One term of the expansion

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Topic: Generalised likelihood-ratio tests · more questions on this topic

‹ 1 of 2 2 of 2

A cell has expected count E=20 and observed count O=24. The one-cell expansion of the likelihood-ratio contribution, before the sum over cells cancels \sum(O_i-E_i), is

2O\log\frac{O}{E}=2(O-E)+\frac{(O-E)^2}{E}+\text{remainder}.

Compute both the left-hand side and 2(O-E)+(O-E)^2/E. How large is the remainder in this cell?

Worked answer

O-E=4, so

2(O-E)+\frac{(O-E)^2}{E}=8+\frac{16}{20}=8.8, \qquad 2O\log\frac{O}{E}=48\log 1.2\approx 8.751.

The remainder in this cell is about 8.8-8.751=0.049. It is the cubic piece the expansion dropped. Summed over a table whose deviations cancel, the 2(O-E) terms disappear and the comparison that remains is between Pearson’s statistic and -2\log\Lambda.