I have one independent variable which has values less than 1, and want to see how the odds of having a disease change when the independent variable increase by 0.1.

I ran a binary logistic regression to predict an outcome (disease vs healthy) and as expected it turned out odds ratio (when the independent variable increases by 1) had a crazy high value and confidence intervals.

Therefore I took the 10th root of odds ratio and CI and now odds are interpretable, however, I don't know if what I did makes sense at all from a mathematical standpoint...

  • 4
    $\begingroup$ I don't see how being less than 1 is a big deal for a predictor (I won't willingly use outdated terminology "independent variable"). Perhaps all you need is a change of units of measurement. 10th root doesn't sound helpful and I don't know why or how that would make results interpretable. I suggest that you show data and/or output from your software. $\endgroup$
    – Nick Cox
    Commented Jun 21, 2018 at 8:44
  • $\begingroup$ In practice, the predictor is always less than 1, it's the norm. FYI I get statistically significant difference between the two groups and a fair effect size ( cliff's delta 0.7, where a two completely separated distributions have delta =1 and complete overlap is 0). Also, I saw an example online where they were using time in seconds and minutes as predictor. To interpret the odds of having a disease for one extra minute of excercise they exponentiated the odds to the 60th power. $\endgroup$
    – frada
    Commented Jun 21, 2018 at 9:21
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    $\begingroup$ Sorry, but I don't think that's made your question clearer. $\endgroup$
    – Nick Cox
    Commented Jun 21, 2018 at 10:24
  • $\begingroup$ If you have as implied just one predictor you should be able to show us (a sample of) your data and certainly software output. $\endgroup$
    – Nick Cox
    Commented Jun 21, 2018 at 10:34
  • $\begingroup$ I'll try my best: I have a set (n=32) of values for a predictor ranging from 0.342 to 0.827 arbitrary units. I have 2 groups (disease vs control) whose medians of said predictor are significantly different. I want to test its predictive ability so I run a binary logistic regression. Odds ratio come out extremely high (thousands). So I want to see how much a change of 0.1(instead of 1) in the predictor value increase/decrease the odds of disease. If I take the 10th root of the odds then I have real-world odds ratio i.e 5 (CI 2-15). The problem is I don't know if this is sound mathematically... $\endgroup$
    – frada
    Commented Jun 21, 2018 at 10:37

2 Answers 2


It appears that your issue is that the independent variable has a small range and that, therefore, the odds ratio for a one unit increase are very high.

Rather than take the 10th root (or any root), it is much more straightforward to simply multiply the IV by 10 or 100. This changes nothing about the meaning of the regression, it just makes the output have fewer digits.

This is equivalent to changing units from (say) kilometers to meters.

There may be some situation where taking the 10th root makes sense, but I can't think of one offhand and this certainly doesn't seem like one. But the problem isn't mathematical, it's substantive. What would the 10th root of your IV be?

  • $\begingroup$ Hi peter thank you. In fact either methods yield the same result. My problem is I want to know the chance of getting a disease for a 0.1 increase in the IV instead of a 1 unit increase. Simply put: a 0.1 increase is clinically meaningful whereas a 1 unit increase is not. I assume it is equivalent for when you have the IV in seconds and want to predict the DV for a 1 minute change (instead of 1 second change). In this case I guess either you divide your IV by 60 or you exponentiate the OR by the 60th power...Same results but different approach. The concern is which one is the most correct? $\endgroup$
    – frada
    Commented Jun 22, 2018 at 8:28
  • $\begingroup$ I would just change the units. $\endgroup$
    – Peter Flom
    Commented Jun 22, 2018 at 12:09

You are right. You can take the 10-th root of the odds ratio (relative change of the odds) for a step of size 1, in order to get the odds ratio for a step of size 0.1

The reason is because a logistic model is modeling the logarithm of the odds as a linear function of the parameter/predictor. And this makes the odds an exponential function of the parameter/predictor.

$$\begin{array}{rrcl}\text{log-odds:} \qquad \qquad & \log\left( \frac{p}{1-p} \right) &=&a+bx\\ \text{odds:} \qquad \qquad & \frac{p}{1-p} &=&e^{a+bx} \end{array}$$

So the changes in the odds are multiplicative when we are adding multiple effects (or increasing the size of an effect).

Consider a change of the parameter/predictor by some step $x_\Delta$. Then this is equivalent to the odds being multiplied by a factor $e^{bx_\Delta}$

$$e^{a+b(x+x_\Delta)} = e^{a+bx} \cdot e^{bx_\Delta}$$

and adding more of those steps (or only a fraction of that step) equates to more multiplications with that factor

$$e^{a+b(x+kx_\Delta)} = e^{a+bx} \cdot e^{bkx_\Delta} = e^{a+bx} \cdot \left(e^{bx_\Delta}\right)^k$$

The image below shows the different representations (probability, odds and log-odds) as function of the parameter $x$. The logistic model $p = (1+\exp(x))^{-1}$ will model the log-odds as a linear function of $x$.



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