The long title says it all.

For example, I have performed linear regression (OLS) with commonly used iris dataset using following formula:

PL ~ SW + Species

Following is the output:

==================== Summary2() ====================
                  Results: Ordinary least squares
Model:               OLS               Adj. R-squared:      0.949   
Dependent Variable:  PL                AIC:                 154.5772
Date:                2020-08-10 05:57  BIC:                 166.6197
No. Observations:    150               Log-Likelihood:      -73.289 
Df Model:            3                 F-statistic:         919.7   
Df Residuals:        146               Prob (F-statistic):  1.45e-94
R-squared:           0.950             Scale:               0.15983 
                       Coef.  Std.Err.    t    P>|t|   [0.025 0.975]
Intercept             -0.1792   0.3375 -0.5309 0.5963 -0.8463 0.4879
Species[T.versicolor]  3.1130   0.1023 30.4196 0.0000  2.9108 3.3153
Species[T.virginica]   4.3074   0.0913 47.1795 0.0000  4.1269 4.4878
SW                     0.4788   0.0971  4.9321 0.0000  0.2869 0.6706
Omnibus:               12.413        Durbin-Watson:           1.889 
Prob(Omnibus):         0.002         Jarque-Bera (JB):        19.064
Skew:                  0.435         Prob(JB):                0.000 
Kurtosis:              4.514         Condition No.:           36    

I now convert the coefficients to exp(coefficients), as is done to get odds ratio in logistic regression. I get following values:

Species[T.versicolor] 22.8
Species[T.virginica]  74.2
SW                    1.61

What do these numbers indicate or how can these values be interpreted?

Edit: The answer to this question states that exponetiation is useful in the setting of Poisson regression. Does it apply to linear regression also?

  • $\begingroup$ It has some uses if you take the natural log of the outcome, which are related to the Poisson. $\endgroup$
    – dimitriy
    Commented Aug 10, 2020 at 5:38
  • $\begingroup$ It will be great if you can explain this more in an answer here. $\endgroup$
    – rnso
    Commented Aug 10, 2020 at 5:50

2 Answers 2


The interpretation of coefficents from logistic regression is due to the formulation, specifically:

$$ ln(\frac{P}{1-P}) = \beta_0 + \beta_1x $$

The log odds is on the left and the linear predictor with your coefficients is on the right. If we exponentiate both sides, we now have exp(linear predictor) related to the odds ratio, or a unit change in exp(linear predictor) gives a unit change in the odds ratio.

$$ \frac{P}{1-P} = e^{\beta_0 + \beta_1x} $$

In linear regression, the relationship is simply:

$$ y = \beta_0 + \beta_1x $$

Exponentiation of the coefficents here does not give you a directly interpretable relationship to the response variable, Y.

Edit to address comment

The Poisson Regression formulation is

$$ln(y) = \beta_0 + \beta_1x $$

Exponentiation of each side gives that the exponentiated coefficients are related to the change in y. Also remember that here, y is assumed to be count data following a Poisson distribution.

The various formulations provide the interpretations. For OLS, there is no exponential in the formulation, so the coefficients don't need to be modified for interpretation. Also, if you exponentiate the coefficients from an OLS model, there is not direct relationship to the endpoint any longer.

  • $\begingroup$ The answer here states it is useful for Poisson regression: stats.stackexchange.com/questions/319015/… $\endgroup$
    – rnso
    Commented Aug 10, 2020 at 4:45
  • $\begingroup$ Not quite. I edited my answer to addresses Poisson Regression. $\endgroup$
    – KirkD_CO
    Commented Aug 10, 2020 at 12:25

Exponentiation of coefficients will generally be useful when the expected value involves an exponential function in some way. This non-exhaustive list includes

Sometimes you will need to go beyond mere exponentiation to get something interpretable.


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