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2

$begingroup$

Generally, both of them can represent the linear relationship between x and y scale to [0,1].
Are they 99% very similar?

regression correlation generalized-linear-model similarities univariate

share|cite|improve this question

edited Jun 20 at 3:22

Tommy Yu

asked Jun 19 at 3:27

Tommy YuTommy Yu

1844 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Not the same thing. See answers.
  $endgroup$
  – BruceET
  Jun 19 at 5:41
  

  
  
  
  

add a comment | 

2

$begingroup$

Generally, both of them can represent the linear relationship between x and y scale to [0,1].
Are they 99% very similar?

regression correlation generalized-linear-model similarities univariate

share|cite|improve this question

edited Jun 20 at 3:22

Tommy Yu

asked Jun 19 at 3:27

Tommy YuTommy Yu

1844 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Not the same thing. See answers.
  $endgroup$
  – BruceET
  Jun 19 at 5:41
  

  
  
  
  

add a comment | 

$begingroup$

Generally, both of them can represent the linear relationship between x and y scale to [0,1].
Are they 99% very similar?

regression correlation generalized-linear-model similarities univariate

share|cite|improve this question

edited Jun 20 at 3:22

Tommy Yu

asked Jun 19 at 3:27

Tommy YuTommy Yu

1844 bronze badges

$endgroup$

share|cite|improve this question

edited Jun 20 at 3:22

Tommy Yu

asked Jun 19 at 3:27

Tommy YuTommy Yu

1844 bronze badges

share|cite|improve this question

edited Jun 20 at 3:22

Tommy Yu

asked Jun 19 at 3:27

Tommy YuTommy Yu

1844 bronze badges

asked Jun 19 at 3:27

Tommy YuTommy Yu

1844 bronze badges

asked Jun 19 at 3:27

Tommy YuTommy Yu

1844 bronze badges

2 Answers
2

active

oldest

votes

4

$begingroup$

Computationally, R-sq in computer printouts is the square $r^2$of the (Pearson) correlation $r.$ Sometimes $r^2$ is called the 'coefficient of determination'. Because $-1 le r le 1$ we have $0 le r^2 le 1.$

Suppose we have the simple linear regression model
$$Y_i = beta_0 + beta_1 x_i + e_i,$$
for $i = 1,2, dots,n$ where $e_i stackreliidsimmathsfNorm(0, sigma_e).$

In regression $r^2$ is sometimes referred to (intuitively) as the proportion of the variability in $Y$ that is explained by regression on $x.$ This interpretation roughly matches the equation
$$S_x^2 = fracn-1n-2S_Y^2(1 - r^2),$$
where $S_Y^2$ is the sample variance of the $Y_i$ and $S_x^2$ is the sum of squared residuals divided by $n-2,$ sometimes referred to as the variance about the regression line.

Thus if $r = pm 1$ so that $r^2 = 1,$ then $S_x^2 = 0$ and all of the $(x_i,Y_i)$-points lie
precisely on the regression line. Also, if $r approx 0,$ then $S_x^2 approx S_Y^2$ and the $x_i$ have no
role to play in 'explaining' the $Y_i.$

Sometimes, the Pearson 'correlation coefficient' $r$ is said to express the 'linear component' of the association of $X_i$ and $Y_i.$

share|cite|improve this answer

edited Jun 19 at 5:50

answered Jun 19 at 5:11

BruceETBruceET

9,74811 gold badge88 silver badges2424 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @mkt. Thanks for fixing typo.
  $endgroup$
  – BruceET
  Jun 19 at 5:59
  

  
  
  
  

add a comment | 

1

$begingroup$

R-squared or coefficient of determination generally has the value between 0 and 1. However cor(x,y) is between -1 and 1.
Definition of R-squared for a multiple linear regression is the square of correlation between output (y) and predicted values(f). For the case of simple linear regression with an intercept and one explanatory variable it happens R-squared is as same as square of cor(x,y).
I hope this helps you.

share|cite|improve this answer

edited Jun 19 at 5:15

answered Jun 19 at 5:03

Masoud Norouzi DarabadMasoud Norouzi Darabad

1122 bronze badges

New contributor

Masoud Norouzi Darabad is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Definitely they are not same, but for simple linear regression : Y=ax+b (x is a single variable and not a vector), value of coefficient of determination is equal to square of correlation between x and y.
  $endgroup$
  – Masoud Norouzi Darabad
  Jun 19 at 5:19
  

  
  
  
  

add a comment | 

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4

$begingroup$

Computationally, R-sq in computer printouts is the square $r^2$of the (Pearson) correlation $r.$ Sometimes $r^2$ is called the 'coefficient of determination'. Because $-1 le r le 1$ we have $0 le r^2 le 1.$

Suppose we have the simple linear regression model
$$Y_i = beta_0 + beta_1 x_i + e_i,$$
for $i = 1,2, dots,n$ where $e_i stackreliidsimmathsfNorm(0, sigma_e).$

In regression $r^2$ is sometimes referred to (intuitively) as the proportion of the variability in $Y$ that is explained by regression on $x.$ This interpretation roughly matches the equation
$$S_x^2 = fracn-1n-2S_Y^2(1 - r^2),$$
where $S_Y^2$ is the sample variance of the $Y_i$ and $S_x^2$ is the sum of squared residuals divided by $n-2,$ sometimes referred to as the variance about the regression line.

Thus if $r = pm 1$ so that $r^2 = 1,$ then $S_x^2 = 0$ and all of the $(x_i,Y_i)$-points lie
precisely on the regression line. Also, if $r approx 0,$ then $S_x^2 approx S_Y^2$ and the $x_i$ have no
role to play in 'explaining' the $Y_i.$

Sometimes, the Pearson 'correlation coefficient' $r$ is said to express the 'linear component' of the association of $X_i$ and $Y_i.$

share|cite|improve this answer

edited Jun 19 at 5:50

answered Jun 19 at 5:11

BruceETBruceET

9,74811 gold badge88 silver badges2424 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @mkt. Thanks for fixing typo.
  $endgroup$
  – BruceET
  Jun 19 at 5:59
  

  
  
  
  

add a comment | 

4

$begingroup$

Computationally, R-sq in computer printouts is the square $r^2$of the (Pearson) correlation $r.$ Sometimes $r^2$ is called the 'coefficient of determination'. Because $-1 le r le 1$ we have $0 le r^2 le 1.$

Suppose we have the simple linear regression model
$$Y_i = beta_0 + beta_1 x_i + e_i,$$
for $i = 1,2, dots,n$ where $e_i stackreliidsimmathsfNorm(0, sigma_e).$

In regression $r^2$ is sometimes referred to (intuitively) as the proportion of the variability in $Y$ that is explained by regression on $x.$ This interpretation roughly matches the equation
$$S_x^2 = fracn-1n-2S_Y^2(1 - r^2),$$
where $S_Y^2$ is the sample variance of the $Y_i$ and $S_x^2$ is the sum of squared residuals divided by $n-2,$ sometimes referred to as the variance about the regression line.

Thus if $r = pm 1$ so that $r^2 = 1,$ then $S_x^2 = 0$ and all of the $(x_i,Y_i)$-points lie
precisely on the regression line. Also, if $r approx 0,$ then $S_x^2 approx S_Y^2$ and the $x_i$ have no
role to play in 'explaining' the $Y_i.$

Sometimes, the Pearson 'correlation coefficient' $r$ is said to express the 'linear component' of the association of $X_i$ and $Y_i.$

share|cite|improve this answer

edited Jun 19 at 5:50

answered Jun 19 at 5:11

BruceETBruceET

9,74811 gold badge88 silver badges2424 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @mkt. Thanks for fixing typo.
  $endgroup$
  – BruceET
  Jun 19 at 5:59
  

  
  
  
  

add a comment | 

$begingroup$

Computationally, R-sq in computer printouts is the square $r^2$of the (Pearson) correlation $r.$ Sometimes $r^2$ is called the 'coefficient of determination'. Because $-1 le r le 1$ we have $0 le r^2 le 1.$

Suppose we have the simple linear regression model
$$Y_i = beta_0 + beta_1 x_i + e_i,$$
for $i = 1,2, dots,n$ where $e_i stackreliidsimmathsfNorm(0, sigma_e).$

In regression $r^2$ is sometimes referred to (intuitively) as the proportion of the variability in $Y$ that is explained by regression on $x.$ This interpretation roughly matches the equation
$$S_x^2 = fracn-1n-2S_Y^2(1 - r^2),$$
where $S_Y^2$ is the sample variance of the $Y_i$ and $S_x^2$ is the sum of squared residuals divided by $n-2,$ sometimes referred to as the variance about the regression line.

Thus if $r = pm 1$ so that $r^2 = 1,$ then $S_x^2 = 0$ and all of the $(x_i,Y_i)$-points lie
precisely on the regression line. Also, if $r approx 0,$ then $S_x^2 approx S_Y^2$ and the $x_i$ have no
role to play in 'explaining' the $Y_i.$

Sometimes, the Pearson 'correlation coefficient' $r$ is said to express the 'linear component' of the association of $X_i$ and $Y_i.$

share|cite|improve this answer

edited Jun 19 at 5:50

answered Jun 19 at 5:11

BruceETBruceET

9,74811 gold badge88 silver badges2424 bronze badges

$endgroup$

share|cite|improve this answer

edited Jun 19 at 5:50

answered Jun 19 at 5:11

BruceETBruceET

9,74811 gold badge88 silver badges2424 bronze badges

answered Jun 19 at 5:11

BruceETBruceET

9,74811 gold badge88 silver badges2424 bronze badges

answered Jun 19 at 5:11

BruceETBruceET

9,74811 gold badge88 silver badges2424 bronze badges

1

$begingroup$

R-squared or coefficient of determination generally has the value between 0 and 1. However cor(x,y) is between -1 and 1.
Definition of R-squared for a multiple linear regression is the square of correlation between output (y) and predicted values(f). For the case of simple linear regression with an intercept and one explanatory variable it happens R-squared is as same as square of cor(x,y).
I hope this helps you.

share|cite|improve this answer

edited Jun 19 at 5:15

answered Jun 19 at 5:03

Masoud Norouzi DarabadMasoud Norouzi Darabad

1122 bronze badges

New contributor

Masoud Norouzi Darabad is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Definitely they are not same, but for simple linear regression : Y=ax+b (x is a single variable and not a vector), value of coefficient of determination is equal to square of correlation between x and y.
  $endgroup$
  – Masoud Norouzi Darabad
  Jun 19 at 5:19
  

  
  
  
  

add a comment | 

1

$begingroup$

R-squared or coefficient of determination generally has the value between 0 and 1. However cor(x,y) is between -1 and 1.
Definition of R-squared for a multiple linear regression is the square of correlation between output (y) and predicted values(f). For the case of simple linear regression with an intercept and one explanatory variable it happens R-squared is as same as square of cor(x,y).
I hope this helps you.

share|cite|improve this answer

edited Jun 19 at 5:15

answered Jun 19 at 5:03

Masoud Norouzi DarabadMasoud Norouzi Darabad

1122 bronze badges

New contributor

Masoud Norouzi Darabad is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Definitely they are not same, but for simple linear regression : Y=ax+b (x is a single variable and not a vector), value of coefficient of determination is equal to square of correlation between x and y.
  $endgroup$
  – Masoud Norouzi Darabad
  Jun 19 at 5:19
  

  
  
  
  

add a comment | 

$begingroup$

R-squared or coefficient of determination generally has the value between 0 and 1. However cor(x,y) is between -1 and 1.
Definition of R-squared for a multiple linear regression is the square of correlation between output (y) and predicted values(f). For the case of simple linear regression with an intercept and one explanatory variable it happens R-squared is as same as square of cor(x,y).
I hope this helps you.

share|cite|improve this answer

edited Jun 19 at 5:15

answered Jun 19 at 5:03

Masoud Norouzi DarabadMasoud Norouzi Darabad

1122 bronze badges

New contributor

Masoud Norouzi Darabad is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this answer

edited Jun 19 at 5:15

answered Jun 19 at 5:03

Masoud Norouzi DarabadMasoud Norouzi Darabad

1122 bronze badges

New contributor

Masoud Norouzi Darabad is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered Jun 19 at 5:03

Masoud Norouzi DarabadMasoud Norouzi Darabad

1122 bronze badges

New contributor

Masoud Norouzi Darabad is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered Jun 19 at 5:03

Masoud Norouzi DarabadMasoud Norouzi Darabad

1122 bronze badges