Development Factor Correlation Test

Development Factor Correlation Test checks whether a triangle's development factors are correlated across ages, that is, whether a year with a high factor at one age tends to have a low (or high) factor at the next. The chain ladder's mean assumption rules this out. Mack (1994) tests the whole triangle with Spearman's rank correlation, averaging over adjacent columns:

$T_k = 1 - \frac{6\sum_{i=1}^{I-k}(r_{ik} - s_{ik})^2}{(I-k)^3 - (I-k)}$

$T = \frac{\sum_{k=2}^{I-2}(I-k-1)\,T_k}{\sum_{k=2}^{I-2}(I-k-1)}$

$\text{Var}(T) = \frac{1}{(I-2)(I-3)/2}$

Why it matters. Under Mack's assumption $E(C_{i,k+1} \mid C_{i1}, \dots, C_{ik}) = C_{ik}f_k$ , the expected next factor is $f_k$ whatever the previous factor was, so subsequent factors are uncorrelated. The chain ladder should not be applied to business where a high factor is usually followed by a low one, or vice versa (Mack Chain Ladder Model).
Mack's procedure. $r_{ik}$ ranks the $I-k$ factors $C_{i,k+1}/C_{ik}$ of column $k$ . $s_{ik}$ ranks the preceding factors $C_{ik}/C_{i,k-1}$ of the same accident years, leaving out the latest year, which has no next factor yet. Ranks make the test distribution-free and blunt the unequal variances of the factors. Under the null hypothesis $E(T_k) = 0$ and $\text{Var}(T_k) = 1/(I-k-1)$ , so the weights $I-k-1$ are inversely proportional to the variances. $T$ is treated as Normal. Because the test is only approximate and should catch correlation in a substantial part of the triangle, Mack uses a 50% interval: reject if $|T| > 0.67\sqrt{\text{Var}(T)}$ .
Venter's pairwise test. Venter (1998) computes the sample correlation $r$ for every pair of columns, over the first $n$ years of both, where $n$ is the length of the shorter column. Then:

$T = r\sqrt{\frac{n-2}{1-r^2}} \sim t_{n-2}$

The 2020 erratum. Correlations can be positive or negative, so the test is two-tailed: $r$ is significant at the 10% level when $|T|$ exceeds the $t$ -statistic for $0.95$ at $n-2$ degrees of freedom. The paper as printed said $T$ greater than the statistic for $0.9$ .
Counting significant pairs. With $m$ pairs, the number significant at 10% is Binomial $(m, 0.1)$ , with standard deviation $0.3\sqrt m$ . More than $0.1m + \sqrt m$ significant pairs (the mean plus $3.33$ standard deviations) strongly suggests real correlation, as would a single pair significant at the 0.1% level. If the factors are correlated, their product is biased; $E[XY] = E[X]E[Y] + \text{Cov}(X,Y)$ can correct it (Chain Ladder Assumptions).

Mack's Spearman Test on a Triangle Paid age-to-age factors:

AY 12–24 24–36 36–48 48–60 60–72

$2019$ $2.10$ $1.25$ $1.06$ $1.05$ $1.01$

$2020$ $1.60$ $1.30$ $1.12$ $1.03$

$2021$ $1.90$ $1.20$ $1.08$

$2022$ $1.75$ $1.38$

$2023$ $2.30$

Test whether adjacent factors are correlated, at Mack's 50% level.
Answer Here $I = 6$ , so $T_k$ is computed for $k = 2, 3, 4$ .

$T_2$ (24–36 against 12–24, AYs 2019–2022; the 2023 factor is left out): the 24–36 ranks are $r = 2, 3, 1, 4$ and the 12–24 ranks are $s = 4, 1, 3, 2$ .
 $\begin{align*} \textstyle\sum d^2 &= 4 + 4 + 4 + 4 \\ &= 16 \\ T_2 &= 1 - \frac{6(16)}{4^3 - 4} \\ &= -0.6 \end{align*}$ 
$T_3$ (36–48 against 24–36, AYs 2019–2021): $r = 1, 3, 2$ and $s = 2, 3, 1$ , so $\sum d^2 = 2$ and $T_3 = 1 - 12/24 = 0.5$ .

$T_4$ (48–60 against 36–48, AYs 2019–2020): $r = 2, 1$ and $s = 1, 2$ , so $\sum d^2 = 2$ and $T_4 = 1 - 12/6 = -1$ .
 $\begin{align*} T &= \frac{3(-0.6) + 2(0.5) + 1(-1)}{3 + 2 + 1} \\ &= -0.30 \\ \text{Var}(T) &= \frac{1}{(4)(3)/2} \\ &= \frac{1}{6} \\ 0.67\sqrt{1/6} &= 0.274 \end{align*}$ 
$|T| = 0.30 > 0.274$ , so uncorrelated factors are rejected. The correlation is negative: a high 12–24 factor tends to be followed by a low 24–36 one. Mack's advice is to apply the chain ladder with reluctance here and to look at the correlations in more detail.

AY	12–24	24–36	36–48	48–60	60–72
$2019$	$2.10$	$1.25$	$1.06$	$1.05$	$1.01$
$2020$	$1.60$	$1.30$	$1.12$	$1.03$
$2021$	$1.90$	$1.20$	$1.08$
$2022$	$1.75$	$1.38$
$2023$	$2.30$

Venter's Pairwise Test With the Two-Tailed Erratum Use the triangle above. Test the 12–24 and 24–36 columns for correlation at the 10% level. Then apply Venter's count to all pairs of columns that share at least three years. The $t$ -statistics for $0.95$ are $6.314$ ( $1$ df) and $2.920$ ( $2$ df).
Answer The two columns share $n = 4$ years. $X$ = 12–24 has mean $1.8375$ and $Y$ = 24–36 has mean $1.2825$ :
 $\begin{align*} \textstyle\sum (x - \bar x)(y - \bar y) &= -0.026375 \\ \textstyle\sum (x - \bar x)^2 &= 0.136875 \\ \textstyle\sum (y - \bar y)^2 &= 0.017675 \\ r &= \frac{-0.026375}{\sqrt{0.136875 \times 0.017675}} \\ &= -0.536 \\ T &= -0.536\sqrt{\frac{2}{1 - 0.2875}} \\ &= -0.898 \end{align*}$ 
$|T| = 0.898 < 2.920$ , so this pair is not significant at 10%.

All pairs. Three pairs share at least three years. 12–24 with 36–48 ( $n = 3$ ) gives $r = -0.997$ and $T = -13.3$ , and $|T| > 6.314$ , so it is significant. 24–36 with 36–48 ( $n = 3$ ) gives $r = 0.655$ and $T = 0.87$ , not significant.

With $m = 3$ , the threshold is $0.1(3) + \sqrt 3 = 2.03$ significant pairs. One is well short of it, so Venter's count does not suggest real correlation.

The two tests disagree because they are built differently. Mack's pools the whole triangle and deliberately uses a 50% interval; Venter's asks each pair for 10% significance. Under the printed (pre-erratum) wording, a negative $T$ could never have exceeded the $0.9$ statistic, so negative correlations such as these would have gone unflagged.