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