I was reading a post on this website and it said:
For linear models the deviance equals the RSS (SSE)
$\ D_i=−2∑n_{ik}~log(p_{ik})$
so,
$\ SSE=Di=∑(y_{j}−μ_{i})^2,$
I can not see why this is true , also is there a good source for this kind of material. Thanks
As @jbowman explains in the comment above, the specific form of deviance $D_i= -2 \sum_k n_{ik} \log(p_{ik})$ is only true for the classification problems discussed in the referenced post.
For linear models with Normal errors, the result $SSE = D = \Sigma (y_i - \mu)^2$ can be derived from the general definition of deviance ($D$), scaled deviance ($D^*$) and the normal pdf as follows:
First remember that
$ f(y; \mu, \sigma^2) = \frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(-\frac{(y-\mu)^2}{2\sigma^2}\right) $
then for a single observation we have
$$ \begin{align} D^{*}(y, \mu) &= 2l(y;y) - 2l(\mu;y) \\ &= 2\log(f(y;y)) - 2\log(f(y;\mu))\\ &= 2 \bigg[-\frac{1}{2}\log(2 \pi \sigma^2) - \frac{(y - y)^2}{2\sigma^2}\bigg] - 2 \bigg[-\frac{1}{2}\log(2 \pi \sigma^2) - \frac{(y - \mu)^2}{2\sigma^2}\bigg]\\ &= \frac{(y - \mu)^2}{\sigma^2} \end{align} $$
and $D^{*}(y; \mu) = \frac{D(y; \mu)}{\phi} = \frac{D(y; \mu)}{\sigma^2}$, where the last equality holds since for a Normal distribution the dispersion parameter equals the variance, i.e. $\phi = \sigma ^ 2$. Therefore
$$ D(y, \mu) = (y - \mu)^2 $$
For a good introduction to deviance, see McCullagh and Nelder (1989). Generalized Linear Models (2nd ed.). Chapman and Hall. pp. 23-25, 33-36
