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For a random vector $\xi_n \in \mathbb{R}^p$, if $\xi_n \rightarrow_d N(\mu, \Sigma)$, we can construct \begin{equation*} \Psi := \xi_n^{\top} \widehat{\Sigma}^{-1} \xi_n \end{equation*} for testing whether $\mu = 0$, where $\widehat{\Sigma}$ is an reasonable estimate of $\Sigma$.

Now assume $\xi_n$ converge to a mixed normal distribution, i.e. \begin{equation*} \xi_n \rightarrow_d \sum_{k = 1}^K w_k Z_k \end{equation*} where $w_1 +\cdots + w_K = 1$ and $Z_1, \cdots, Z_K$ are independently distributed as $N(0, \Sigma_k)$. Since the elements in one $\Sigma_k$ are quite different from those in the other, we cannot write this mixed normal distribution as a single normal distribution. In this case, I try to construct $$\tilde{\Psi} = \sum_{k = 1}^K w_k \xi_n^{\top} \widehat{\Sigma}_k^{-1} \xi_n$$ as the statistic. However, it seems to be difficult to analyze the properties of this statistic since the components on the right side are NOT independent from each other.

Do anyone know how to construct a proper statistic in this case. Thanks so much!

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    $\begingroup$ I don't understand. $\sum_{k=1}^K w_kZ_k$ is again $N(0,\Sigma)$-distributed with some appropriate $\Sigma$, if $w_i$ are constant. So what is your assumption about $w_i$? $\endgroup$
    – Dieter Kadelka
    Sep 10, 2020 at 12:42
  • $\begingroup$ @DieterKadelka Thank you! $w_k$ indeed are constant. Do you mean the weighted sum is again normal distribution, even the elements in each $\Sigma_k$ are quite different from those in another one? If so, is there a possible way to calculate the elements in $\Sigma$ exactly? $\endgroup$
    – 香结丁
    Sep 10, 2020 at 13:16
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    $\begingroup$ The distribution is $N(0,\Sigma)$ with $\Sigma = \sum_{k=1}^K w_k^2 \Sigma_k$. This holds if the $Z_k$ are independent. $\endgroup$
    – Dieter Kadelka
    Sep 10, 2020 at 13:31
  • $\begingroup$ @DieterKadelka I see, I deemed only one-dimensional normal distributions have this property before. Thanks so much! $\endgroup$
    – 香结丁
    Sep 11, 2020 at 1:45

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