Why is it futile to use the deviance as a goodness-of-fit measure for Bernoulli data? In Ordinal Data Modelling by Johson & Albert, page 102-103:

For Bernoulli observations [...] the asymptotic chi-squared
  distribution of the deviance statistic may not pertain. Indeed, for
  the linear logistic models with Bernoulli observations, the deviance
  function can be expresses solely as a function of the MLE of the
  regression parameter, which demonstrates the futility of using this
  statistic to measure goodness of fit for such data.

Could someone kindly explain what this means, please? 
 A: I am reiterating from lecture notes I have on the subject. 
The issue is the log-likelihood function for the saturated model is always zero when dealing with a binary distribution. 
In the case of that $m_j$ = 1, the log likelihood is
log(L) = $\Sigma_i$ =[$y_i$log($\theta$($x_i$)) + (1 - $y_i$)log(1-$\theta$($x_i$)) + log(${1\choose y_i}$))
If $y_i$ = 0: then you can input it into the above equation and find that the sum is equal to zero
If $y_i$ = 1: then you can input it into the above equation and fine that the sum is also equal to zero.
So in the case where $m_i$ = 1, the deviance between the saturated model and the current model depends only on log($L_m$). Recalling that deviance is defined as 
$G^2$ = 2[log($L_s$) - log($L_m$)].
Deviance doesn't proved an assessment of the goodness-of-fit of the model! (The exclamation is in the notes). It also does not have a $\chi^2$ distribution. 
However, we can use deviance to compare two models; the difference between two deviance still has an approximate $\chi^2$ distribution.
