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I heard that "for ungrouped binary data, the deviance cannot be approximated by a $\chi^2$-distribution." Is it true? Why?

kjetil b halvorsen
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Randel
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    Who said so/where was this? Do you have some context? I'm not at all convinced that the claim is true in sufficiently large samples (indeed, I believe I have some reason to think it's false), but it would be most invigorating to be convinced of it. – Glen_b Jun 24 '13 at 04:42
  • I got this from an instructor's slides. It is a remark comes after one says that "This approximation is not good when some of $n_i$s are very small, or the fitted probabilities are near zero or unity." I would like to hear your reason. – Randel Jun 24 '13 at 05:05
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    If you start adding conditions like those, there may be an argument for the claim... but your original statement doesn't include those conditions. That's like saying "The binomial cannot be approximated by a normal" without adding that you actually mean "... when $n$ is small and $p$ is near $0$ or $1$". – Glen_b Jun 24 '13 at 05:51
  • Thanks! In my mind, "ungrouped binary data" suggests that data are listed by subject number and $m_i=1$ for each $i$. I found a good reference for this question, Page 120 of McCullagh and Nelder (1989). For this extremely sparse instance, the deviance has a conditionally exactly degenerate distribution given the estimate $\beta$, thus fails to have the properties required for goodness-of-fit statistics. – Randel Jun 24 '13 at 16:27
  • It looks like I took that phrase to mean something else, but your interpretation makes sense. But under that interpretation how does the likelihood change if you group or don't group the observations? – Glen_b Jun 24 '13 at 21:21
  • I think the likelihood function would be the same, so the ML estimates and SE values are the same for either type of data. But the full models would change, and the deviances would be different. They both use the same formula, $D=2\Sigma\left{ylog\left(y/\hat{\mu}\right)+\left(m-y\right)log\left[\left(m-y\right)/\left(m-\hat{\mu}\right)\right]\right}$. For ungrouped binary data, $m=1$ and $y=0,1$, so the deviance would be different and not reliable. – Randel Jun 25 '13 at 00:37
  • Ah, I believe I get it, finally. Perhaps you should write your comments up as an answer. – Glen_b Jun 25 '13 at 00:39

2 Answers2

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The context for this question is that, it is a remark comes after one says that "This ($\chi^2$) approximation is not good when some of $n_i$s are very small, or the fitted probabilities are near zero or unity."

The phrase "ungrouped binary data" suggests that data are listed by subject number and $m_i=1$ for each $i$. I found a good reference for this question, Page 120 of McCullagh and Nelder (1989). For this extremely sparse instance, the deviance has a conditionally exactly degenerate distribution given $\hat{\beta}$, thus fails to have the properties required for goodness-of-fit statistics.

The likelihood function would be the same for either type of data, so the ML estimates and SE values are the same. But the full models would change, and the deviances would be different. They both use the same formula, $2\Sigma\left\{ y\mathrm{log}\left(y/\hat{\mu}\right)+\left(m-y\right)\mathrm{log}\left[\left(m-y\right)/\left(m-\hat{\mu}\right)\right]\right\}$. For ungrouped binary data, $m=1$ and $y=0,1$, so the deviance would be different and not reliable.

Randel
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Build a standard two way contingency table where the rows are the ungrouped observations and columns are success/failure. You can prove that the chi-squared statistic is exactly equal to your number of observations. The fact that the test statistic depends only on the volume of observations suggests that it is not useful statistic for ungrouped data.

See Agresti’s, Categorical Data Analysis, problem 4.18.

Chandler
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