How can this adjustment of sum be correct?












2












$begingroup$


In our Probability and statistics materials, I run into this equation:equation



$X_i$ is one result (I hope)



$n$ is a count of the results



$overline{X_n}$ is sample mean of the results



It is connected with point estimations, we are trying to estimate variance, and this adjustment should help. But I don't understand how we can do it.



I expect it has something to do with this adjustment:




$(a-b)^2=a^2-2ab+b^2$




But I just don't see how (especially the minus sign confuses me). I understand that the $(overline{X_n})^2$ was removed from the sum. But where is the $2ab$ part?



So, How can the adjustment be accurate?






algebra-precalculus statistics estimation-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    If you expand the sum you will see how.
    $endgroup$
    – John Douma
    Jan 12 at 20:36
















2












$begingroup$


In our Probability and statistics materials, I run into this equation:equation



$X_i$ is one result (I hope)



$n$ is a count of the results



$overline{X_n}$ is sample mean of the results



It is connected with point estimations, we are trying to estimate variance, and this adjustment should help. But I don't understand how we can do it.



I expect it has something to do with this adjustment:




$(a-b)^2=a^2-2ab+b^2$




But I just don't see how (especially the minus sign confuses me). I understand that the $(overline{X_n})^2$ was removed from the sum. But where is the $2ab$ part?



So, How can the adjustment be accurate?






algebra-precalculus statistics estimation-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    If you expand the sum you will see how.
    $endgroup$
    – John Douma
    Jan 12 at 20:36














2












2








2





$begingroup$


In our Probability and statistics materials, I run into this equation:equation



$X_i$ is one result (I hope)



$n$ is a count of the results



$overline{X_n}$ is sample mean of the results



It is connected with point estimations, we are trying to estimate variance, and this adjustment should help. But I don't understand how we can do it.



I expect it has something to do with this adjustment:




$(a-b)^2=a^2-2ab+b^2$




But I just don't see how (especially the minus sign confuses me). I understand that the $(overline{X_n})^2$ was removed from the sum. But where is the $2ab$ part?



So, How can the adjustment be accurate?






algebra-precalculus statistics estimation-theory







share|cite|improve this question











$endgroup$




In our Probability and statistics materials, I run into this equation:equation



$X_i$ is one result (I hope)



$n$ is a count of the results



$overline{X_n}$ is sample mean of the results



It is connected with point estimations, we are trying to estimate variance, and this adjustment should help. But I don't understand how we can do it.



I expect it has something to do with this adjustment:




$(a-b)^2=a^2-2ab+b^2$




But I just don't see how (especially the minus sign confuses me). I understand that the $(overline{X_n})^2$ was removed from the sum. But where is the $2ab$ part?



So, How can the adjustment be accurate?






algebra-precalculus statistics estimation-theory




algebra-precalculus statistics estimation-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 12 at 20:39







TGar

















asked Jan 12 at 20:27









TGarTGar

1558




1558












  • $begingroup$
    If you expand the sum you will see how.
    $endgroup$
    – John Douma
    Jan 12 at 20:36


















  • $begingroup$
    If you expand the sum you will see how.
    $endgroup$
    – John Douma
    Jan 12 at 20:36
















$begingroup$
If you expand the sum you will see how.
$endgroup$
– John Douma
Jan 12 at 20:36




$begingroup$
If you expand the sum you will see how.
$endgroup$
– John Douma
Jan 12 at 20:36










1 Answer
1






active

oldest

votes


















4












$begingroup$

It's actually pretty simple but the brackets in your picture might be a bit confusing. For simplicity reasons I'll only look at the sum.



We know that $$sum^{n}_{i=1}(X_i - bar{X_{n}})^2=(sum^{n}_{i=1}X_i^2)-2bar{X_n}sum^{n}_{i=1}X_i+sum^{n}_{i=1}bar{X_n}^2$$



Which in turn reduces to $$(sum^{n}_{i=1}X_i^2)-2bar{X_n}sum^{n}_{i=1}X_i+sum^{n}_{i=1}bar{X_n}^2=(sum^{n}_{i=1}X_i^2)-2bar{X_n}nbar{X_{n}}+nbar{X_n}^2$$



Because ofcourse we know that $frac{1}{n}sum^{n}_{i=1}X_i=bar{X_n}$ so $sum^{n}_{i=1}X_i=nbar{X_n}$. Here is now the thing we want because $$(sum^{n}_{i=1}X_i^2)-2bar{X_n}nbar{X_{n}}+nbar{X_n}^2=(sum^{n}_{i=1}X_i^2)-nbar{X_n}^2$$



Put the constant of $frac{1}{1-n}$ that we left out for simplicity, back in front, and you're done!






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How can you put $2bar{Xn}$ outside the sum in the first part?
    $endgroup$
    – TGar
    Jan 12 at 20:42










  • $begingroup$
    Because there is no subscript $i$ anymore, so you are not summing over anything. Instead, try looking at the $bar{X_n}$ as just a constant $c$ (ofcourse it isn't actually constant, but for simplicity). Do you agree that you could then take this constant $c$ out of the sum?
    $endgroup$
    – S. Crim
    Jan 12 at 20:43












  • $begingroup$
    Sure, of course, you're right! I understand. Thanks!
    $endgroup$
    – TGar
    Jan 12 at 20:45











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',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3071349%2fhow-can-this-adjustment-of-sum-be-correct%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

It's actually pretty simple but the brackets in your picture might be a bit confusing. For simplicity reasons I'll only look at the sum.



We know that $$sum^{n}_{i=1}(X_i - bar{X_{n}})^2=(sum^{n}_{i=1}X_i^2)-2bar{X_n}sum^{n}_{i=1}X_i+sum^{n}_{i=1}bar{X_n}^2$$



Which in turn reduces to $$(sum^{n}_{i=1}X_i^2)-2bar{X_n}sum^{n}_{i=1}X_i+sum^{n}_{i=1}bar{X_n}^2=(sum^{n}_{i=1}X_i^2)-2bar{X_n}nbar{X_{n}}+nbar{X_n}^2$$



Because ofcourse we know that $frac{1}{n}sum^{n}_{i=1}X_i=bar{X_n}$ so $sum^{n}_{i=1}X_i=nbar{X_n}$. Here is now the thing we want because $$(sum^{n}_{i=1}X_i^2)-2bar{X_n}nbar{X_{n}}+nbar{X_n}^2=(sum^{n}_{i=1}X_i^2)-nbar{X_n}^2$$



Put the constant of $frac{1}{1-n}$ that we left out for simplicity, back in front, and you're done!






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How can you put $2bar{Xn}$ outside the sum in the first part?
    $endgroup$
    – TGar
    Jan 12 at 20:42










  • $begingroup$
    Because there is no subscript $i$ anymore, so you are not summing over anything. Instead, try looking at the $bar{X_n}$ as just a constant $c$ (ofcourse it isn't actually constant, but for simplicity). Do you agree that you could then take this constant $c$ out of the sum?
    $endgroup$
    – S. Crim
    Jan 12 at 20:43












  • $begingroup$
    Sure, of course, you're right! I understand. Thanks!
    $endgroup$
    – TGar
    Jan 12 at 20:45
















4












$begingroup$

It's actually pretty simple but the brackets in your picture might be a bit confusing. For simplicity reasons I'll only look at the sum.



We know that $$sum^{n}_{i=1}(X_i - bar{X_{n}})^2=(sum^{n}_{i=1}X_i^2)-2bar{X_n}sum^{n}_{i=1}X_i+sum^{n}_{i=1}bar{X_n}^2$$



Which in turn reduces to $$(sum^{n}_{i=1}X_i^2)-2bar{X_n}sum^{n}_{i=1}X_i+sum^{n}_{i=1}bar{X_n}^2=(sum^{n}_{i=1}X_i^2)-2bar{X_n}nbar{X_{n}}+nbar{X_n}^2$$



Because ofcourse we know that $frac{1}{n}sum^{n}_{i=1}X_i=bar{X_n}$ so $sum^{n}_{i=1}X_i=nbar{X_n}$. Here is now the thing we want because $$(sum^{n}_{i=1}X_i^2)-2bar{X_n}nbar{X_{n}}+nbar{X_n}^2=(sum^{n}_{i=1}X_i^2)-nbar{X_n}^2$$



Put the constant of $frac{1}{1-n}$ that we left out for simplicity, back in front, and you're done!






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How can you put $2bar{Xn}$ outside the sum in the first part?
    $endgroup$
    – TGar
    Jan 12 at 20:42










  • $begingroup$
    Because there is no subscript $i$ anymore, so you are not summing over anything. Instead, try looking at the $bar{X_n}$ as just a constant $c$ (ofcourse it isn't actually constant, but for simplicity). Do you agree that you could then take this constant $c$ out of the sum?
    $endgroup$
    – S. Crim
    Jan 12 at 20:43












  • $begingroup$
    Sure, of course, you're right! I understand. Thanks!
    $endgroup$
    – TGar
    Jan 12 at 20:45














4












4








4





$begingroup$

It's actually pretty simple but the brackets in your picture might be a bit confusing. For simplicity reasons I'll only look at the sum.



We know that $$sum^{n}_{i=1}(X_i - bar{X_{n}})^2=(sum^{n}_{i=1}X_i^2)-2bar{X_n}sum^{n}_{i=1}X_i+sum^{n}_{i=1}bar{X_n}^2$$



Which in turn reduces to $$(sum^{n}_{i=1}X_i^2)-2bar{X_n}sum^{n}_{i=1}X_i+sum^{n}_{i=1}bar{X_n}^2=(sum^{n}_{i=1}X_i^2)-2bar{X_n}nbar{X_{n}}+nbar{X_n}^2$$



Because ofcourse we know that $frac{1}{n}sum^{n}_{i=1}X_i=bar{X_n}$ so $sum^{n}_{i=1}X_i=nbar{X_n}$. Here is now the thing we want because $$(sum^{n}_{i=1}X_i^2)-2bar{X_n}nbar{X_{n}}+nbar{X_n}^2=(sum^{n}_{i=1}X_i^2)-nbar{X_n}^2$$



Put the constant of $frac{1}{1-n}$ that we left out for simplicity, back in front, and you're done!






share|cite|improve this answer









$endgroup$



It's actually pretty simple but the brackets in your picture might be a bit confusing. For simplicity reasons I'll only look at the sum.



We know that $$sum^{n}_{i=1}(X_i - bar{X_{n}})^2=(sum^{n}_{i=1}X_i^2)-2bar{X_n}sum^{n}_{i=1}X_i+sum^{n}_{i=1}bar{X_n}^2$$



Which in turn reduces to $$(sum^{n}_{i=1}X_i^2)-2bar{X_n}sum^{n}_{i=1}X_i+sum^{n}_{i=1}bar{X_n}^2=(sum^{n}_{i=1}X_i^2)-2bar{X_n}nbar{X_{n}}+nbar{X_n}^2$$



Because ofcourse we know that $frac{1}{n}sum^{n}_{i=1}X_i=bar{X_n}$ so $sum^{n}_{i=1}X_i=nbar{X_n}$. Here is now the thing we want because $$(sum^{n}_{i=1}X_i^2)-2bar{X_n}nbar{X_{n}}+nbar{X_n}^2=(sum^{n}_{i=1}X_i^2)-nbar{X_n}^2$$



Put the constant of $frac{1}{1-n}$ that we left out for simplicity, back in front, and you're done!







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 12 at 20:38









S. CrimS. Crim

404212




404212












  • $begingroup$
    How can you put $2bar{Xn}$ outside the sum in the first part?
    $endgroup$
    – TGar
    Jan 12 at 20:42










  • $begingroup$
    Because there is no subscript $i$ anymore, so you are not summing over anything. Instead, try looking at the $bar{X_n}$ as just a constant $c$ (ofcourse it isn't actually constant, but for simplicity). Do you agree that you could then take this constant $c$ out of the sum?
    $endgroup$
    – S. Crim
    Jan 12 at 20:43












  • $begingroup$
    Sure, of course, you're right! I understand. Thanks!
    $endgroup$
    – TGar
    Jan 12 at 20:45


















  • $begingroup$
    How can you put $2bar{Xn}$ outside the sum in the first part?
    $endgroup$
    – TGar
    Jan 12 at 20:42










  • $begingroup$
    Because there is no subscript $i$ anymore, so you are not summing over anything. Instead, try looking at the $bar{X_n}$ as just a constant $c$ (ofcourse it isn't actually constant, but for simplicity). Do you agree that you could then take this constant $c$ out of the sum?
    $endgroup$
    – S. Crim
    Jan 12 at 20:43












  • $begingroup$
    Sure, of course, you're right! I understand. Thanks!
    $endgroup$
    – TGar
    Jan 12 at 20:45
















$begingroup$
How can you put $2bar{Xn}$ outside the sum in the first part?
$endgroup$
– TGar
Jan 12 at 20:42




$begingroup$
How can you put $2bar{Xn}$ outside the sum in the first part?
$endgroup$
– TGar
Jan 12 at 20:42












$begingroup$
Because there is no subscript $i$ anymore, so you are not summing over anything. Instead, try looking at the $bar{X_n}$ as just a constant $c$ (ofcourse it isn't actually constant, but for simplicity). Do you agree that you could then take this constant $c$ out of the sum?
$endgroup$
– S. Crim
Jan 12 at 20:43






$begingroup$
Because there is no subscript $i$ anymore, so you are not summing over anything. Instead, try looking at the $bar{X_n}$ as just a constant $c$ (ofcourse it isn't actually constant, but for simplicity). Do you agree that you could then take this constant $c$ out of the sum?
$endgroup$
– S. Crim
Jan 12 at 20:43














$begingroup$
Sure, of course, you're right! I understand. Thanks!
$endgroup$
– TGar
Jan 12 at 20:45




$begingroup$
Sure, of course, you're right! I understand. Thanks!
$endgroup$
– TGar
Jan 12 at 20:45


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3071349%2fhow-can-this-adjustment-of-sum-be-correct%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Human spaceflight

Image
Human spaceflight From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Space traveler" redirects here. For other uses, see Space traveler (disambiguation). @media all and (max-width:720px){.mw-parser-output .tmulti>.thumbinner{width:100%!important;max-width:none!important}.mw-parser-output .tmulti .tsingle{float:none!important;max-width:none!important;width:100%!important;text-align:center}} Apollo 11 crewmember Buzz Aldrin on the Moon, 1969 International Space Station crewmember Tracy Caldwell Dyson views the Earth, 2010 Space Shuttle Discovery heads into space with a crew aboard, STS-121 in 2006 Inside a space suit on the Canadarm, 1993 Human spaceflight (also referred to as crewed spaceflight or manned spaceflight ) is space travel with a crew or passengers aboard the spacecraft. Spacecraft carrying people may be operated directly, by human crew, or it may be e
Read more

Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

Image
0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB]
Read more

File:DeusFollowingSea.jpg

Image
File:DeusFollowingSea.jpg From Wikipedia, the free encyclopedia Jump to navigation Jump to search File File history File usage No higher resolution available. DeusFollowingSea.jpg ‎ (238 × 211 pixels, file size: 12 KB, MIME type: image/jpeg ) Summary [ edit ] Media data and Non-free use rationale Description Cover art of the Deus CD Following Sea Author or copyright owner Deus Source (WP:NFCC#4) rocksucker.co.uk Date of publication 2012 Use in article (WP:NFCC#7) Following Sea Purpose of use in article (WP:NFCC#8) to serve as the primary means of visual identification at the top of the article dedicated to the work in question. Not replaceable with free media because (WP:NFCC#1) n.a. Minimal use (WP:NFCC#3) Low res image used in infobox of the article on the album Respect for commercial opportunities (WP:NFCC#2) n.a. Fair use Fair use of co
Read more