Kolmogorov-Smirnov Test

Study guide: Exam MAS-I

The Kolmogorov-Smirnov Test (K–S) checks whether individual data came from a hypothesized continuous distribution $F^*$ by measuring the largest vertical gap $D$ between $F^*$ and the empirical distribution function $F_n$ . A large gap is evidence against the model, so $H_0$ is rejected when $D$ exceeds its critical value.

$D = \max_{x}\,\big|F_n(x) - F^*(x)\big|$

$D = \max_{1 \le i \le n}\left\{F^*(x_{(i)}) - \frac{i-1}{n},\ \ \frac{i}{n} - F^*(x_{(i)})\right\}$

$x_{(1)} \le \cdots \le x_{(n)}$ are the ordered observations. $F_n$ jumps at each data point, so the gap must be checked on both sides of every jump: against the step's height before it, $(i-1)/n$ , and after it, $i/n$ . With tied values, compare $F^*(x)$ with $F_n(x^-)$ and $F_n(x)$ .
Critical values. The distribution of $D$ does not depend on $F^*$ when $F^*$ is fully specified, so one table serves every model. The SOA's Exam C sample questions supply $1.22/\sqrt{n}$ , $1.36/\sqrt{n}$ , $1.48/\sqrt{n}$ and $1.63/\sqrt{n}$ at the 10%, 5%, 2.5% and 1% levels (for $n = 5$ : $0.546, 0.608, 0.662, 0.729$ ). These are large-sample values: for small $n$ an exact table gives a slightly smaller critical value — at $n = 20$ the exact 5% value is $0.294$ , against $1.36/\sqrt{20} = 0.304$ . The CAS MAS-I tables do not include them.
Limitations. It applies only to continuous distributions and is most sensitive near the centre of the distribution, less so in the tails. If parameters are estimated from the same data, the tabled critical values are no longer valid (they must be found by simulation). For grouped data or a discrete model, use the Chi-Square Goodness-of-Fit Test.
Anderson–Darling test. A refinement that weights squared differences by $1/[F^*(1-F^*)]$ , so it puts more emphasis on the tails — where large losses live. With $u_i = F^*(x_{(i)})$ :

$A^2 = -n - \frac{1}{n}\sum_{i=1}^{n}(2i-1)\left[\ln u_i + \ln\big(1 - u_{n+1-i}\big)\right]$

Unlike $D$ , the critical values of $A^2$ depend on which distribution is being tested, so they are tabulated separately for the normal, lognormal, exponential, Weibull and a few others.

K–S Test of an Exponential Severity Model Eight claims are observed: $45, 110, 180, 260, 390, 520, 800, 1{,}350$ . Test at the 5% level whether they come from an exponential distribution with mean $400$ .

Answer With $F^*(x) = 1 - e^{-x/400}$ , compare $F^*$ with the empirical step just below and just at each point:

$x_{(i)}$ $F^*(x_{(i)})$ $\frac{i-1}{8}$ $\frac{i}{8}$ larger gap

45 0.1064 0.000 0.125 0.1064

110 0.2404 0.125 0.250 0.1154

180 0.3624 0.250 0.375 0.1124

260 0.4780 0.375 0.500 0.1030

390 0.6228 0.500 0.625 0.1228

520 0.7275 0.625 0.750 0.1025

800 0.8647 0.750 0.875 0.1147

1,350 0.9658 0.875 1.000 0.0908

So $D = 0.1228$ , at $x = 390$ . The large-sample 5% critical value is $1.36/\sqrt{8} = 0.481$ ; since $0.123 < 0.481$ , do not reject the exponential model.

$x_{(i)}$	$F^*(x_{(i)})$	$\frac{i-1}{8}$	$\frac{i}{8}$	larger gap
45	0.1064	0.000	0.125	0.1064
110	0.2404	0.125	0.250	0.1154
180	0.3624	0.250	0.375	0.1124
260	0.4780	0.375	0.500	0.1030
390	0.6228	0.500	0.625	0.1228
520	0.7275	0.625	0.750	0.1025
800	0.8647	0.750	0.875	0.1147
1,350	0.9658	0.875	1.000	0.0908

Anderson–Darling Statistic for a Small Sample Four losses, $20, 70, 130, 260$ , are tested against an exponential distribution with mean $100$ . Compute $A^2$ .
Answer The fitted CDF values are $u = (0.1813,\ 0.5034,\ 0.7275,\ 0.9257)$ , so $\ln u_i = (-1.7078, -0.6863, -0.3182, -0.0772)$ and $\ln(1 - u_{5-i}) = (-2.6, -1.3, -0.7, -0.2)$ , i.e. $-x_{(5-i)}/100$ . The weighted terms $(2i-1)[\ln u_i + \ln(1 - u_{5-i})]$ are $-4.3078, -5.9590, -5.0909, -1.9402$ , which sum to $-17.2979$ :
 $\begin{align*} A^2 &= -4 - \frac{1}{4}(-17.2979) \\ &= -4 + 4.3245 \\ &= 0.3245 \end{align*}$ 
A small $A^2$ means the fitted CDF tracks the data closely, including at the extremes.