PlanetMath: odds ratio

(more info)

Math for the people, by the people.

Main Menu

sections Encyclopædia
Papers
Books
Expositions

meta Requests (211)
Orphanage
Unclass'd (1)
Unproven (472)
Corrections (36)
Classification

talkback Polls
Forums
Feedback
Bug Reports

downloads Snapshots
PM Book

information News
Docs
Wiki
ChangeLog
TODO List
Legalese
About

Entry average rating: No information on entry rating

odds ratio

(Definition)

Suppose the probability of an event is not 1. The odds of the event is the ratio

$\displaystyle \operatorname{odds}(A):=\frac{P(A)}{P(A^c)}=\frac{P(A)}{1-P(A)}.$

For example, if the odds of landing a head in a coin flip is 2, then the probability of landing a head is twice as likely as that of landing a tail.

In a $2\times2$ contingency table, with a dichotomous explanatory variable having levels 1 and 2, and a binary response variable with success and failure as two possible levels of outcomes:

	success	failure
1	$n_{11}$	$n_{12}$
2	$n_{21}$	$n_{22}$

where, given the

th level of

, $n_{i1}$ and $n_{i2}$ are the counts of success and failure, respectively. Two levels of odds for success can be formed:

$\displaystyle \operatorname{odds}(Y=success \mid X=i)=\frac{n_{i1}/(n_{i1}+n_{i2})}{n_{i2}/(n_{i1}+n_{i2})}=\frac{n_{i1}}{n_{i2}}$ , $\displaystyle i=1,2.$

We can form the ratio of these two odds, called the odds ratio:

$\displaystyle OR_{XY}=\frac{\operatorname{odds}(Y=success \mid X=1)}{\operatorname{odds}(Y=success \mid X=2)}.$

To interpret odds ratio, we look at some hypothetical examples:

Suppose that, during an average year, 35 out of 100 young drivers get involved in traffic violations, while 10 out of 100 adult drivers are involved. The odds of a young driver getting involved in traffic violations in a typical year is 35/65, and for the adults, it is 10/90. Calculating the odds ratio, $(35/65)/(10/90)\approx4.85$ , and we find that a young driver is almost 5 times as likely to get a traffic ticket as an adult. This indicates that the driver's age (the explanatory variable) and the chance of getting a traffic ticket (the response variable) might be associated.
Another example. Suppose a die is tossed 100 times and the face with one dot is observed 20 times. Another die twice the size of the first die is tossed 100 times and one is observed 16 times. The odds ratio of getting one between the smaller die and the bigger die is $(20/80)/(16/84)\approx1.31$ , which shows that the odds of getting a one is about the same for the smaller die as for the bigger die.

From the two examples above, we see that the odds ratio can be used to test the association of two dichotomous variables. The further away OR is from 1, the higher the association is between the two variables.

On the other hand, the closer to 1 the odds ratio is, the closer to independence between the variables. In fact, odds ratio = 1 iff the two dichotomous variables are independent, as can be readily seen in the following argument:

$\displaystyle OR_{XY}=1$ iff $\displaystyle \operatorname{odds}(Y=success \mid X=1)=\operatorname{odds}(Y=success \mid X=2).$

From the last equation, we see that

$\displaystyle P(Y=success \mid X=1)$	$\displaystyle =$	$\displaystyle \frac{\operatorname{odds}(Y=success \mid X=1)}{1+\operatorname{odds}(Y=success \mid X=1)}$
	$\displaystyle =$	$\displaystyle \frac{\operatorname{odds}(Y=success \mid X=2)}{1+\operatorname{odds}(Y=success \mid X=2)}$
	$\displaystyle =$	$\displaystyle P(Y=success \mid X=2).$

Because the random variable

is dichotomous, we see that

$\displaystyle P(Y=success)=P(Y=success \mid X=i)$ , i=1,2 $\displaystyle .$

Remarks

Since the odds ratio lies in the interval $\lbrack 0,\infty)$ , 1 is highly skewed towards 0. So we see that even though 10 and 0.1 both indicate similar degrees of association, 0.1 is a lot closer to 1 than 10 is to 1.
By taking the natural log of the odds ratio, $\operatorname{ln}(OR_{XY})$ , we see that the left boundary value 0 of the odds ratio is now stretched to $-\infty$ , and testing of independence is now boiled down to testing whether the log-odds ratio, is 0 or not. It turns out that the log-odds ratio has an asymptotic normal distribution, which can be used to find the confidence interval of OR.
Another way of transforming the odds ratio so as to get a more symmetrical distribution is:
$\displaystyle \frac{OR-1}{OR+1}.$
The transformed odds ratio lies in the interval $\lbrack -1,1 \rbrack$ . This transformed value is also known as the Yule's Q.
The use of odds ratios can be generalized to applications of by 2-way contingency tables, where and $N\ge2$ , as well as higher way contingency tables involving more than one explanatory variable.