Prove that the odds ratio is always greater than the risk ratio.

Question

I am trying to compare the odds and risk ratios of any given study; I have often read that odds ratio is always an overestimate of the risk ratio, but cannot find any literature that goes through and proves this relationship.

For context, this is how the odds ratio and risk ratio are found:

	drug given	no drug given
Developed disease	a	b
Did not develop disease	c	d

Where the odds ratio is $\frac{ad}{bc}$ and the risk ratio is $\frac{a(b+d)}{b(a+c)}$. Assume both of these are greater than 1. Also assume that a, b, c, and d are all greater than or equal to 1. Is there any way I can prove that $\frac{ad}{bc} > \frac{a(b+d)}{b(a+c)}$ ? I tried graphing the relationship and it holds true as long as the greater than 1 conditions are met, but I would like to see how it works out algebraically. I've been finicking around with this and moving things around in the inequality, but cannot seem to get the expression I want. Any help would be hugely appreciated!

Does this answer your question? Given 4 integers, $a, b, c, d > 0$, does $\frac{a}{b} < \frac{c}{d}$ imply $\frac{a}{b} < \frac{a+c}{b+d} < \frac{c}{d}$? — Martin R, Oct 04 '21 at 06:55
$\frac{a(b+d)}{b(a+c)} = \frac{ab +ad}{ab+bc}$ and clearly $\frac{ab}{ab}=1$ — Henry, Oct 04 '21 at 10:02
Thank you for the help! I didn't realize that this was just a mediant inequality, sorry about that. — chimera, Oct 05 '21 at 12:09

score 0 · Accepted Answer · answered Oct 04 '21 at 06:16

0

For positives $a$, $b$, $c$ and $d$ we have $$\frac{ad}{bc}-\frac{a(b+d)}{b(a+c)}=\frac{a}{b}\left(\frac{d}{c}-\frac{b+d}{a+c}\right)=\frac{a(ad-bc)}{bc(a+c)}>0$$ because $ad>bc$.

answered Oct 04 '21 at 06:16

Michael Rozenberg

194,933

1

That is the mediant inequality, and it as been asked and answered more than a dozen times. – Martin R Oct 04 '21 at 06:56

Prove that the odds ratio is always greater than the risk ratio.

1 Answers1