Coefficient of determination, why?












1












$begingroup$


I mean it is written in a book "Statistics for Management and Economics", that coefficient of determination is coefficient of correlation squared. Well, am I the only one to whom this is surprising fact as he expected something more clear or natural?? I mean, if someone can present me the proof why, or why some other, more natural things do not work, like I don't know, absolute value of the coefficient of correlation or something similar to Chebyshev theorem ($1-text{coefficient of correlation}$)?






statistics







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    The title of the question is very confusing, could you please edit it so it is more clear?
    $endgroup$
    – Yuriy S
    Jan 17 at 11:35
















1












$begingroup$


I mean it is written in a book "Statistics for Management and Economics", that coefficient of determination is coefficient of correlation squared. Well, am I the only one to whom this is surprising fact as he expected something more clear or natural?? I mean, if someone can present me the proof why, or why some other, more natural things do not work, like I don't know, absolute value of the coefficient of correlation or something similar to Chebyshev theorem ($1-text{coefficient of correlation}$)?






statistics







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    The title of the question is very confusing, could you please edit it so it is more clear?
    $endgroup$
    – Yuriy S
    Jan 17 at 11:35














1












1








1


1



$begingroup$


I mean it is written in a book "Statistics for Management and Economics", that coefficient of determination is coefficient of correlation squared. Well, am I the only one to whom this is surprising fact as he expected something more clear or natural?? I mean, if someone can present me the proof why, or why some other, more natural things do not work, like I don't know, absolute value of the coefficient of correlation or something similar to Chebyshev theorem ($1-text{coefficient of correlation}$)?






statistics







share|cite|improve this question











$endgroup$




I mean it is written in a book "Statistics for Management and Economics", that coefficient of determination is coefficient of correlation squared. Well, am I the only one to whom this is surprising fact as he expected something more clear or natural?? I mean, if someone can present me the proof why, or why some other, more natural things do not work, like I don't know, absolute value of the coefficient of correlation or something similar to Chebyshev theorem ($1-text{coefficient of correlation}$)?






statistics




statistics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 17 at 11:03









Christoph

12k1642




12k1642










asked Jan 17 at 10:54









nikolanikola

641314




641314








  • 1




    $begingroup$
    The title of the question is very confusing, could you please edit it so it is more clear?
    $endgroup$
    – Yuriy S
    Jan 17 at 11:35














  • 1




    $begingroup$
    The title of the question is very confusing, could you please edit it so it is more clear?
    $endgroup$
    – Yuriy S
    Jan 17 at 11:35








1




1




$begingroup$
The title of the question is very confusing, could you please edit it so it is more clear?
$endgroup$
– Yuriy S
Jan 17 at 11:35




$begingroup$
The title of the question is very confusing, could you please edit it so it is more clear?
$endgroup$
– Yuriy S
Jan 17 at 11:35










1 Answer
1






active

oldest

votes


















2












$begingroup$

Suppose you have $n$ paired observations $(x_i,y_i)$ on $(x,y)$ and you want to predict $y$ on the basis of $x$.



You consider the prediction model $$y=phi(x)+e$$ where $phi$ is the part of $y$ explained by $x$ through $phi$ and $e$ is the unexplained part.



Suppose you see from the scatter plot of $x$ and $y$ that $phi$ is more or less linear.



So you choose $$phi(x)=a+bx$$



Then the least square linear predictor of $y$ obtained on the basis of $x$ is $$hat y=hat a+hat b x$$, where
$$hat a=bar y-hat bbar xquad,quad hat b=frac{operatorname{cov}(x,y)}{operatorname{var}(x)}$$



It can be shown that



begin{align}
operatorname{var}(hat y)&=operatorname{var}(hat a+hat b x)
\&=hat b^2operatorname{var}(x)
\&=r^2 operatorname{var}(y)
end{align}



, where $r$ is the correlation coefficient between $x$ and $y$.




A measure of efficacy of the predictor $phi$ is given by the proportion of variation in $y$ explained by $phi$, i.e., $$frac{operatorname{var}(hat y)}{operatorname{var}(y)}=r^2$$, which is termed as coefficient of determination.




Of course, the coefficient of determination is numerically equal to the square of the correlation coefficient, but that is hardly a definition or a motivation for the former.



When $r^2=0$ the linear prediction of $y$ obtained on the basis of $x$ is worst, and when $r^2=1$ the prediction is perfect as $phi$ explains the variability in $y$ completely.



For more details, the following threads might be helpful:




  • Correlation Coefficient and Determination Coefficient


  • https://stats.stackexchange.com/questions/123651/geometric-interpretation-of-multiple-correlation-coefficient-r-and-coefficient?noredirect=1&lq=1


  • https://stats.stackexchange.com/questions/1447/coefficient-of-determination-r2-i-have-never-fully-grasped-the-interpretat?noredirect=1&lq=1.







share|cite|improve this answer









$endgroup$













  • $begingroup$
    Excellent answer
    $endgroup$
    – lux
    Jan 18 at 1:07











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%2f3076842%2fcoefficient-of-determination-why%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









2












$begingroup$

Suppose you have $n$ paired observations $(x_i,y_i)$ on $(x,y)$ and you want to predict $y$ on the basis of $x$.



You consider the prediction model $$y=phi(x)+e$$ where $phi$ is the part of $y$ explained by $x$ through $phi$ and $e$ is the unexplained part.



Suppose you see from the scatter plot of $x$ and $y$ that $phi$ is more or less linear.



So you choose $$phi(x)=a+bx$$



Then the least square linear predictor of $y$ obtained on the basis of $x$ is $$hat y=hat a+hat b x$$, where
$$hat a=bar y-hat bbar xquad,quad hat b=frac{operatorname{cov}(x,y)}{operatorname{var}(x)}$$



It can be shown that



begin{align}
operatorname{var}(hat y)&=operatorname{var}(hat a+hat b x)
\&=hat b^2operatorname{var}(x)
\&=r^2 operatorname{var}(y)
end{align}



, where $r$ is the correlation coefficient between $x$ and $y$.




A measure of efficacy of the predictor $phi$ is given by the proportion of variation in $y$ explained by $phi$, i.e., $$frac{operatorname{var}(hat y)}{operatorname{var}(y)}=r^2$$, which is termed as coefficient of determination.




Of course, the coefficient of determination is numerically equal to the square of the correlation coefficient, but that is hardly a definition or a motivation for the former.



When $r^2=0$ the linear prediction of $y$ obtained on the basis of $x$ is worst, and when $r^2=1$ the prediction is perfect as $phi$ explains the variability in $y$ completely.



For more details, the following threads might be helpful:




  • Correlation Coefficient and Determination Coefficient


  • https://stats.stackexchange.com/questions/123651/geometric-interpretation-of-multiple-correlation-coefficient-r-and-coefficient?noredirect=1&lq=1


  • https://stats.stackexchange.com/questions/1447/coefficient-of-determination-r2-i-have-never-fully-grasped-the-interpretat?noredirect=1&lq=1.







share|cite|improve this answer









$endgroup$













  • $begingroup$
    Excellent answer
    $endgroup$
    – lux
    Jan 18 at 1:07
















2












$begingroup$

Suppose you have $n$ paired observations $(x_i,y_i)$ on $(x,y)$ and you want to predict $y$ on the basis of $x$.



You consider the prediction model $$y=phi(x)+e$$ where $phi$ is the part of $y$ explained by $x$ through $phi$ and $e$ is the unexplained part.



Suppose you see from the scatter plot of $x$ and $y$ that $phi$ is more or less linear.



So you choose $$phi(x)=a+bx$$



Then the least square linear predictor of $y$ obtained on the basis of $x$ is $$hat y=hat a+hat b x$$, where
$$hat a=bar y-hat bbar xquad,quad hat b=frac{operatorname{cov}(x,y)}{operatorname{var}(x)}$$



It can be shown that



begin{align}
operatorname{var}(hat y)&=operatorname{var}(hat a+hat b x)
\&=hat b^2operatorname{var}(x)
\&=r^2 operatorname{var}(y)
end{align}



, where $r$ is the correlation coefficient between $x$ and $y$.




A measure of efficacy of the predictor $phi$ is given by the proportion of variation in $y$ explained by $phi$, i.e., $$frac{operatorname{var}(hat y)}{operatorname{var}(y)}=r^2$$, which is termed as coefficient of determination.




Of course, the coefficient of determination is numerically equal to the square of the correlation coefficient, but that is hardly a definition or a motivation for the former.



When $r^2=0$ the linear prediction of $y$ obtained on the basis of $x$ is worst, and when $r^2=1$ the prediction is perfect as $phi$ explains the variability in $y$ completely.



For more details, the following threads might be helpful:




  • Correlation Coefficient and Determination Coefficient


  • https://stats.stackexchange.com/questions/123651/geometric-interpretation-of-multiple-correlation-coefficient-r-and-coefficient?noredirect=1&lq=1


  • https://stats.stackexchange.com/questions/1447/coefficient-of-determination-r2-i-have-never-fully-grasped-the-interpretat?noredirect=1&lq=1.







share|cite|improve this answer









$endgroup$













  • $begingroup$
    Excellent answer
    $endgroup$
    – lux
    Jan 18 at 1:07














2












2








2





$begingroup$

Suppose you have $n$ paired observations $(x_i,y_i)$ on $(x,y)$ and you want to predict $y$ on the basis of $x$.



You consider the prediction model $$y=phi(x)+e$$ where $phi$ is the part of $y$ explained by $x$ through $phi$ and $e$ is the unexplained part.



Suppose you see from the scatter plot of $x$ and $y$ that $phi$ is more or less linear.



So you choose $$phi(x)=a+bx$$



Then the least square linear predictor of $y$ obtained on the basis of $x$ is $$hat y=hat a+hat b x$$, where
$$hat a=bar y-hat bbar xquad,quad hat b=frac{operatorname{cov}(x,y)}{operatorname{var}(x)}$$



It can be shown that



begin{align}
operatorname{var}(hat y)&=operatorname{var}(hat a+hat b x)
\&=hat b^2operatorname{var}(x)
\&=r^2 operatorname{var}(y)
end{align}



, where $r$ is the correlation coefficient between $x$ and $y$.




A measure of efficacy of the predictor $phi$ is given by the proportion of variation in $y$ explained by $phi$, i.e., $$frac{operatorname{var}(hat y)}{operatorname{var}(y)}=r^2$$, which is termed as coefficient of determination.




Of course, the coefficient of determination is numerically equal to the square of the correlation coefficient, but that is hardly a definition or a motivation for the former.



When $r^2=0$ the linear prediction of $y$ obtained on the basis of $x$ is worst, and when $r^2=1$ the prediction is perfect as $phi$ explains the variability in $y$ completely.



For more details, the following threads might be helpful:




  • Correlation Coefficient and Determination Coefficient


  • https://stats.stackexchange.com/questions/123651/geometric-interpretation-of-multiple-correlation-coefficient-r-and-coefficient?noredirect=1&lq=1


  • https://stats.stackexchange.com/questions/1447/coefficient-of-determination-r2-i-have-never-fully-grasped-the-interpretat?noredirect=1&lq=1.







share|cite|improve this answer









$endgroup$



Suppose you have $n$ paired observations $(x_i,y_i)$ on $(x,y)$ and you want to predict $y$ on the basis of $x$.



You consider the prediction model $$y=phi(x)+e$$ where $phi$ is the part of $y$ explained by $x$ through $phi$ and $e$ is the unexplained part.



Suppose you see from the scatter plot of $x$ and $y$ that $phi$ is more or less linear.



So you choose $$phi(x)=a+bx$$



Then the least square linear predictor of $y$ obtained on the basis of $x$ is $$hat y=hat a+hat b x$$, where
$$hat a=bar y-hat bbar xquad,quad hat b=frac{operatorname{cov}(x,y)}{operatorname{var}(x)}$$



It can be shown that



begin{align}
operatorname{var}(hat y)&=operatorname{var}(hat a+hat b x)
\&=hat b^2operatorname{var}(x)
\&=r^2 operatorname{var}(y)
end{align}



, where $r$ is the correlation coefficient between $x$ and $y$.




A measure of efficacy of the predictor $phi$ is given by the proportion of variation in $y$ explained by $phi$, i.e., $$frac{operatorname{var}(hat y)}{operatorname{var}(y)}=r^2$$, which is termed as coefficient of determination.




Of course, the coefficient of determination is numerically equal to the square of the correlation coefficient, but that is hardly a definition or a motivation for the former.



When $r^2=0$ the linear prediction of $y$ obtained on the basis of $x$ is worst, and when $r^2=1$ the prediction is perfect as $phi$ explains the variability in $y$ completely.



For more details, the following threads might be helpful:




  • Correlation Coefficient and Determination Coefficient


  • https://stats.stackexchange.com/questions/123651/geometric-interpretation-of-multiple-correlation-coefficient-r-and-coefficient?noredirect=1&lq=1


  • https://stats.stackexchange.com/questions/1447/coefficient-of-determination-r2-i-have-never-fully-grasped-the-interpretat?noredirect=1&lq=1.








share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 17 at 13:02









StubbornAtomStubbornAtom

5,98311238




5,98311238












  • $begingroup$
    Excellent answer
    $endgroup$
    – lux
    Jan 18 at 1:07


















  • $begingroup$
    Excellent answer
    $endgroup$
    – lux
    Jan 18 at 1:07
















$begingroup$
Excellent answer
$endgroup$
– lux
Jan 18 at 1:07




$begingroup$
Excellent answer
$endgroup$
– lux
Jan 18 at 1:07


















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%2f3076842%2fcoefficient-of-determination-why%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

Mario Kart Wii

Image
Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro
Read more

What does “Dominus providebit” mean?

Image
5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1
Read more

Antonio Litta Visconti Arese

Image
Antonio Litta Visconti Arese I duca Litta, VII marchese di Gambolò In carica 1797 – 1820 Predecessore Pompeo Litta Visconti Arese Successore Alfonso Litta Visconti Arese Trattamento Sua grazia Altri titoli Duca Litta Conte del Regno napoleonico d'Italia Patrizio milanese Don Nascita Milano, 2 maggio 1748 Morte Milano, 14 agosto 1820 Dinastia Litta Visconti Arese Padre Pompeo Litta Visconti Arese Madre Maria Elisabetta Visconti Borromeo Consorte Barbara Barbiano di Belgioioso Religione cattolicesimo Antonio Litta Visconti Arese , I duca Litta, VII marchese di Gambolò (Milano, 2 maggio 1748 – Milano, 14 agosto 1820), è stato un nobile e politico italiano. Composizione "Il pastore svizzero" di Pietro Morlacchi, con dedica Indice 1 Biografia 2 Onorificenze 3 Araldica 4 Albero genealogico 5 Note 6 Altri progetti Biografia | Antonio Litta Visco
Read more