Logistic regression - the hat matrix In logistic regression with binary outcomes $y_{i},...y_{n}$, the hat matrix is 
$$H = \hat{V}^{1/2}X(X^{T}\hat{V}X)^{-1}X^{T}\hat{V}^{1/2}$$
where $\hat{V}$ is diagonal matrix with $n_{n}\hat{\pi}_{n}(1-\hat{\pi}_{n})$ and the diagonal elements of the hat matrix are $h_{ii} = n_{i}\hat{\pi}_{i}(1-\hat{\pi}_{n})x^{T}_{n}(X^{T}\hat{V}X)^{-1}x_{n}$
Now when I compare this hat matrix with the one in linear regression, there is not the $\hat{V}$ matrix. I know that the diagonal elements on $\hat{V}$ are actually variance of Binomial distribution.
So my question is, what is the purpose of the $\hat{V}$ matrix? Why the variance of Binomial distribution is used in order to estimate the parameters?
 A: In regular multiple regression $\hat{V}=I\hat{\sigma}^2$ where $I$ is the identity matrix, since $(X^TVX)^{-1}=(X^TI\hat{\sigma}^2X)^{-1}=\hat{\sigma}^2(X^TIX)^{-1}=\hat{\sigma}^2(X^TX)^{-1}$.  So the $\hat{V}$ matrix is there, it's just that you can't see it since it's essentially $\sigma^2$ times the identity matrix.  Without the $\hat{V}$, you will not get proper estimates of the variables of each $\hat{\beta}$. A binomial model is used in logistic regression because the outcomes in both a binomial model with $n=1$ is either zero or 1 as is each error term associated with each observation in a logistic regression.
A: After almost a month, it struck me (and the comment from @John Madden helped as well). Logistic regression (the MLE estimates to be precise) is calculated by Newton-Raphson algorithm. And this algorithm can be actually rewritten as Iterative Reweighted Least Squares. 
In WLS, the weights are $w =  1/\sigma^2$. Which is same as the $V$ matrix, which contains variance a well.
