Deviance Residual

The standardized deviance residual is the signed square root of one observation's contribution to a GLM's deviance, after that contribution is divided by the estimated dispersion parameter. Taylor and McGuire (2016) use it for every residual diagnostic of a reserving model, because it is much closer to normal than the Pearson residual.

$R_i^D = \operatorname{sgn}(Y_i - \hat Y_i)\left(\frac{d_i}{\hat\phi}\right)^{1/2}$

$d_i = 2\left[Y_i\ln\frac{Y_i}{\hat Y_i} - (Y_i - \hat Y_i)\right]$

$d_i$ is observation $i$ 's contribution to the unscaled deviance $D^* = \sum_i d_i$ , and $\hat\phi = D^*/(n - p)$ . The second line is $d_i$ for the over-dispersed Poisson, from its log-density $[y\ln\mu - \mu]/\phi$ compared with the saturated fit $\mu = y$ (with $y\ln y = 0$ at $y = 0$ ). See Residual for the unstandardized version.
Why not Pearson. The Pearson residual $(Y_i - \hat Y_i)/\hat\sigma_i$ is a linear transform of the observation, so it reproduces any skewness in the data. Pierce and Schafer (1986) showed that deviance residuals are normal up to an error of order $m^{-1/2}$ . Taylor and McGuire's histograms from a gamma model of bodily injury claims (CoV $1.16$ ) show highly skewed Pearson residuals and nearly normal deviance residuals.
What to look for. Plotted against accident, development and calendar period or the fitted value, the residuals should scatter evenly about zero with constant spread. A trend means the cell means are wrong, such as a missing effect. Fanning means the dispersion assumption is wrong, the heteroscedasticity a spread plot measures: the standard deviation of standardized residuals should vary randomly about $1$ . The cure is weights roughly inversely proportional to the residuals' variance (see Residual Plot).
Checking the distribution. A Q-Q plot needs observations from one distribution, which raw GLM observations with different means are not. Standardized deviance residuals are asymptotically $N(0,1)$ , so they are plotted against standard normal quantiles, alongside a histogram and the P-P plot.
In the bootstrap. The semi-parametric bootstrap may resample Pearson, deviance or any residual whose inverse exists. Because deviance residuals are asymptotically normal, a parametric bootstrap in Shibata's sense can replace resampling the actual residuals with draws from a normal distribution (see GLM Loss Reserving).

Pearson and Deviance Residuals for Two ODP Cells An over-dispersed Poisson reserving GLM has $\hat\phi = 50$ . Two cells each have a fitted value of $400$ ($000); one came in at $520$ and the other at $280$ .

Compute the standardized Pearson and deviance residuals for each.
Answer Under the ODP, $\hat\sigma = \sqrt{\hat\phi\,\hat Y} = \sqrt{50 \times 400} = 141.42$ , so the Pearson residuals are $\pm 120/141.42 = \pm 0.849$ .
 $\begin{align*} d_{520} &= 2\left[520\ln\frac{520}{400} - 120\right] \\ &= 32.859 \\ R^D_{520} &= +\sqrt{32.859/50} \\ &= 0.811 \\ d_{280} &= 2\left[280\ln\frac{280}{400} + 120\right] \\ &= 40.262 \\ R^D_{280} &= -\sqrt{40.262/50} \\ &= -0.897 \end{align*}$ 
The Pearson residuals are symmetric because they are linear in $Y$ . The deviance residuals are not: the ODP is right-skewed, so a shortfall of $120$ is less likely than an excess of $120$ , and the deviance residual marks it as the more extreme outcome.

From a Spread Plot to Weights In a spread plot of standardized deviance residuals by development year, the standard deviation is about $1.6$ in development years 1–2 and about $0.8$ in later years. After the refit below, $\hat\phi = 100$ . A development-year-1 cell has fitted value $40{,}000$ and actual $46{,}000$ .

Set weights to remove the heteroscedasticity, and show the effect on this cell's variance and standardized residual.
Answer The standard deviation is twice as large in years 1–2, so their dispersion is about $2^2 = 4$ times as large. Since $\phi_i = \phi/w_i$ , set $w = 1/4$ in years 1–2 and $w = 1$ elsewhere.
 $\begin{align*} \mathrm{SD}_{w = 1} &= \sqrt{100 \times 40{,}000} \\ &= 2{,}000 \\ \mathrm{SD}_{w = 1/4} &= \sqrt{4 \times 100 \times 40{,}000} \\ &= 4{,}000 \\ d &= 2\left[46{,}000\ln\frac{46{,}000}{40{,}000} - 6{,}000\right] \\ &= 858.1 \end{align*}$ 
Standardized by $\hat\phi$ alone the residual is $\sqrt{858.1/100} = 2.93$ , which looks like an outlier. Standardized by the cell's own dispersion $\hat\phi/w = 400$ it is $\sqrt{858.1/400} = 1.46$ , an ordinary value. The weights put the early years' residuals back on the same unit spread as the rest, and they also give those cells the larger process variance they really have.