Why Yates correction in McNemar's Test subtract 1?

Question

McNemar's test statistic is given by:

Yates's correction defines something like this:

However, the correction for the McNemar's test statistic requires 1 (not 0.5) be subtracted, which is not the same as other chi-square tests. Why?

Answer 1

The uncorrected McNemar statistic is the square of a standardized* difference in two counts; the continuity correction is $\frac12$ to both, but it's $-\frac12$ for the larger of the two and $+\frac12$ for the smaller of the two[1].

* under the null

As a result, when you take $|B-C|$ the total of the two continuity corrections is always $-1$.

(This post may be of some possible interest as well.)

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Edit in response to question in comment

You need to think about what is "as, or more extreme" here.

The calculation effectively conditions on the discordant-pair total $B+C=N_{d}$ and does an approximation to the binomial test that $p_B=0.5$. Imagine you have observed say $b=4$ and $c=12$:

However, the continuity correction reasoning itself doesn't rely on the binomial -- it only relies on the fact that we want to look at the cases that are at least as extreme as $(4,12)$, which are $b=4,3,2,1,0$ plus the "other tail" values $12,13,14,15,16$. Then the usual continuity correction reasoning would say you want the approximating normal from 4.5 down and from 11.5 up.

So more generaly we're always taking both values toward the middle, $(b+c)/2$ by 0.5, which means the larger values reduce by 0.5 and the smaller values increase by 0.5. This reasoning follows through to the squared form used in the chi-squared version.

[1] Edwards, A.L. (1948),
"Note on the correction for continuity in testing the significance of the difference between correlated proportions."
Psychometrika, 13:3 (Sept), pp185-187.