Noncentral Chi-Squared Distribution

Study guide: Exam MAS-I

The Noncentral Chi-Squared Distribution $\chi^2(r, \theta)$ is the distribution of $\sum X_i^2/\sigma^2$ for $r$ independent normal variables $X_i \sim N(\mu_i, \sigma^2)$ whose means are not all zero. When the null hypothesis is false, a chi-square statistic built from normal data follows it and an F statistic follows the noncentral F-distribution built from it, so the two give the power of those tests.

$Y = \frac{1}{\sigma^2}\sum_{i=1}^{r} X_i^2 \sim \chi^2(r, \theta)$

$\theta = \frac{1}{\sigma^2}\sum_{i=1}^{r} \mu_i^2$

$M_Y(t) = \frac{e^{\theta t/(1 - 2t)}}{(1 - 2t)^{r/2}}, \quad t < \tfrac{1}{2}$

$f(y) = \sum_{j=0}^{\infty} \frac{e^{-\theta/2}\,(\theta/2)^j}{j!}\, f_{r + 2j}(y)$

$E[Y] = r + \theta$

$\text{Var}(Y) = 2r + 4\theta$

$r$ is the degrees of freedom and $\theta \geq 0$ the noncentrality parameter. $f_v$ is the density of the central chi-squared distribution with $v$ degrees of freedom, so $f$ is a mixture of central chi-squared densities with Poisson $(\theta/2)$ weights. With $\theta = 0$ it is the central $\chi^2(r)$ .
Convention: $\theta$ here is the full sum $\sum \mu_i^2/\sigma^2$ , not half of it. Dobson & Barnett write $\chi^2(n, \lambda)$ with $\lambda = \sum \mu_i^2$ for unit-variance normals, mean $n + \lambda$ and variance $2n + 4\lambda$ . Some references define the parameter as half that sum, or its square root; under the halved parameter $\lambda'$ the mean is $r + 2\lambda'$ . Check which one a source uses before reading off its mean.
Additivity: independent $\chi^2(r_i, \theta_i)$ variables sum to $\chi^2\!\left(\sum r_i,\ \sum \theta_i\right)$ .
Quadratic forms: when $Q/\sigma^2$ is chi-squared for normal $X_i$ , its noncentrality parameter is found by replacing each $X_i$ in $Q/\sigma^2$ by its mean, $\theta = Q(\mu_1, \dots, \mu_n)/\sigma^2$ . For $\mathbf{y} \sim \text{MVN}(\boldsymbol\mu, \mathbf{V})$ , $\mathbf{y}^\top\mathbf{V}^{-1}\mathbf{y} \sim \chi^2(n,\ \boldsymbol\mu^\top\mathbf{V}^{-1}\boldsymbol\mu)$ .
Noncentral F: for independent $U \sim \chi^2(r_1, \theta)$ and $V \sim \chi^2(r_2)$ , $F = \dfrac{U/r_1}{V/r_2}$ has the noncentral F-distribution $F(r_1, r_2, \theta)$ , with mean $\dfrac{r_2(r_1 + \theta)}{r_1(r_2 - 2)}$ for $r_2 > 2$ , larger than the central F-Distribution's $r_2/(r_2 - 2)$ .
Power of the ANOVA F test: with $k$ groups of $n$ observations and group means $\mu_j$ , the between-groups sum of squares over $\sigma^2$ is $\chi^2(k - 1, \theta)$ with $\theta = n\sum_j(\mu_j - \bar\mu)^2/\sigma^2$ , so the ANOVA statistic is $F(k - 1,\ k(n - 1),\ \theta)$ . It is central only when $H_0$ (all $\mu_j$ equal) is true. The power is $P\big(F(k - 1, k(n - 1), \theta) > c\big)$ , where $c$ is the central critical value; $\theta$ grows with $n$ and the spread of the true means and falls as $\sigma^2$ grows.
The CAS MAS-I tables give only central chi-squared and F percentiles, so a noncentral probability needs software.

Mean and Power for a Loss-Ratio Monitoring Statistic Each of four lines' loss-ratio deviation from plan is independent and normal with standard deviation $\sigma = 0.05$ . The monitoring statistic is $Y = \sum_{i=1}^{4} X_i^2/\sigma^2$ , which is $\chi^2(4)$ when every true deviation is 0. This year the true mean deviations are $0.05$ , $-0.05$ , $0.10$ and $0$ . Find the distribution of $Y$ , its mean and variance, and the chance it exceeds the 5% critical value.
Answer
 $\begin{align*} \theta &= \frac{0.05^2 + 0.05^2 + 0.10^2 + 0^2}{0.05^2} \\ &= \frac{0.015}{0.0025} \\ &= 6 \end{align*}$ 
So $Y \sim \chi^2(4, 6)$ , with
 $\begin{align*} E[Y] &= 4 + 6 \\ &= 10 \\ \text{Var}(Y) &= 2(4) + 4(6) \\ &= 32 \end{align*}$ 
The test rejects when $Y > 9.49$ , the $0.950$ entry in the $df = 4$ row of the CAS chi-square table. Software gives $P\big(\chi^2(4, 6) > 9.49\big) = 0.470$ . Even with one line's true deviation at two standard deviations, the test flags the drift less than half the time.

Power of a Territory Severity ANOVA Mean log severity in three territories is truly $8.0$ , $8.3$ and $8.6$ , with $\sigma = 0.4$ . A one-way ANOVA at the 5% level uses $n = 6$ claims per territory. Find the noncentrality parameter, the expected F-statistic and the power.
Answer $\bar\mu = 8.3$ , so $\sum(\mu_j - \bar\mu)^2 = 0.09 + 0 + 0.09 = 0.18$ :
 $\begin{align*} \theta &= \frac{6(0.18)}{0.4^2} \\ &= 6.75 \end{align*}$ 
The statistic is $F(2, 15, 6.75)$ , with $k - 1 = 2$ and $k(n - 1) = 15$ :
 $\begin{align*} E[F] &= \frac{15(2 + 6.75)}{2(15 - 2)} \\ &= 5.05 \end{align*}$ 
against $15/13 = 1.15$ when $H_0$ is true. The CAS table's upper 5% point of $F(2, 15)$ is $3.682$ , and software gives $P\big(F(2, 15, 6.75) > 3.682\big) = 0.546$ . Raising $n$ to 10 claims per territory gives $\theta = 11.25$ on $(2, 27)$ df and a power of $0.817$ .