t-Test for Mixture Normal Data
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Let $X$ be a random variable having a compound normal distribution with pdf given by
\[f(x) = \pi \phi(x; \mu_1, \sigma_1^2) + (1-\pi) \phi(x; \mu_2, \sigma_2^2)\]where $0 < \pi < 1$.
Suppose $X_1,\ldots, X_n$ is a random sample from the above distribution, and let $\bar X$ and $S^2$ be the mean and variance of the sample. The cdf of the statistic $t^2 = n\bar X^2/S^2$ can be written as
\[G(x) = \sum_{m=0}^n H(x\mid m) \binom{n}{m} \pi^m (1-\pi)^{n-m}, x > 0\]where $H(x\mid m), m=0,1,\ldots, n$ is the conditional cdf of $t^2$, given that $m$ observations are drawn from $N(\mu_1,\sigma_1^2)$.
The conditional cdf $H(x\mid m)$ of $t^2$ can be expressed as
\[H(x\mid m) = P[d_1\chi_1^2(\delta_1^2) / (d_2\chi_1^2(\delta_2^2) + d_3\chi_{m-1}^2 + d_4\chi_{n-m-1}^2) < 1], m=1,2,\ldots, n-1;\] \[H(x\mid 0) = P[\chi_1^2(\delta^2)/\chi_{n-1}^2 < x/(n-1)]\quad \text{with } ]\delta^2 = n\mu_2^2/\sigma_2^2;\]and
\[H(x\mid n) = P[\chi_1^2(\delta^2)/\chi_{n-1}^2 < x/(n-1)]\quad \text{with }] \delta^2 = n\mu_1^2/\sigma_1^2\]where $\chi_\nu^2(\delta^2)$ denotes a chi-square random variable with degrees of freedom $\nu$ and noncentrality parameter $\delta^2$
Let $r = \sigma_2/\sigma_1$ and
\[B = \begin{bmatrix} (n-m)x/(n-1) - m & -r(m(n-m))^{1/2}(1+x/(n-1))\\ -r(m(n-m))^{1/2}(1+x/(n-1)) & r^2(mx/(n-1) - (n-m)) \end{bmatrix}\]Note $\vert B\vert < 0$, the two characteristic roots of $B$ are opposite in sign. Then $-d_1, d_2$ are these roots with $d_1 > 0, d_2 > 0; d_3 = nx/(n-1); d_4 = r^2nx/(n-1)$. Furthermore,
\[\delta_i = (u_{i1}m^{1/2}\mu_1 + u_{i2}(n-m)^{1/2}\mu_2/r)/\sigma_1\,, i=1,2\]where $(u_{11}, u_{12})$ and $(u_{21}, u_{22})$ are the normalized characteristic vectors of $B$ corresponding to $-d_1$ and $d_2$, respectively.
For the particular case $r=1$, it becomes much easier
\[H(x\mid m) = P[\chi_1^2(\delta_1^2)/\chi^2_{n-1}(\delta^2_2) < x / (n-1)], m=0,1,\ldots, n\]where
\[\delta_1^2 = (m\mu_1 + (n-m)\mu_2)^2 / (n\sigma_1^2)\]and
\[\delta_2^2 = m(n-m)(\mu_2 - \mu_1)^2 / (n\sigma_1^2)\]Let $X$ be a normal variate with mean $\delta$ and unit variance, and $Y$ be independently distributed as a noncentral chi-square with $\nu$ degrees of freedom and noncentrality parameter $\lambda$. Then the ratio
\[t'' = X \sqrt{\nu/Y}\]follows a doubly noncentral $t$-distributioon with $\nu$ degrees of freedom and noncentrality parameters $\delta$ and $\lambda, (\lambda \ge 0)$