. Most authors use the limit from 0 to infinity, whereas Wikipedia definition shows minus infinity to plus infinity as limits for central moments. The choice of minus infinity to plus infinity makes more sense, what is the logic of using 0 to infinity? Please see the attached equation or what is the assumption behind this choice.

Generalized central moments

Thanks.

calculus statistics probability-distributions definite-integrals

share|cite|improve this question

asked Jul 29 at 18:26

M. Farooq

83

add a comment | 

up vote
0
down vote

favorite

Could anyone clarify as to why central moments have integration limits from 0 to infinity rather than minus infinity to positive infinity? I am asking with particular reference to moment analysis chromatographic peaks (a distribution of a compound's concentration as a function of time). The chromatographic peaks are slightly asymmetric Gaussians which appear at a certain time "t" (>0) as an output of a detector. This characteristic time is the first moment M1. The area under the peak is taken as the zeroth moment. Most authors use the limit from 0 to infinity, whereas Wikipedia definition shows minus infinity to plus infinity as limits for central moments. The choice of minus infinity to plus infinity makes more sense, what is the logic of using 0 to infinity? Please see the attached equation or what is the assumption behind this choice.

Generalized central moments

Thanks.

calculus statistics probability-distributions definite-integrals

share|cite|improve this question

asked Jul 29 at 18:26

M. Farooq

83

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Could anyone clarify as to why central moments have integration limits from 0 to infinity rather than minus infinity to positive infinity? I am asking with particular reference to moment analysis chromatographic peaks (a distribution of a compound's concentration as a function of time). The chromatographic peaks are slightly asymmetric Gaussians which appear at a certain time "t" (>0) as an output of a detector. This characteristic time is the first moment M1. The area under the peak is taken as the zeroth moment. Most authors use the limit from 0 to infinity, whereas Wikipedia definition shows minus infinity to plus infinity as limits for central moments. The choice of minus infinity to plus infinity makes more sense, what is the logic of using 0 to infinity? Please see the attached equation or what is the assumption behind this choice.

Generalized central moments

Thanks.

calculus statistics probability-distributions definite-integrals

share|cite|improve this question

asked Jul 29 at 18:26

M. Farooq

83

Could anyone clarify as to why central moments have integration limits from 0 to infinity rather than minus infinity to positive infinity? I am asking with particular reference to moment analysis chromatographic peaks (a distribution of a compound's concentration as a function of time). The chromatographic peaks are slightly asymmetric Gaussians which appear at a certain time "t" (>0) as an output of a detector. This characteristic time is the first moment M1. The area under the peak is taken as the zeroth moment. Most authors use the limit from 0 to infinity, whereas Wikipedia definition shows minus infinity to plus infinity as limits for central moments. The choice of minus infinity to plus infinity makes more sense, what is the logic of using 0 to infinity? Please see the attached equation or what is the assumption behind this choice.

Generalized central moments

Thanks.

calculus statistics probability-distributions definite-integrals

share|cite|improve this question

asked Jul 29 at 18:26

M. Farooq

83

share|cite|improve this question

share|cite|improve this question

asked Jul 29 at 18:26

M. Farooq

83

asked Jul 29 at 18:26

M. Farooq

83

asked Jul 29 at 18:26

M. Farooq

83

83

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
0
down vote



accepted

Assuming that your "chromatographic peaks" are represented by the $c(t)$ in your screenshot, then both formulas are equivalent. If you integrated with boundaries $pm infty$, the whole domain from $-infty$ to $0$ would just give you zero, as your peaks do not show up until $t > 0$.
In general, if you had probability density functions $c$ which can also take values for $t leq 0$, you need the formula with boundaries $pm infty$, because to calculate the moments, you really need all the information that your density function contains. But in your case, the formula from your screenshot does the job.

share|cite|improve this answer

answered Jul 29 at 18:39

Lukas Miristwhisky

325111

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks. Yes, c(t) represents the concentration profile and the condition is that t>0. That makes sense given that minus infinity to zero by default is all zero, so minus infinity is not needed. I sometimes use the summation version of this screenshot, and my limits are t1 to t2, where t1 is the start of the peak and t2 is the end of the peak (where detector noise is the same on both ends of the peak, t2>t1, t>0). This should be a good approximation as well.
  – M. Farooq
  Jul 29 at 19:04
  

  
  
  
  

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

 

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2866286%2fintegration-limits-of-central-moments%23new-answer', 'question_page');

);

Post as a guest

Name Email

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote



accepted

Assuming that your "chromatographic peaks" are represented by the $c(t)$ in your screenshot, then both formulas are equivalent. If you integrated with boundaries $pm infty$, the whole domain from $-infty$ to $0$ would just give you zero, as your peaks do not show up until $t > 0$.
In general, if you had probability density functions $c$ which can also take values for $t leq 0$, you need the formula with boundaries $pm infty$, because to calculate the moments, you really need all the information that your density function contains. But in your case, the formula from your screenshot does the job.

share|cite|improve this answer

answered Jul 29 at 18:39

Lukas Miristwhisky

325111

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks. Yes, c(t) represents the concentration profile and the condition is that t>0. That makes sense given that minus infinity to zero by default is all zero, so minus infinity is not needed. I sometimes use the summation version of this screenshot, and my limits are t1 to t2, where t1 is the start of the peak and t2 is the end of the peak (where detector noise is the same on both ends of the peak, t2>t1, t>0). This should be a good approximation as well.
  – M. Farooq
  Jul 29 at 19:04
  

  
  
  
  

add a comment | 

up vote
0
down vote



accepted

Assuming that your "chromatographic peaks" are represented by the $c(t)$ in your screenshot, then both formulas are equivalent. If you integrated with boundaries $pm infty$, the whole domain from $-infty$ to $0$ would just give you zero, as your peaks do not show up until $t > 0$.
In general, if you had probability density functions $c$ which can also take values for $t leq 0$, you need the formula with boundaries $pm infty$, because to calculate the moments, you really need all the information that your density function contains. But in your case, the formula from your screenshot does the job.

share|cite|improve this answer

answered Jul 29 at 18:39

Lukas Miristwhisky

325111

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks. Yes, c(t) represents the concentration profile and the condition is that t>0. That makes sense given that minus infinity to zero by default is all zero, so minus infinity is not needed. I sometimes use the summation version of this screenshot, and my limits are t1 to t2, where t1 is the start of the peak and t2 is the end of the peak (where detector noise is the same on both ends of the peak, t2>t1, t>0). This should be a good approximation as well.
  – M. Farooq
  Jul 29 at 19:04
  

  
  
  
  

add a comment | 

up vote
0
down vote



accepted

up vote
0
down vote



accepted

Assuming that your "chromatographic peaks" are represented by the $c(t)$ in your screenshot, then both formulas are equivalent. If you integrated with boundaries $pm infty$, the whole domain from $-infty$ to $0$ would just give you zero, as your peaks do not show up until $t > 0$.
In general, if you had probability density functions $c$ which can also take values for $t leq 0$, you need the formula with boundaries $pm infty$, because to calculate the moments, you really need all the information that your density function contains. But in your case, the formula from your screenshot does the job.

share|cite|improve this answer

answered Jul 29 at 18:39

Lukas Miristwhisky

325111

Assuming that your "chromatographic peaks" are represented by the $c(t)$ in your screenshot, then both formulas are equivalent. If you integrated with boundaries $pm infty$, the whole domain from $-infty$ to $0$ would just give you zero, as your peaks do not show up until $t > 0$.
In general, if you had probability density functions $c$ which can also take values for $t leq 0$, you need the formula with boundaries $pm infty$, because to calculate the moments, you really need all the information that your density function contains. But in your case, the formula from your screenshot does the job.

share|cite|improve this answer

answered Jul 29 at 18:39

Lukas Miristwhisky

325111

share|cite|improve this answer

share|cite|improve this answer

answered Jul 29 at 18:39

Lukas Miristwhisky

325111

answered Jul 29 at 18:39

Lukas Miristwhisky

325111

answered Jul 29 at 18:39

Lukas Miristwhisky

325111

325111

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks. Yes, c(t) represents the concentration profile and the condition is that t>0. That makes sense given that minus infinity to zero by default is all zero, so minus infinity is not needed. I sometimes use the summation version of this screenshot, and my limits are t1 to t2, where t1 is the start of the peak and t2 is the end of the peak (where detector noise is the same on both ends of the peak, t2>t1, t>0). This should be a good approximation as well.
  – M. Farooq
  Jul 29 at 19:04
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks. Yes, c(t) represents the concentration profile and the condition is that t>0. That makes sense given that minus infinity to zero by default is all zero, so minus infinity is not needed. I sometimes use the summation version of this screenshot, and my limits are t1 to t2, where t1 is the start of the peak and t2 is the end of the peak (where detector noise is the same on both ends of the peak, t2>t1, t>0). This should be a good approximation as well.
  – M. Farooq
  Jul 29 at 19:04
  

  
  
  
  

Thanks. Yes, c(t) represents the concentration profile and the condition is that t>0. That makes sense given that minus infinity to zero by default is all zero, so minus infinity is not needed. I sometimes use the summation version of this screenshot, and my limits are t1 to t2, where t1 is the start of the peak and t2 is the end of the peak (where detector noise is the same on both ends of the peak, t2>t1, t>0). This should be a good approximation as well.
– M. Farooq
Jul 29 at 19:04

Thanks. Yes, c(t) represents the concentration profile and the condition is that t>0. That makes sense given that minus infinity to zero by default is all zero, so minus infinity is not needed. I sometimes use the summation version of this screenshot, and my limits are t1 to t2, where t1 is the start of the peak and t2 is the end of the peak (where detector noise is the same on both ends of the peak, t2>t1, t>0). This should be a good approximation as well.
– M. Farooq
Jul 29 at 19:04

add a comment | 

 

draft saved

draft discarded

 

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2866286%2fintegration-limits-of-central-moments%23new-answer', 'question_page');

);

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Name Email

Name

Email

Name Email

Name

Email