Chi-Squared Distribution

Study guide: Exam MAS-I

The Chi-Squared Distribution $\chi^2(n)$ with $n$ degrees of freedom is the distribution of the sum of squares of $n$ independent standard normal variables. It is the sampling distribution behind tests and intervals for a normal variance and behind likelihood ratio tests.

$V = \sum_{i=1}^{n} Z_i^2 \sim \chi^2(n), \quad Z_i \sim N(0, 1)$

$f(x) = \frac{x^{n/2 - 1}\,e^{-x/2}}{2^{n/2}\,\Gamma(n/2)}, \quad x > 0$

$E[V] = n$

$\text{Var}(V) = 2n$

$f$ is the density of $V$ . It is the Gamma distribution with shape $n/2$ and scale 2; with $n = 2$ it is the exponential with mean 2.
Additivity: a sum of independent chi-squared variables is chi-squared, with the degrees of freedom added.
Noncentral: if the $Z_i$ are replaced by independent $N(\mu_i, 1)$ variables with at least one $\mu_i \neq 0$ , the sum of squares has the noncentral chi-squared distribution $\chi^2(n, \lambda)$ , $\lambda = \sum \mu_i^2$ .
Normal samples: for a sample of size $n$ from $N(\mu, \sigma^2)$ with sample variance $S^2$ , $(n - 1)S^2/\sigma^2 \sim \chi^2(n - 1)$ . This gives the confidence interval and the test for $\sigma^2$ .
Likelihood ratio tests: for nested models with $p_1 < p_2$ parameters, $-2\ln\lambda = 2(\ell_2 - \ell_1)$ is approximately $\chi^2(p_2 - p_1)$ in large samples. For GLMs this is the drop in deviance divided by the dispersion parameter.
The CAS MAS-I table gives lower-tail areas for $n = 1$ to 50: the entry in column $p$ is the value $c$ with $P(V \leq c) = p$ , for $p = 0.005,\ 0.010,\ 0.025,\ 0.050,\ 0.950,\ 0.975,\ 0.990,\ 0.995$ . An upper-tail 5% critical value is read from the $0.950$ column. For large $n$ the distribution is approximately normal with mean $n$ and variance $2n$ .
On MAS-I it supplies the critical values for tests of a normal variance, chi-square tests (Hogg §4.7), likelihood ratio tests (Hogg §8.3) and GLM model comparisons. The Student's t-Distribution and the F-Distribution are both built from it.

Confidence Interval for the Variance of Log-Losses The logarithms of 10 claim amounts are treated as a sample from a normal distribution. Their sample variance is $s^2 = 0.36$ . Using the CAS chi-square table, find a 95% confidence interval for $\sigma^2$ .
Answer $(n - 1)S^2/\sigma^2 \sim \chi^2(9)$ , and $(n - 1)s^2 = 9(0.36) = 3.24$ . From the $df = 9$ row, $P(\chi^2_9 \leq 2.70) = 0.025$ and $P(\chi^2_9 \leq 19.02) = 0.975$ . Solving $2.70 \leq 3.24/\sigma^2 \leq 19.02$ for $\sigma^2$ :
 $\begin{align*} \text{lower} &= \frac{3.24}{19.02} \\ &= 0.170 \\ \text{upper} &= \frac{3.24}{2.70} \\ &= 1.200 \end{align*}$ 
The interval $(0.170,\ 1.200)$ is not centred on $0.36$ : the chi-squared distribution is skewed right, so the large quantile sets the lower bound.

Testing an Added Rating Variable A claim-frequency GLM has maximised log-likelihood $-1{,}250.6$ . Adding a rating variable with 4 levels (3 extra parameters) raises it to $-1{,}245.9$ . Test at the 5% level whether the variable is significant.
Answer
 $\begin{align*} 2(\ell_2 - \ell_1) &= 2\bigl(-1{,}245.9 - (-1{,}250.6)\bigr) \\ &= 9.4 \end{align*}$ 
Compare with $\chi^2(3)$ . The $df = 3$ row gives $7.81$ in the $0.950$ column and $9.35$ in the $0.975$ column. Since $9.4 > 7.81$ , reject the smaller model at 5%: the variable is significant. Because $9.4$ also exceeds $9.35$ but not $11.34$ (the $0.990$ column), the p-value lies between 0.01 and 0.025.