Why are exponentiated logistic regression coefficients considered “odds ratios”?

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Logistic regression models the log odds of an event as some set of predictors. That is, log(p/(1-p)) where p is the probability of some outcome. Thus, the interpretation of the raw logistic regression coefficients for some variable (x) has to be on the log odds scale. That is, if the coefficient for x = 5 then we know that a 1 unit change in x correspondents to 5 unit change on the log odds scale that an outcome will occur.

However, I often see people interpret exponentiated logistic regression coefficients as odds ratios. However, clearly exp(log(p/(1-p))) = p/(1-p), which is an odds. As far as I understand it, an odds ratio is the odds of one event occurring (e.g., p/(1-p) for event A) over the odds of another event occurring (e.g., p/(1-p) for event B).

What am I missing here? Is seems like this common interpretation of exponentiated logistic regression coefficients is incorrect.

regression logistic logit

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asked Aug 9 at 18:02

jack

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up vote
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Logistic regression models the log odds of an event as some set of predictors. That is, log(p/(1-p)) where p is the probability of some outcome. Thus, the interpretation of the raw logistic regression coefficients for some variable (x) has to be on the log odds scale. That is, if the coefficient for x = 5 then we know that a 1 unit change in x correspondents to 5 unit change on the log odds scale that an outcome will occur.

However, I often see people interpret exponentiated logistic regression coefficients as odds ratios. However, clearly exp(log(p/(1-p))) = p/(1-p), which is an odds. As far as I understand it, an odds ratio is the odds of one event occurring (e.g., p/(1-p) for event A) over the odds of another event occurring (e.g., p/(1-p) for event B).

What am I missing here? Is seems like this common interpretation of exponentiated logistic regression coefficients is incorrect.

regression logistic logit

share|cite|improve this question

asked Aug 9 at 18:02

jack

482

add a comment | 

up vote
9
down vote

favorite

6

up vote
9
down vote

favorite

6

6

Logistic regression models the log odds of an event as some set of predictors. That is, log(p/(1-p)) where p is the probability of some outcome. Thus, the interpretation of the raw logistic regression coefficients for some variable (x) has to be on the log odds scale. That is, if the coefficient for x = 5 then we know that a 1 unit change in x correspondents to 5 unit change on the log odds scale that an outcome will occur.

However, I often see people interpret exponentiated logistic regression coefficients as odds ratios. However, clearly exp(log(p/(1-p))) = p/(1-p), which is an odds. As far as I understand it, an odds ratio is the odds of one event occurring (e.g., p/(1-p) for event A) over the odds of another event occurring (e.g., p/(1-p) for event B).

What am I missing here? Is seems like this common interpretation of exponentiated logistic regression coefficients is incorrect.

regression logistic logit

share|cite|improve this question

asked Aug 9 at 18:02

jack

482

Logistic regression models the log odds of an event as some set of predictors. That is, log(p/(1-p)) where p is the probability of some outcome. Thus, the interpretation of the raw logistic regression coefficients for some variable (x) has to be on the log odds scale. That is, if the coefficient for x = 5 then we know that a 1 unit change in x correspondents to 5 unit change on the log odds scale that an outcome will occur.

However, I often see people interpret exponentiated logistic regression coefficients as odds ratios. However, clearly exp(log(p/(1-p))) = p/(1-p), which is an odds. As far as I understand it, an odds ratio is the odds of one event occurring (e.g., p/(1-p) for event A) over the odds of another event occurring (e.g., p/(1-p) for event B).

What am I missing here? Is seems like this common interpretation of exponentiated logistic regression coefficients is incorrect.

regression logistic logit

share|cite|improve this question

asked Aug 9 at 18:02

jack

482

share|cite|improve this question

share|cite|improve this question

asked Aug 9 at 18:02

jack

482

asked Aug 9 at 18:02

jack

482

asked Aug 9 at 18:02

jack

482

482

add a comment | 

add a comment | 

2 Answers
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8
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accepted

@Laconic's answer is great and complete, in my opinion. Something I wanted to add is that the original coefficients describe a difference in the log odds for two units who differ by 1 in the predictor. E.g., for a coefficient on $X$ of 5, we can say that the difference in log odds between two units who differ on $X$ by 1 is 5. Mathematically,

$$beta = log(textodds(p|X=x_0+1))-log(textodds(p|X=x_0)) $$

When you exponentiate $beta$, you get

$$exp(beta) = exp(log(textodds(p|X=x_0+1))-log(textodds(p|X=x_0))) \
= fracX=x_0+1)))X=x_0))) \
= fractextodds(ptextodds(p$$

which is a ratio of odds, an odds ratio.

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edited Aug 10 at 14:14

answered Aug 9 at 18:48

Noah

2,3811314

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  This is extremely clear to me. My question is resolved.
  – jack
  Aug 9 at 18:51
  

  
  
  
  

add a comment | 

up vote
9
down vote

Consider two set of conditions, the first described by the vector of independent variables $X$, and the second described by the vector $X'$, which differs only in the ith variable $x_i$, and by one unit. Let $beta$ be the vector of model parameters as usual.

According to the logistic regression model, the probability of the event occurring in the first case is $p_1 = frac11 + exp(-X beta)$, so that the odds of the event occurring is $fracp_11-p_1 = exp(X beta)$.

The probability of the event occurring in the second case is $p_2 = frac11 + exp(-X' beta)$, so that the odds of the event occurring is $fracp_21-p_2 = exp(X' beta) = exp(X beta + beta_i)$.

The ratio of the odds in the second case to the odds in the first case is therefore $exp(beta_i)$. Hence the interpretation of the exponential of the parameter as an odds ratio.

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answered Aug 9 at 18:25

The Laconic

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
8
down vote



accepted

@Laconic's answer is great and complete, in my opinion. Something I wanted to add is that the original coefficients describe a difference in the log odds for two units who differ by 1 in the predictor. E.g., for a coefficient on $X$ of 5, we can say that the difference in log odds between two units who differ on $X$ by 1 is 5. Mathematically,

$$beta = log(textodds(p|X=x_0+1))-log(textodds(p|X=x_0)) $$

When you exponentiate $beta$, you get

$$exp(beta) = exp(log(textodds(p|X=x_0+1))-log(textodds(p|X=x_0))) \
= fracX=x_0+1)))X=x_0))) \
= fractextodds(ptextodds(p$$

which is a ratio of odds, an odds ratio.

share|cite|improve this answer

edited Aug 10 at 14:14

answered Aug 9 at 18:48

Noah

2,3811314

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  This is extremely clear to me. My question is resolved.
  – jack
  Aug 9 at 18:51
  

  
  
  
  

add a comment | 

up vote
8
down vote



accepted

@Laconic's answer is great and complete, in my opinion. Something I wanted to add is that the original coefficients describe a difference in the log odds for two units who differ by 1 in the predictor. E.g., for a coefficient on $X$ of 5, we can say that the difference in log odds between two units who differ on $X$ by 1 is 5. Mathematically,

$$beta = log(textodds(p|X=x_0+1))-log(textodds(p|X=x_0)) $$

When you exponentiate $beta$, you get

$$exp(beta) = exp(log(textodds(p|X=x_0+1))-log(textodds(p|X=x_0))) \
= fracX=x_0+1)))X=x_0))) \
= fractextodds(ptextodds(p$$

which is a ratio of odds, an odds ratio.

share|cite|improve this answer

edited Aug 10 at 14:14

answered Aug 9 at 18:48

Noah

2,3811314

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  This is extremely clear to me. My question is resolved.
  – jack
  Aug 9 at 18:51
  

  
  
  
  

add a comment | 

up vote
8
down vote



accepted

up vote
8
down vote



accepted

@Laconic's answer is great and complete, in my opinion. Something I wanted to add is that the original coefficients describe a difference in the log odds for two units who differ by 1 in the predictor. E.g., for a coefficient on $X$ of 5, we can say that the difference in log odds between two units who differ on $X$ by 1 is 5. Mathematically,

$$beta = log(textodds(p|X=x_0+1))-log(textodds(p|X=x_0)) $$

When you exponentiate $beta$, you get

$$exp(beta) = exp(log(textodds(p|X=x_0+1))-log(textodds(p|X=x_0))) \
= fracX=x_0+1)))X=x_0))) \
= fractextodds(ptextodds(p$$

which is a ratio of odds, an odds ratio.

share|cite|improve this answer

edited Aug 10 at 14:14

answered Aug 9 at 18:48

Noah

2,3811314

@Laconic's answer is great and complete, in my opinion. Something I wanted to add is that the original coefficients describe a difference in the log odds for two units who differ by 1 in the predictor. E.g., for a coefficient on $X$ of 5, we can say that the difference in log odds between two units who differ on $X$ by 1 is 5. Mathematically,

$$beta = log(textodds(p|X=x_0+1))-log(textodds(p|X=x_0)) $$

When you exponentiate $beta$, you get

$$exp(beta) = exp(log(textodds(p|X=x_0+1))-log(textodds(p|X=x_0))) \
= fracX=x_0+1)))X=x_0))) \
= fractextodds(ptextodds(p$$

which is a ratio of odds, an odds ratio.

share|cite|improve this answer

edited Aug 10 at 14:14

answered Aug 9 at 18:48

Noah

2,3811314

share|cite|improve this answer

share|cite|improve this answer

edited Aug 10 at 14:14

edited Aug 10 at 14:14

edited Aug 10 at 14:14

answered Aug 9 at 18:48

Noah

2,3811314

answered Aug 9 at 18:48

Noah

2,3811314

answered Aug 9 at 18:48

Noah

2,3811314

2,3811314

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  This is extremely clear to me. My question is resolved.
  – jack
  Aug 9 at 18:51
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  This is extremely clear to me. My question is resolved.
  – jack
  Aug 9 at 18:51
  

  
  
  
  

1

1

This is extremely clear to me. My question is resolved.
– jack
Aug 9 at 18:51

This is extremely clear to me. My question is resolved.
– jack
Aug 9 at 18:51

add a comment | 

up vote
9
down vote

Consider two set of conditions, the first described by the vector of independent variables $X$, and the second described by the vector $X'$, which differs only in the ith variable $x_i$, and by one unit. Let $beta$ be the vector of model parameters as usual.

According to the logistic regression model, the probability of the event occurring in the first case is $p_1 = frac11 + exp(-X beta)$, so that the odds of the event occurring is $fracp_11-p_1 = exp(X beta)$.

The probability of the event occurring in the second case is $p_2 = frac11 + exp(-X' beta)$, so that the odds of the event occurring is $fracp_21-p_2 = exp(X' beta) = exp(X beta + beta_i)$.

The ratio of the odds in the second case to the odds in the first case is therefore $exp(beta_i)$. Hence the interpretation of the exponential of the parameter as an odds ratio.

share|cite|improve this answer

answered Aug 9 at 18:25

The Laconic

777414

add a comment | 

up vote
9
down vote

Consider two set of conditions, the first described by the vector of independent variables $X$, and the second described by the vector $X'$, which differs only in the ith variable $x_i$, and by one unit. Let $beta$ be the vector of model parameters as usual.

According to the logistic regression model, the probability of the event occurring in the first case is $p_1 = frac11 + exp(-X beta)$, so that the odds of the event occurring is $fracp_11-p_1 = exp(X beta)$.

The probability of the event occurring in the second case is $p_2 = frac11 + exp(-X' beta)$, so that the odds of the event occurring is $fracp_21-p_2 = exp(X' beta) = exp(X beta + beta_i)$.

The ratio of the odds in the second case to the odds in the first case is therefore $exp(beta_i)$. Hence the interpretation of the exponential of the parameter as an odds ratio.

share|cite|improve this answer

answered Aug 9 at 18:25

The Laconic

777414

add a comment | 

up vote
9
down vote

up vote
9
down vote

Consider two set of conditions, the first described by the vector of independent variables $X$, and the second described by the vector $X'$, which differs only in the ith variable $x_i$, and by one unit. Let $beta$ be the vector of model parameters as usual.

According to the logistic regression model, the probability of the event occurring in the first case is $p_1 = frac11 + exp(-X beta)$, so that the odds of the event occurring is $fracp_11-p_1 = exp(X beta)$.

The probability of the event occurring in the second case is $p_2 = frac11 + exp(-X' beta)$, so that the odds of the event occurring is $fracp_21-p_2 = exp(X' beta) = exp(X beta + beta_i)$.

The ratio of the odds in the second case to the odds in the first case is therefore $exp(beta_i)$. Hence the interpretation of the exponential of the parameter as an odds ratio.

share|cite|improve this answer

answered Aug 9 at 18:25

The Laconic

777414

Consider two set of conditions, the first described by the vector of independent variables $X$, and the second described by the vector $X'$, which differs only in the ith variable $x_i$, and by one unit. Let $beta$ be the vector of model parameters as usual.

According to the logistic regression model, the probability of the event occurring in the first case is $p_1 = frac11 + exp(-X beta)$, so that the odds of the event occurring is $fracp_11-p_1 = exp(X beta)$.

The probability of the event occurring in the second case is $p_2 = frac11 + exp(-X' beta)$, so that the odds of the event occurring is $fracp_21-p_2 = exp(X' beta) = exp(X beta + beta_i)$.

The ratio of the odds in the second case to the odds in the first case is therefore $exp(beta_i)$. Hence the interpretation of the exponential of the parameter as an odds ratio.

share|cite|improve this answer

answered Aug 9 at 18:25

The Laconic

777414

share|cite|improve this answer

share|cite|improve this answer

answered Aug 9 at 18:25

The Laconic

777414

answered Aug 9 at 18:25

The Laconic

777414

answered Aug 9 at 18:25

The Laconic

777414

777414

add a comment | 

add a comment | 

 

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