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