1
$\begingroup$

I was reading a Wikipedia article on the logit function, and I came across the following in the definition section:

If $p$ is a probability, then $p/(1 - p)$ is the corresponding odds; the logit of the probability is the logarithm of the odds. Similarly, the difference between the logits of two probabilities is the logarithm of the odds ratio (R), ...

$log(R) = log\left( \dfrac{ p_1/(1 - p_1)}{p_2/(1 - p_2)} \right) = log \left( \dfrac{p_1}{1 - p_1} \right) - log \left( \dfrac{p_2}{1 - p_2} \right) = logit(p_1) - logit(p_2)$

However, the Wikipedia article on the odds ratio states that the odds ratio can be computed in the following steps:

  1. For a given individual that has "B" compute the odds that the same individual has "A".
  2. For a given individual that does not have "B" compute the odds that the same individual has "A".
  3. Divide the odds from step 1 by the odds from step 2 to obtain the odds ratio (OR).

Therefore, based on the article on the odds ratio, the claim that the difference between the logits of any two probabilities $p_1$ and $p_2$ is the logarithm of the odds ratio must be false; rather, the difference between the logits of two probabilities that satisfy the aforementioned conditions outlined in the Wikipedia article on the on the odds ratio, and would hence qualify as an "odds ratio", is the logarithm of the odds ratio. The claim that this is true for any two probabilities $p_1$ and $p_2$ seems to be in contradiction with the odds ratio article, which states that the probabilities must satisfy certain conditions (that is, there must exist a certain conditional relationship between them)?

Or am I misunderstanding something?

EDIT: The definition of the odds ratio from Introduction to Probability by Blitzstein and Hwang illustrates my point:

enter image description here

We have $odds(D|C)$ in the numerator and $odds(D|C^c)$ in the denominator. As such, the two odds are not simply any two odds from any two probabilities $p_1$ and $p_2$; rather, they are both odds in terms of the events $D$ and $C$. The issue is that the Wikipedia article does not state that this must be the case and, instead, based on my reading of it, incorrectly states that we can have any two probabilities $p_1$ and $p_2$. And if $p_1$ and $p_2$ can be any probabilities, then can we not have $OR = \dfrac{odds(D|C)}{odds(A|B)}$, which would not be an odds ratio, since the events in the numerator and denominator are arbitrary and, therefore, do not satisfy the definition of an odds ratio?

$\endgroup$

1 Answer 1

0
$\begingroup$

Odds is one probability divided by another. Odds ratio is odds divided by odds. Obviously, in both case the probabilities and odds need to be related in some way so that the odds or odds ratio have any meaningful interpretation. But do formally, odds ratio is ratio of two odds, same as you can divide any two numbers by themselves and the result will be another number.

$\endgroup$
1
  • $\begingroup$ Ahh, ok, that makes sense. The odds ratio Wikipedia article made it seem like the odds ratio couldn't simply be the ratio of any two odds but, rather, that it had to be the ratio of two odds that satisfied the provided definition. Thank you for the clarification. $\endgroup$ Commented May 27, 2018 at 6:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.