Actuarial Notes

Pearson Residual

Pearson Residual is an actual incremental loss minus its fitted value, divided by the standard deviation the model implies for that cell. For the over-dispersed Poisson model the divisor is \sqrt{m}. In the ODP bootstrap these are the residuals that are resampled, and Shapland (2016) defines three versions of them: unscaled, scaled and standardized.

r_{w,d} = \frac{q(w,d) - m_{w,d}}{\sqrt{m_{w,d}^{z}}}

r^{S}_{w,d} = r_{w,d} \times \sqrt{\frac{N}{N - p}}

r^{H}_{w,d} = r_{w,d} \times \sqrt{\frac{1}{1 - H_{i,i}}}

  • q(w,d) is the actual and m_{w,d} the fitted incremental loss, and z = 1 for the ODP. N is the number of incremental cells, and p the number of parameters, which is 2n - 1 for an n \times n triangle (per Shapland's errata). H_{i,i} is the cell's diagonal element of the hat matrix H = X(X^{T}WX)^{-1}X^{T}W, where X is the design matrix and W the diagonal matrix of fitted values.
  • Why Pearson. England and Verrall (1999) considered the deviance, Pearson and Anscombe residuals; the Pearson residuals are preferred because they are calculated consistently with the scale parameter, \phi = \sum r_{w,d}^2/(N - p). Dividing by \sqrt{m} also makes them "exposure independent" (Shapland's Section 4.7.1). (See Residual and Deviance Residual for the other kinds.)
  • The three versions. The unscaled residual (Shapland's 3.16) gives \phi. England and Verrall (2002) multiplied by the degrees-of-freedom factor \sqrt{N/(N-p)} to include the over-dispersion, which gives the scaled residual (3.20). Pinheiro et al. noted that this does not create standardized residuals, which is what makes sure they all have the same variance. The hat-matrix factor does, and it replaces and improves on the degrees-of-freedom factor, giving the standardized residual (3.24). \phi is still calculated from the unscaled residuals, though \sum (r^{H})^2/N approximates it.
  • Zero residuals are dropped. Each corner cell of the triangle has a parameter of its own, so it is fitted exactly. Its residual is zero, and its H_{i,i} = 1. It is left out of the pool that is resampled.
  • In the bootstrap, a sampled residual r^* makes a pseudo value q^*(w,d) = r^* \sqrt{m_{w,d}} + m_{w,d}. For a negative fitted value both formulas use \sqrt{|m_{w,d}|}. The residuals are assumed independent and identically distributed, which is what heteroscedasticity violates; they need not be normal.

Unscaled, Scaled and Standardized Residuals for One Cell A 4 \times 4 triangle of incremental losses ($000s) has actual q(2,2) = 700 and fitted m_{2,2} = 657.85 under the ODP model. The cell's hat-matrix diagonal is H = 0.558. Compute its three residuals. If the standardized residual is then sampled for cell (3,1), whose fitted value is 1{,}158.8, what pseudo value results?

Answer

\begin{align*}
r_{2,2} &= \frac{700 - 657.85}{\sqrt{657.85}} \\
&= \frac{42.15}{25.649} \\
&= 1.643
\end{align*}

The triangle has N = 10 cells and p = 2(4) - 1 = 7 parameters:

\begin{align*}
r^{S}_{2,2} &= 1.643 \times \sqrt{\tfrac{10}{3}} \\
&= 1.643 \times 1.8257 \\
&= 3.000 \\
r^{H}_{2,2} &= 1.643 \times \sqrt{\tfrac{1}{1 - 0.558}} \\
&= 1.643 \times 1.5042 \\
&= 2.472
\end{align*}

The degrees-of-freedom factor inflates every residual by the same 1.826. The hat-matrix factor, 1.504 here, is specific to the cell: it is larger for a cell with more leverage over its own fitted value.

Resampled into cell (3,1):

\begin{align*}
q^*(3,1) &= 2.472\sqrt{1{,}158.8} + 1{,}158.8 \\
&= 84.1 + 1{,}158.8 \\
&= 1{,}242.9
\end{align*}

The residual moves the value by 2.472 of that cell's standard deviations, \sqrt{1{,}158.8} = 34.0, not by the 42.15 seen in cell (2,2).

The Hat Matrix Adjustment Across a Triangle The same 4 \times 4 ODP fit has these unscaled residuals r and hat-matrix diagonals H, computed from X (one column for each \alpha_w and each cumulative \beta_d) and W = \mathrm{diag}(m):

Cell r H
(1,1) -0.694 0.696
(2,1) -0.569 0.712
(3,1) 1.211 0.760
(4,1) 0 1
(1,2) -0.038 0.537
(2,2) 1.643 0.558
(3,2) -1.579 0.591
(1,3) 1.389 0.554
(2,3) -1.328 0.592
(1,4) 0 1

Find the standardized residuals that go into the sampling pool, and compare \sum (r^H)^2/N with \phi.

Answer The H values sum to 7.000, the number of parameters, as the trace of a hat matrix must. The two corners have H = 1: each is fitted exactly by its own parameter, its factor 1/\sqrt{1 - H} is undefined, and its zero residual is excluded from the pool.

The other eight are r^H = r/\sqrt{1 - H}. For example, r^H_{1,1} = -0.694/\sqrt{0.304} = -1.26. The table carries three decimals, computed before rounding:

Cell r^H
(1,1) -1.260
(2,1) -1.059
(3,1) 2.471
(1,2) -0.055
(2,2) 2.472
(3,2) -2.471
(1,3) 2.079
(2,3) -2.079 
\begin{align*}
\phi &= \frac{\sum r^2}{N - p} \\
&= \frac{11.160}{3} \\
&= 3.720 \\
\phi^{H} &= \frac{\sum (r^{H})^2}{N} \\
&= \frac{29.675}{10} \\
&= 2.967
\end{align*}

The standardized residuals are much larger than the unscaled ones, because a small triangle's fit absorbs most of each cell's deviation. Shapland keeps \phi from the unscaled residuals; \phi^H is only an approximation to it, and on a triangle this small the two differ noticeably.