I am making a table from results of an analysis using generalised linear model which involves detecting association of a categorical predictor variable over multiple outcome variables. Of those multiple outcome variables, few are binary where I display the odds ratio for each category of the predictor (as we do in logistic regression); while few are continuous outcome variables, in which case I can display the beta estimate for each category of the predictor. My question is will it be ok if exponentiate the beta value and express it as odds ratios. Can I do that?
Solved – Expressing beta estimate in terms of odds ratio for a continuous variable
generalized linear modellogisticodds-ratiorregression
Related Solutions
It's easiest to illustrate what is going on with a simple example with a single predictor that is dichotomous (e.g., to distinguish two groups). Suppose these are the data (using R for illustration):
y <- c(0,0,0,1,1,0,0,0,0,0)
grp <- c(0,0,0,0,0,1,1,1,1,1)
cbind(grp, y)
So:
grp y
[1,] 0 0
[2,] 0 0
[3,] 0 0
[4,] 0 1
[5,] 0 1
[6,] 1 0
[7,] 1 0
[8,] 1 0
[9,] 1 0
[10,] 1 0
There are 5 observations for each group. In group 0 (the reference group), there are 2 events, so the odds of the event are $2/3$. So, the log odds of the event happening are $\ln(2/3) = -0.4055$. In the second group, the are 0 events, so the odds of the event happening are $0/5$. And the log odds of the event are $\ln(0/5) = -\infty$. So, the odds ratio of the event happening in group 1 versus 0 is $(0/5)/(2/3) = 0$. So, the log odds ratio is $\ln((0/5)/(2/3)) = -\infty$ or, equivalently, $\ln(0/5) - \ln(2/3) = -\infty$.
Now let's actually fit the model:
res <- glm(y ~ grp, family=binomial)
summary(res)
This yields:
Call:
glm(formula = y ~ grp, family = binomial)
Deviance Residuals:
Min 1Q Median 3Q Max
-1.01077 -0.75810 -0.00008 -0.00008 1.35373
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) -0.4055 0.9129 -0.444 0.657
grp -19.1606 4809.3409 -0.004 0.997
(Dispersion parameter for binomial family taken to be 1)
Null deviance: 10.0080 on 9 degrees of freedom
Residual deviance: 6.7301 on 8 degrees of freedom
AIC: 10.73
Number of Fisher Scoring iterations: 18
So, the estimated intercept is $-0.4055$, which is the log odds in group 0. The coefficient for grp is the log odds ratio, which is estimated to be $-19.1606$. Hmmm, that's not quite $-\infty$. But after exponentiation, we get the odds ratio, which we can round to, let's say, 8 digits:
round(exp(coef(res)[2]), 8)
And that is in essence zero. The coefficient for grp is not $-\infty$ due to numerical issues when fitting the model when there is complete separation in the data (and to answer that part of your question: that is indeed exactly what is going on here). But for all practical purposes, the model implies an odds ratio that is in essence zero.
The one that has been adjusted by including other covariates and confounding variables is the so-called better one. Just the magnitude and width of 95% CI tell very little as the original estimate could have been biased upward or downward.
Also, don't be distracted by the "broader" interval. OR operates on a log scale and the CI will appear much wider just because the point estimate is higher. When the actual mean and its 95% CI are expressed in logit (by taking a natural logarithmic transformation), the widths of the CI actually aren't terribly different:
1.53 (0.33, 2.82), the width is about 2.49
2.10 (0.44, 4.03), the width is about 3.59
Best Answer
There are two issues with that. First, you are assuming that a one-unit change in $x$ is meaningful. Second, you are restricting yourself to the case where $x$ is linear. In general think of odds ratios as anti-logs of differences in predicted logits ($X\hat{\beta}$). That way you can handle nonlinearities and meaningful ranges. The R
rmspackage by default produces inter-quartile-range odds ratios for continuous predictors.