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Why is the intercept typed in as a 1 in stats packages (R, python)

Advice on a good production level SPC (statistical process control) package?Understanding what Dassault iSight is doing?What languages are commonly used in medical statistics?What statistical software does NOT provide for Classification and Regression modelsHow must one handle ordinal independent variables when modeling interactions?Predicting with GLM using Gaussian distributed dataStandard error for rolling regressionSlope interpretation in simple linear regressionAlternatives to Journal of Statistical Software?Non-straight lines in random intercept random slope plots

7

2

$begingroup$

When using statistics software, When defining your linear models, why is the intercept typed in as a 1, rather than "const" or "intercept" or something. What significance does 1 have?

Is there some historic reason? Or is this logical in some way I am failing to grasp? The intercept could very well be any number.

Example from statsmodels library in python:

model = smf.ols('Height ~ 1', data = height_sample_data)

I know lmer package for R is very similar.

regression software intercept

share|cite|improve this question

edited 13 hours ago

kjetil b halvorsen

31.3k984224

asked 19 hours ago

Adam BAdam B

21418

$endgroup$

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  The intercept is the coefficient (which indeed could have any value), but what you enter into the regression program when you fit the model are not the coefficients, but the things you multiply the coefficients by in the regression equation (the $x$'s). What do you multiply the intercept by in the regression equation? (Note that $beta_0 times 1 = beta_0$.)
  $endgroup$
  – Glen_b♦
  19 hours ago
  
  

  
  
  
  

add a comment | 

7

2

$begingroup$

When using statistics software, When defining your linear models, why is the intercept typed in as a 1, rather than "const" or "intercept" or something. What significance does 1 have?

Is there some historic reason? Or is this logical in some way I am failing to grasp? The intercept could very well be any number.

Example from statsmodels library in python:

model = smf.ols('Height ~ 1', data = height_sample_data)

I know lmer package for R is very similar.

regression software intercept

share|cite|improve this question

edited 13 hours ago

kjetil b halvorsen

31.3k984224

asked 19 hours ago

Adam BAdam B

21418

$endgroup$

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  The intercept is the coefficient (which indeed could have any value), but what you enter into the regression program when you fit the model are not the coefficients, but the things you multiply the coefficients by in the regression equation (the $x$'s). What do you multiply the intercept by in the regression equation? (Note that $beta_0 times 1 = beta_0$.)
  $endgroup$
  – Glen_b♦
  19 hours ago
  
  

  
  
  
  

add a comment | 

$begingroup$

When using statistics software, When defining your linear models, why is the intercept typed in as a 1, rather than "const" or "intercept" or something. What significance does 1 have?

Is there some historic reason? Or is this logical in some way I am failing to grasp? The intercept could very well be any number.

Example from statsmodels library in python:

model = smf.ols('Height ~ 1', data = height_sample_data)

I know lmer package for R is very similar.

regression software intercept

share|cite|improve this question

edited 13 hours ago

kjetil b halvorsen

31.3k984224

asked 19 hours ago

Adam BAdam B

21418

$endgroup$

model = smf.ols('Height ~ 1', data = height_sample_data)

share|cite|improve this question

edited 13 hours ago

kjetil b halvorsen

31.3k984224

asked 19 hours ago

Adam BAdam B

21418

share|cite|improve this question

edited 13 hours ago

kjetil b halvorsen

31.3k984224

asked 19 hours ago

Adam BAdam B

21418

edited 13 hours ago

kjetil b halvorsen

31.3k984224

edited 13 hours ago

kjetil b halvorsen

31.3k984224

asked 19 hours ago

Adam BAdam B

21418

asked 19 hours ago

Adam BAdam B

21418

1 Answer
1

active

oldest

votes

16

$begingroup$

It is logical, once you consider the matrix notation that your formula will be translated into internally. In the matrix, the non-constant predictors will be translated into (one or more) columns, and the intercept will be translated into a column consisting entirely of ones.

For instance, in R you would write a very simple OLS as:

lm(z~1+x+y)

In matrix notation, this would be translated into a model

$$ beginpmatrix z_1 \ z_2 \ vdots \ z_n endpmatrix =
beginpmatrix 1 & x_1 & y_1 \ 1 & x_2 & y_2 \ vdots & vdots & vdots \ 1 & x_n & y_n endpmatrix
beginpmatrix beta_0 \ beta_x \ beta_z endpmatrix
+beginpmatrix epsilon_1 \ epsilon_2 \ vdots \ epsilon_n endpmatrix,
$$

and now you see where the $1$ comes from.

---

Actually, you could leave the 1+ out, since R will always presume you want to include an intercept, so this is completely equivalent to

lm(z~x+y).

However, if you want to suppress the intercept, you would write something like

lm(z~x+y-1),

which would be translated into a matrix without a 1 column:

$$ beginpmatrix z_1 \ z_2 \ vdots \ z_n endpmatrix =
beginpmatrix x_1 & y_1 \ x_2 & y_2 \ vdots & vdots \ x_n & y_n endpmatrix
beginpmatrix beta_x \ beta_z endpmatrix
+beginpmatrix epsilon_1 \ epsilon_2 \ vdots \ epsilon_n endpmatrix,
$$

share|cite|improve this answer

edited 19 hours ago

answered 19 hours ago

Stephan KolassaStephan Kolassa

46.9k799174

$endgroup$

add a comment | 

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16

$begingroup$

It is logical, once you consider the matrix notation that your formula will be translated into internally. In the matrix, the non-constant predictors will be translated into (one or more) columns, and the intercept will be translated into a column consisting entirely of ones.

For instance, in R you would write a very simple OLS as:

lm(z~1+x+y)

In matrix notation, this would be translated into a model

$$ beginpmatrix z_1 \ z_2 \ vdots \ z_n endpmatrix =
beginpmatrix 1 & x_1 & y_1 \ 1 & x_2 & y_2 \ vdots & vdots & vdots \ 1 & x_n & y_n endpmatrix
beginpmatrix beta_0 \ beta_x \ beta_z endpmatrix
+beginpmatrix epsilon_1 \ epsilon_2 \ vdots \ epsilon_n endpmatrix,
$$

and now you see where the $1$ comes from.

---

Actually, you could leave the 1+ out, since R will always presume you want to include an intercept, so this is completely equivalent to

lm(z~x+y).

However, if you want to suppress the intercept, you would write something like

lm(z~x+y-1),

which would be translated into a matrix without a 1 column:

$$ beginpmatrix z_1 \ z_2 \ vdots \ z_n endpmatrix =
beginpmatrix x_1 & y_1 \ x_2 & y_2 \ vdots & vdots \ x_n & y_n endpmatrix
beginpmatrix beta_x \ beta_z endpmatrix
+beginpmatrix epsilon_1 \ epsilon_2 \ vdots \ epsilon_n endpmatrix,
$$

share|cite|improve this answer

edited 19 hours ago

answered 19 hours ago

Stephan KolassaStephan Kolassa

46.9k799174

$endgroup$

add a comment | 

16

$begingroup$

It is logical, once you consider the matrix notation that your formula will be translated into internally. In the matrix, the non-constant predictors will be translated into (one or more) columns, and the intercept will be translated into a column consisting entirely of ones.

For instance, in R you would write a very simple OLS as:

lm(z~1+x+y)

In matrix notation, this would be translated into a model

$$ beginpmatrix z_1 \ z_2 \ vdots \ z_n endpmatrix =
beginpmatrix 1 & x_1 & y_1 \ 1 & x_2 & y_2 \ vdots & vdots & vdots \ 1 & x_n & y_n endpmatrix
beginpmatrix beta_0 \ beta_x \ beta_z endpmatrix
+beginpmatrix epsilon_1 \ epsilon_2 \ vdots \ epsilon_n endpmatrix,
$$

and now you see where the $1$ comes from.

---

Actually, you could leave the 1+ out, since R will always presume you want to include an intercept, so this is completely equivalent to

lm(z~x+y).

However, if you want to suppress the intercept, you would write something like

lm(z~x+y-1),

which would be translated into a matrix without a 1 column:

$$ beginpmatrix z_1 \ z_2 \ vdots \ z_n endpmatrix =
beginpmatrix x_1 & y_1 \ x_2 & y_2 \ vdots & vdots \ x_n & y_n endpmatrix
beginpmatrix beta_x \ beta_z endpmatrix
+beginpmatrix epsilon_1 \ epsilon_2 \ vdots \ epsilon_n endpmatrix,
$$

share|cite|improve this answer

edited 19 hours ago

answered 19 hours ago

Stephan KolassaStephan Kolassa

46.9k799174

$endgroup$

add a comment | 

$begingroup$

It is logical, once you consider the matrix notation that your formula will be translated into internally. In the matrix, the non-constant predictors will be translated into (one or more) columns, and the intercept will be translated into a column consisting entirely of ones.

For instance, in R you would write a very simple OLS as:

lm(z~1+x+y)

In matrix notation, this would be translated into a model

$$ beginpmatrix z_1 \ z_2 \ vdots \ z_n endpmatrix =
beginpmatrix 1 & x_1 & y_1 \ 1 & x_2 & y_2 \ vdots & vdots & vdots \ 1 & x_n & y_n endpmatrix
beginpmatrix beta_0 \ beta_x \ beta_z endpmatrix
+beginpmatrix epsilon_1 \ epsilon_2 \ vdots \ epsilon_n endpmatrix,
$$

and now you see where the $1$ comes from.

---

Actually, you could leave the 1+ out, since R will always presume you want to include an intercept, so this is completely equivalent to

lm(z~x+y).

However, if you want to suppress the intercept, you would write something like

lm(z~x+y-1),

which would be translated into a matrix without a 1 column:

$$ beginpmatrix z_1 \ z_2 \ vdots \ z_n endpmatrix =
beginpmatrix x_1 & y_1 \ x_2 & y_2 \ vdots & vdots \ x_n & y_n endpmatrix
beginpmatrix beta_x \ beta_z endpmatrix
+beginpmatrix epsilon_1 \ epsilon_2 \ vdots \ epsilon_n endpmatrix,
$$

share|cite|improve this answer

edited 19 hours ago

answered 19 hours ago

Stephan KolassaStephan Kolassa

46.9k799174

$endgroup$

lm(z~1+x+y)

lm(z~x+y).

lm(z~x+y-1),

share|cite|improve this answer

edited 19 hours ago

answered 19 hours ago

Stephan KolassaStephan Kolassa

46.9k799174

answered 19 hours ago

Stephan KolassaStephan Kolassa

46.9k799174

answered 19 hours ago

Stephan KolassaStephan Kolassa

46.9k799174