Pearson VS Deviance Residuals in logistic regression

I know that standardized Pearson Residuals are obtained in a traditional probabilistic way:

ri=yiˆπiˆπi(1ˆπi)

and Deviance Residuals are obtained through a more statistical way (the contribution of each point to the likelihood):

di=si2[yilog^πi+(1yi)log(1ˆπi)]

where si = 1 if yi = 1 and si = -1 if yi = 0.

Can you explain to me, intuitively, how to interpret the formula for deviance residuals?

Moreover, if I want to choose one, which one is more suitable and why?

BTW, some references claim that we derive the deviance residuals based on the term

12ri2

where ri is mentioned above.

Answer

Logistic regression seeks to maximize the log likelihood function

LL=kln(Pi)+rln(1Pi)

where Pi is the predicted probability that case i is ˆY=1; k is the number of cases observed as Y=1 and r is the number of (the rest) cases observed as Y=0.

That expression is equal to

LL=(kd2i+rd2i)/2

because a case’s deviance residual is defined as:

di={2ln(Pi)if Yi=12ln(1Pi)if Yi=0

Thus, binary logistic regression seeks directly to minimize the sum of squared deviance residuals. It is the deviance residuals which are implied in the ML algorithm of the regression.

The Chi-sq statistic of the model fit is 2(LLfull modelLLreduced model), where full model contains predictors and reduced model does not.

Attribution
Source : Link , Question Author : Jack Shi , Answer Author : ttnphns

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