逻辑回归中的Pearson VS Deviance残差

我知道标准化的Pearson残差是以传统的概率方式获得的：

r_{i} = \frac{y_{i} - π_{i}}{\sqrt{π_{i} (1 - π_{i})}}

$r_i = \frac{y_i-\pi_i}{\sqrt{\pi_i(1-\pi_i)}}$

和偏差残差通过更统计的方式获得（每个点对可能性的贡献）：

d_{i} = s_{i} \sqrt{- 2 [y_{i} \log \hat{π_{i}} + (1 - y_{i}) \log (1 - π_{i})]}

$d_i = s_i \sqrt{-2[y_i \log \hat{\pi_i} + (1 - y_i)\log(1-\pi_i)]}$

其中 = 1，如果 = 1和 = -1，如果 = 0。 $s_i$ $y_i$ $s_i$ $y_i$

您能直观地向我解释如何解释偏差残差的公式吗？

此外，如果我要选择一个，那一个更合适，为什么呢？

顺便说一句，一些参考文献声称我们基于以下项得出偏差残差

- \frac{1}{2} {r_{i}}^{2}

$-\frac{1}{2}{r_i}^2$

其中是上面提到的。 $r_i$

— 杰克·史
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任何想法都将不胜感激

— Jack Shi

当您说“一些参考”时...哪些参考以及如何进行？

— Glen_b-恢复莫妮卡

Logistic回归寻求最大化对数似然函数

$LL = \sum^k \ln(P_i) + \sum^r \ln(1-P_i)$

where $P_i$ is the predicted probability that case i is $\hat Y=1$ ; $k$ 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

$LL = ({\sum^k d_i^2} + {\sum^r d_i^2})/-2$

because a case's deviance residual is defined as:

$d_i = \begin{cases} \sqrt{-2\ln(P_i)} &\text{if } Y_i=1\\ -\sqrt{-2\ln(1-P_i)} &\text{if } Y_i=0\\ \end{cases}$

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 $2(LL_\text{full model} - LL_\text{reduced model})$ , where full model contains predictors and reduced model does not.

— ttnphns
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