Beta parameter estimation in least squares method by partial derivative

Clash Royale CLAN TAG#URR8PPP

up vote
0
down vote

favorite

I was reading a particular document on linear regression and I just can't understand how the guy got to the estimation of the beta parameter. His results are as follows:

$SQ(alpha, beta)=sumlimits_i=1^nepsilon_i^2=sumlimits_i=1^n(y_i-alpha-beta x_i)^2$

$fracpartial SQ(alpha, beta)partial alphavert_alpha= hatalpha = 0 implies sumlimits_i=1^n(y_i - hatalpha - hatbeta x_i) = 0 $

$fracpartial SQ(alpha, beta)partial betavert_beta= hatbeta = 0 implies sumlimits_i=1^n x_i(y_i - hatalpha - hatbeta x_i) = 0$

$sumlimits_i=1^n y_i = n hatalpha + hatbeta sumlimits_i=1^n x_i implies hatalpha = bary - hatbetabarx$

$sumlimits_i=1^n x_i y_i = hatalpha sumlimits_i=1^n x_i + hatbeta sumlimits_i=1^n x_i^2 implies hatbeta = fracsumlimits_i=1^n x_i y_i - nbarxbarysumlimits_i=1^n x_i^2 - nbarx^2$

However, my result of the partial derivative with beta fixed is $hatbeta = fracsumlimits_i=1^nx_iy_i - barxbarysumlimits_i=1^nx_i^2-barx^2$

What is going on? What did I miss?

least-squares linear-regression

share|cite|improve this question

edited Jul 26 at 14:08

asked Jul 25 at 23:55

Luís Muniz

1012

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
  – José Carlos Santos
  Jul 26 at 0:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  No prob. I'll edit the question!
  – Luís Muniz
  Jul 26 at 13:37
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

I was reading a particular document on linear regression and I just can't understand how the guy got to the estimation of the beta parameter. His results are as follows:

$SQ(alpha, beta)=sumlimits_i=1^nepsilon_i^2=sumlimits_i=1^n(y_i-alpha-beta x_i)^2$

$fracpartial SQ(alpha, beta)partial alphavert_alpha= hatalpha = 0 implies sumlimits_i=1^n(y_i - hatalpha - hatbeta x_i) = 0 $

$fracpartial SQ(alpha, beta)partial betavert_beta= hatbeta = 0 implies sumlimits_i=1^n x_i(y_i - hatalpha - hatbeta x_i) = 0$

$sumlimits_i=1^n y_i = n hatalpha + hatbeta sumlimits_i=1^n x_i implies hatalpha = bary - hatbetabarx$

$sumlimits_i=1^n x_i y_i = hatalpha sumlimits_i=1^n x_i + hatbeta sumlimits_i=1^n x_i^2 implies hatbeta = fracsumlimits_i=1^n x_i y_i - nbarxbarysumlimits_i=1^n x_i^2 - nbarx^2$

However, my result of the partial derivative with beta fixed is $hatbeta = fracsumlimits_i=1^nx_iy_i - barxbarysumlimits_i=1^nx_i^2-barx^2$

What is going on? What did I miss?

least-squares linear-regression

share|cite|improve this question

edited Jul 26 at 14:08

asked Jul 25 at 23:55

Luís Muniz

1012

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
  – José Carlos Santos
  Jul 26 at 0:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  No prob. I'll edit the question!
  – Luís Muniz
  Jul 26 at 13:37
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

I was reading a particular document on linear regression and I just can't understand how the guy got to the estimation of the beta parameter. His results are as follows:

$SQ(alpha, beta)=sumlimits_i=1^nepsilon_i^2=sumlimits_i=1^n(y_i-alpha-beta x_i)^2$

$fracpartial SQ(alpha, beta)partial alphavert_alpha= hatalpha = 0 implies sumlimits_i=1^n(y_i - hatalpha - hatbeta x_i) = 0 $

$fracpartial SQ(alpha, beta)partial betavert_beta= hatbeta = 0 implies sumlimits_i=1^n x_i(y_i - hatalpha - hatbeta x_i) = 0$

$sumlimits_i=1^n y_i = n hatalpha + hatbeta sumlimits_i=1^n x_i implies hatalpha = bary - hatbetabarx$

$sumlimits_i=1^n x_i y_i = hatalpha sumlimits_i=1^n x_i + hatbeta sumlimits_i=1^n x_i^2 implies hatbeta = fracsumlimits_i=1^n x_i y_i - nbarxbarysumlimits_i=1^n x_i^2 - nbarx^2$

However, my result of the partial derivative with beta fixed is $hatbeta = fracsumlimits_i=1^nx_iy_i - barxbarysumlimits_i=1^nx_i^2-barx^2$

What is going on? What did I miss?

least-squares linear-regression

share|cite|improve this question

edited Jul 26 at 14:08

asked Jul 25 at 23:55

Luís Muniz

1012

I was reading a particular document on linear regression and I just can't understand how the guy got to the estimation of the beta parameter. His results are as follows:

$SQ(alpha, beta)=sumlimits_i=1^nepsilon_i^2=sumlimits_i=1^n(y_i-alpha-beta x_i)^2$

$fracpartial SQ(alpha, beta)partial alphavert_alpha= hatalpha = 0 implies sumlimits_i=1^n(y_i - hatalpha - hatbeta x_i) = 0 $

$fracpartial SQ(alpha, beta)partial betavert_beta= hatbeta = 0 implies sumlimits_i=1^n x_i(y_i - hatalpha - hatbeta x_i) = 0$

$sumlimits_i=1^n y_i = n hatalpha + hatbeta sumlimits_i=1^n x_i implies hatalpha = bary - hatbetabarx$

$sumlimits_i=1^n x_i y_i = hatalpha sumlimits_i=1^n x_i + hatbeta sumlimits_i=1^n x_i^2 implies hatbeta = fracsumlimits_i=1^n x_i y_i - nbarxbarysumlimits_i=1^n x_i^2 - nbarx^2$

However, my result of the partial derivative with beta fixed is $hatbeta = fracsumlimits_i=1^nx_iy_i - barxbarysumlimits_i=1^nx_i^2-barx^2$

What is going on? What did I miss?

least-squares linear-regression

share|cite|improve this question

edited Jul 26 at 14:08

asked Jul 25 at 23:55

Luís Muniz

1012

share|cite|improve this question

share|cite|improve this question

edited Jul 26 at 14:08

edited Jul 26 at 14:08

edited Jul 26 at 14:08

asked Jul 25 at 23:55

Luís Muniz

1012

asked Jul 25 at 23:55

Luís Muniz

1012

asked Jul 25 at 23:55

Luís Muniz

1012

1012

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
  – José Carlos Santos
  Jul 26 at 0:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  No prob. I'll edit the question!
  – Luís Muniz
  Jul 26 at 13:37
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
  – José Carlos Santos
  Jul 26 at 0:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  No prob. I'll edit the question!
  – Luís Muniz
  Jul 26 at 13:37
  

  
  
  
  

1

1

Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
– José Carlos Santos
Jul 26 at 0:03

Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
– José Carlos Santos
Jul 26 at 0:03

No prob. I'll edit the question!
– Luís Muniz
Jul 26 at 13:37

No prob. I'll edit the question!
– Luís Muniz
Jul 26 at 13:37

add a comment | 

1 Answer
1

active

oldest

votes

up vote
0
down vote

It might be due to missing parentheses in your attempt: $$sum_i=1^n (x_i^2 - barx^2) = left(sum_i=1^n x_i^2right) - n barx^2.$$

---

By substituting $hatalpha = bary - hatbeta barx$ and $sum_i x_i = n barx$ into the last equation we have
beginalign
sum_i x_i y_i &= (bary - hatbeta barx) sum_i x_i + hatbeta sum_i x_i^2
\
sum_i x_i y_i &= n barxbary + hatbeta ((sum_i x_i^2) - n barx^2)
endalign

share|cite|improve this answer

edited Jul 26 at 16:37

answered Jul 25 at 23:59

angryavian

34.6k12874

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Man, it really looks like this is the problem. Thank's for calling my attention to this. However, still when I do the math, originally the summation including $x_i^2$ does not include $barx^2$. It is $(sumlimits_i=1^n x_i^2) - barx^2$. Am I missing something here?
  – Luís Muniz
  Jul 26 at 14:00
  
  

  
  
  
  

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%2f2862946%2fbeta-parameter-estimation-in-least-squares-method-by-partial-derivative%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

It might be due to missing parentheses in your attempt: $$sum_i=1^n (x_i^2 - barx^2) = left(sum_i=1^n x_i^2right) - n barx^2.$$

---

By substituting $hatalpha = bary - hatbeta barx$ and $sum_i x_i = n barx$ into the last equation we have
beginalign
sum_i x_i y_i &= (bary - hatbeta barx) sum_i x_i + hatbeta sum_i x_i^2
\
sum_i x_i y_i &= n barxbary + hatbeta ((sum_i x_i^2) - n barx^2)
endalign

share|cite|improve this answer

edited Jul 26 at 16:37

answered Jul 25 at 23:59

angryavian

34.6k12874

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Man, it really looks like this is the problem. Thank's for calling my attention to this. However, still when I do the math, originally the summation including $x_i^2$ does not include $barx^2$. It is $(sumlimits_i=1^n x_i^2) - barx^2$. Am I missing something here?
  – Luís Muniz
  Jul 26 at 14:00
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

It might be due to missing parentheses in your attempt: $$sum_i=1^n (x_i^2 - barx^2) = left(sum_i=1^n x_i^2right) - n barx^2.$$

---

By substituting $hatalpha = bary - hatbeta barx$ and $sum_i x_i = n barx$ into the last equation we have
beginalign
sum_i x_i y_i &= (bary - hatbeta barx) sum_i x_i + hatbeta sum_i x_i^2
\
sum_i x_i y_i &= n barxbary + hatbeta ((sum_i x_i^2) - n barx^2)
endalign

share|cite|improve this answer

edited Jul 26 at 16:37

answered Jul 25 at 23:59

angryavian

34.6k12874

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Man, it really looks like this is the problem. Thank's for calling my attention to this. However, still when I do the math, originally the summation including $x_i^2$ does not include $barx^2$. It is $(sumlimits_i=1^n x_i^2) - barx^2$. Am I missing something here?
  – Luís Muniz
  Jul 26 at 14:00
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

It might be due to missing parentheses in your attempt: $$sum_i=1^n (x_i^2 - barx^2) = left(sum_i=1^n x_i^2right) - n barx^2.$$

---

By substituting $hatalpha = bary - hatbeta barx$ and $sum_i x_i = n barx$ into the last equation we have
beginalign
sum_i x_i y_i &= (bary - hatbeta barx) sum_i x_i + hatbeta sum_i x_i^2
\
sum_i x_i y_i &= n barxbary + hatbeta ((sum_i x_i^2) - n barx^2)
endalign

share|cite|improve this answer

edited Jul 26 at 16:37

answered Jul 25 at 23:59

angryavian

34.6k12874

It might be due to missing parentheses in your attempt: $$sum_i=1^n (x_i^2 - barx^2) = left(sum_i=1^n x_i^2right) - n barx^2.$$

---

By substituting $hatalpha = bary - hatbeta barx$ and $sum_i x_i = n barx$ into the last equation we have
beginalign
sum_i x_i y_i &= (bary - hatbeta barx) sum_i x_i + hatbeta sum_i x_i^2
\
sum_i x_i y_i &= n barxbary + hatbeta ((sum_i x_i^2) - n barx^2)
endalign

share|cite|improve this answer

edited Jul 26 at 16:37

answered Jul 25 at 23:59

angryavian

34.6k12874

share|cite|improve this answer

share|cite|improve this answer

edited Jul 26 at 16:37

edited Jul 26 at 16:37

edited Jul 26 at 16:37

answered Jul 25 at 23:59

angryavian

34.6k12874

answered Jul 25 at 23:59

angryavian

34.6k12874

answered Jul 25 at 23:59

angryavian

34.6k12874

34.6k12874

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Man, it really looks like this is the problem. Thank's for calling my attention to this. However, still when I do the math, originally the summation including $x_i^2$ does not include $barx^2$. It is $(sumlimits_i=1^n x_i^2) - barx^2$. Am I missing something here?
  – Luís Muniz
  Jul 26 at 14:00
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Man, it really looks like this is the problem. Thank's for calling my attention to this. However, still when I do the math, originally the summation including $x_i^2$ does not include $barx^2$. It is $(sumlimits_i=1^n x_i^2) - barx^2$. Am I missing something here?
  – Luís Muniz
  Jul 26 at 14:00
  
  

  
  
  
  

Man, it really looks like this is the problem. Thank's for calling my attention to this. However, still when I do the math, originally the summation including $x_i^2$ does not include $barx^2$. It is $(sumlimits_i=1^n x_i^2) - barx^2$. Am I missing something here?
– Luís Muniz
Jul 26 at 14:00

Man, it really looks like this is the problem. Thank's for calling my attention to this. However, still when I do the math, originally the summation including $x_i^2$ does not include $barx^2$. It is $(sumlimits_i=1^n x_i^2) - barx^2$. Am I missing something here?
– Luís Muniz
Jul 26 at 14:00

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%2f2862946%2fbeta-parameter-estimation-in-least-squares-method-by-partial-derivative%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