How to get approximative confidence interval for Gini and AUC? I found an interesting way to calculate a confidence interval for the Gini and respectively AUC coefficient for credit risk scoring. 
Question: Can anyone explain me, why the sum 
$$
 AUC = \frac{1}{n \cdot m}\sum_{i=1}^n \sum_{j=1}^m S(X_i,y_j), \text{with}
$$
$$
  S(x_i,y_j) = \begin{cases} 1 & \text{if } x_i > y_j \\ 0.5 & \text{if } x_i = y_j \\ 0 & \text{if } x_i < y_j
\end{cases}
$$
has got the standard deviation 
$$
SE(AUC)= $$ $$\sqrt{\frac{AUC(1-AUC) + (m-1)\left(\frac{AUC}{2-AUC}- AUC^2\right) + (n-1)\left(\frac{2AUC^2}{1+AUC}-AUC^2\right)}{n \cdot m}} $$
and especially why the AUC is accepted as standard normal distributed?
In the book "Kreditrisikomessung" written by Henking, Bluhm and Fahrmeier (ISBN-10 3-540-32145-4 on page 223) is then a confidence interval given by
$$
AUC \pm z_{\alpha/2} SE(AUC)
$$
 A: Assumptions addressed
The paper that proposed your formula (Hanely and MacNeil 1982) explicitly states that a key assumption is that the ratings are derived from a continuous scale that does not produce ‘ties’. 
A typical explanation of "AUC is the probability that a sample randomly taken from the positive cases will rank higher than a randomly chosen negative case". If you have a tie then the positive case is not ranked higher and so S should = 0 to fit this explanation. If you include 0.5 for ties then the definition should be "AUC is the probability that a sample randomly taken from the positive cases will rank equal or higher than a randomly chosen negative case"
This means that 1-AUC is the probability that a sample randomly taken from the positive cases will rank lower than a randomly chosen negative case (for 0.5 when tied) or equal or lower for 0 when tied case. 
If you compare tie behaviour of S= 0 with S = 0.5 then AUC and 1-AUC will swap values everytime a tie occurs, so if there are a large proportion of ties the SE will diverge significantly between the two definitions. Since using S=0.5 when tied increases AUC it will therefore lead to a decrease in the calculated SE. 
I can see why Hanley and MacNeil ignored ties, maybe other readers will know of another source that explicitly details how ties impact on the SE calculation and can fill in that gap better.
Meaning of Elements in the equation
I'll include all elements for completeness, even the obvious ones.
AUC(1-AUC) is self explanatory, the AUC times its inverse. It is at a maximum when AUC = 0.5, i.e. it is a squared value and becomes smaller as AUC deviates from 0.5. Since AUC = 0.5 is equivalent to random chance then you would expect there to be large uncertainty at this value and for it to diminish as the value increases.
M is the number of positive cases
Hanley and MacNeil had compared gaussian, gamma and negative exponential distribution assumptions and chose the latter as it was the most conservative of that set and also provided the easiest terms to use in the equation. This assumption  is the basis of using AUC/(2-AUC) and 2AUC^2 / (1+AUC) in the formula.
AUC/(2-AUC) is the probability of ranking two randomly chosen positive sample higher than a negative one based on an underlying assumption of a negative exponential distribution
Thus $$ \frac{AUC}{2-AUC}- AUC^2 $$ is the difference between the probability of two positive samples being ranked higher and the square of the probability of one sample being ranked higher than a random negative one.
N is the number of negative cases 
2AUC^2 / (1+AUC) is the probability of ranking one randomly chosen positive samples higher than two randomly chosen negative ones based on an underlying assumption of  negative exponential distribution 
Thus $$ \frac{2AUC^2}{1+AUC}-AUC^2\ $$ is the difference between the probability of one positive samples being ranked higher than two negative ones and the square of the probability of one positive sample being ranked higher than one random negative sample.
Basically the equation is taking second order effects into account as well as first order effects (the SE for a Bernoulli only uses first order effects)
The SE equation you quote assumes negative exponential rather than standard normal distribution. I’ll add a comment to the question to get more details on what you mean and what sources. Without understanding the background to this I can’t really address it. 
