I have something strange going on. My phi coefficient of two binary variables is .07 while my odds ratio of the same two binary variables is 1.80. How is this possible?

  • 2
    $\begingroup$ Why not? Binary Phi is a symmetric measure. Odds ratio is asymmetric, either (ad)/(bc) or (bc)/(ad). That is, either 0.56 or 1.80. $\endgroup$
    – ttnphns
    Commented Dec 16, 2013 at 21:07

1 Answer 1


Note that an odds ratio of 1.80 is in the same direction as a phi of 0.07 - they both represent positive association. But they're not monotonically associated.

Binary phi may be either positive or negative:

$\phi = \frac{ad-bc}{\sqrt{(a+b)(c+d)(a+c)(b+d)}}$

It will be in the same direction as log(OR) (log(OR) is symmetric) -- at least when all your counts are >0. This is easy to see, because when $bc>0$:

$\phi = (OR-1)\frac{bc}{\sqrt{(a+b)(c+d)(a+c)(b+d)}} = k(OR-1)$

log(OR) and OR-1 are monotonically related and if $bc>0$ then $k>0$.

There's no particular basis on which to be surprised at the combination of an odds-ratio of 1.8 and a phi of 0.07; it's perfectly easy to get a result like that. Indeed it's trivial to construct cases where phi is lower and the odds ratio is higher or vice-versa.

You can hold OR roughly constant and yet manipulate $\phi$, pushing it up or down. Consider:

1) a=12,b=156,c=4,d=94: \begin{array}{cc} 12 &156\\4 &94 \end{array}

2) a=80,b=60,c=59,d=80: \begin{array}{cc} 80 &60\\59 &80 \end{array}

both have odds ratios close to 1.81, but their phis are quite different.

We can play with $\phi$ while holding ad/bc close to constant by manipulating $k$ above, which we can do fairly readily by changing (at least) two of the numbers a,b,c,d at a time.


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