# Deviance Residuals

Recall that the likelihood of a model is the probability of the data set given the model ($$P(D|\theta)$$).

The deviance of a model is defined by

$D(\theta,D) = 2(\log(P(D|\theta_s)) - \log(P(D|\theta)))$

where $$\theta_s$$ is the saturated model which is so named because it perfectly fits the data.

In the case of normally distributed errors the likelihood for a single prediction ($$\mu_i$$) and data point ($$y_i$$) is given by

$P(y_i|\mu_i) = \frac{1}{\sigma\sqrt{2\pi}} \exp\bigg(-\frac{1}{2}\bigg(\frac{y_i - \mu_i}{\sigma}\bigg)^2\bigg)$ and the log-likelihood by

$\log(P(y_i|\mu_i)) = -\log(\sigma) - \frac{1}{2}\big(\log(2\pi)\big) -\frac{1}{2}\bigg(\frac{y_i - \mu_i}{\sigma}\bigg)^2$

The log-likelihood for the saturated model, which is when $$\mu_i = y_i$$, is therefore simply

$\log(P(y_i|\mu_{s_i})) = -\log(\sigma) - \frac{1}{2}\big(\log(2\pi)\big)$

It follows that the unit deviance is

$d_i = 2(\log(P(y_i|\mu_{s_i})) - \log(P(y_i|\mu_i)))$

$d_i = 2\bigg(\frac{1}{2}\bigg(\frac{y_i - \mu_i}{\sigma}\bigg)^2\bigg)$

$d_i = \bigg(\frac{y_i - \mu_i}{\sigma}\bigg)^2$

As the deviance residual is the signed squared root of the unit deviance,

$r_i = \text{sign}(y_i - \mu_i) \sqrt{d_i}$ in the case of normally distributed errors we arrive at $r_i = \frac{y_i - \mu_i}{\sigma}$ which is the Pearson residual.

To confirm this consider a normal distribution with a $$\hat{\mu} = 2$$ and $$\sigma = 0.5$$ and a value of 1.

library(extras)
mu <- 2
sigma <- 0.5
y <- 1

(y - mu) / sigma
#> [1] -2
dev_norm(y, mu, sigma, res = TRUE)
#> [1] -2
sign(y - mu) * sqrt(dev_norm(y, mu, sigma))
#> [1] -2
sign(y - mu) * sqrt(2 * (log(dnorm(y, y, sigma)) - log(dnorm(y, mu, sigma))))
#> [1] -2