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