13
$\begingroup$

Suppose I have a scenario like this :

For every 1-point increase in X, odds ratio of event Y happening is 0.80

Does that mean the same as 'For every 1-point increase in X, odds of event Y happening is reduced by 20%'?

Whereas 'For every 1-point increase in X, event Y is 20% less likely to happen' is an incorrect interpretation of odds ratio (that's interpreting it as relative risk), am I correct?

$\endgroup$

5 Answers 5

9
$\begingroup$

As other answers have clearly articulated, you can't represent an odds ratio as a simple percent increase or decrease of an event happening, as this value depends on the baserate. However, if you have a meaningful baserate, you can calculate the percent success (or failure) relative to that rate.

For example, if we have an odds ratio of 0.75 for the effect of an intervention and we know that the baserate for failure (failure in the control group, for example) is 20%, then the failure rate for the treatment group based on an odds ratio of 0.75 is: $$ p_{treatment} = \frac{OR \times p_{control}}{1 + OR \times p_{control} - p_{control}} = \frac{.75 \times .2}{1 + (.75 \times .2) - .2} = .158 $$

Thus, an odds ratio of .75 translates into a failure rate of 15.8% in the treatment group relative to an assumed failure rate of 20% in the control group.

This translation of odds ratios into an easily understand metric is commonly used in meta-analyses of odds ratios.

This simplifies if we assume a baserate of .50 to: $$ p_{treatment} = \frac{OR}{1+OR} $$

$\endgroup$
1
$\begingroup$

Both the probability and the odds measure how likely it is that something happens. The odds by showing the expected number of success per failure and the probability by showing the expected number of success per trial (yes, I am a frequentist). So strictly speaking both interpretations are correct. However, the danger is that a reader might equate likely with probability, and thus misinterpret the results. Your first sentence avoids that danger.

$\endgroup$
1
$\begingroup$

It would be nice if statistical concepts could all be reduced to simple percentages and retain the correct information. However, this is not the case.

Framing the odds ratio effect as a percentage reduction or increase completely dissociates the OR measure from an interpretation aligned with users’ expectations of analyses.

In the colloquial interpretation, a 20% reduction does not correlate to, but implies a decrease from 100% to 80% or 20% reduction from some baseline. However, OR is a measure of relative reduction relatable but not intrinsically descriptive of the absolute reduction. For an interpretation anchored in the absolute, which is the interpretation we expect and observe more readily, one needs to calculate the mean probability of an event under the mean or meaningful conditions and then modify this probability with the risk factor or treatment to get the new, absolute probability. Only then can the importance of the factor be weighed and judged.

Thus, your first interpretation is the safest, but neither answer is well grounded in absolute probability that the human mind operates upon, albeit tenuously.

$\endgroup$
0
$\begingroup$

You can interpret odds ratio as a conditional probabilities. So as I see your case, the odds ratio of event $Y$ to happen is $P(Y=True|X=x)= 0.8$. If I understand you correctly, this seems to be the case for all $x \in X$ since you say that after a 1-point increase in $X$ the probability of $Y$ happening is still $0.8$. But that would be the notion of statistical independence, so $P(Y|X)=P(Y)=0.8$ which means the odds of your event $Y$ happening is $0.8$ for all cases and so, both of your statements are false.

$\endgroup$
0
$\begingroup$

Yes, you are right in both of your observations. with one unit increase in X, the odds ratio of event Y happening is 0.80

Does that mean the same as 'For every 1-point increase in X, odds of event Y happening is reduced by 20%'?

Odds ratio (for one unit increase in X) 

                = (Odds in favor of Y at X = x + 1) / (Odds in favor of Y at X = x)

0.8 = (Odds in favor of Y at X = x + 1) / (Odds in favor of Y at X = x)

0.8*(Odds in favor of Y at X = x) = (Odds in favor of Y at X = x + 1)

Now,

change in odds (increase/reduction) 
                = (Odds in favor of Y at X = x + 1) - (Odds in favor of Y at X = x)

**change in odds** = 0.8*(Odds in favor of Y at X = x) - (Odds in favor of Y at X = x)
               = - 0.2 (Odds in favor of Y at X = x) 
               = **20% reduction**

*In short, (Odds Ratio - 1)100 gives the percentage change in Odds

Whereas 'For every 1-point increase in X, event Y is 20% less likely to happen' is an incorrect interpretation of odds ratio (that's interpreting it as relative risk), am I correct?

Yes, that's an incorrect statement as odds are different from probabilities. In the same fashion, the odds ratio is different for relative risk.

$\endgroup$

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