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Under standard assumptions, Hotelling's $t$-squared statistic, \begin{align*} t^2=(\overline{x}-\mu)'\hat{\Sigma}_\overline{x}^{-1} (\overline{x}-\mu), \end{align*} is distributed as \begin{align*} t^2 \sim T^2_{p,n-1}=\frac{p(n-1)}{n-p} F_{p,n-p}. \end{align*}

All references I can find assume a point null hypothesis, however I am almost certain this is not necessary. For a point null hypothesis, $p=\dim\left(\Theta_1\right)$, where $\Theta_1$ is the alternative parameter space. However, if the null hypothesis is not a point but composite where some parameters must be estimated then I am pretty sure $p$ can simply be replaced with $p=\dim\left(\Theta_1\right)-\dim\left(\Theta_0\right)$.

This would be the same as the degrees of freedom in a $\chi^2$ test when $\Sigma_\overline{x}$ is known.

I have done simulations that strongly support this.

Edit: A proof of Hotelling's $t$-squared statistic being distributed as described for point null hypotheses would be helpful too.

Edit: I think I worked it out, but haven't proved it. If $\left(\overline{x}-\mu\right)\sim{}N\left(0,\Sigma\right)$ then the Mahalanobis distance is distributed as $\left(\overline{x}-\mu\right)'\Sigma\left(\overline{x}-\mu\right)\sim\chi_p^2$, where $p=\dim\left(\Theta_1\right)-\dim\left(\Theta_0\right)$ as described in the question. This is true even for examples such as the one I provided. There is an affine transformation of the parameter space based on the Fisher information matrix. This transformation puts the null parameter space in a lower dimensional affine space with dimension equal to the number of parameters required to be estimated. (I think) the distribution of $t^2$ would then fall out as usual, but with $p$ being the number of parameters to estimate under $H_1$ minus the number of parameters to estimate under $H_0$.

Edit: Suppose $\Theta$ is an open subset of $\mathbb{R}^k$ such that $H_0:\mu\in\Theta_0$ and $H_1:\mu\in\Theta_1$. Then it follows that \begin{align*} \Sigma_\overline{x}^{-\frac{1}{2}}\left(\overline{x}-\mu\right)\sim{}N\left(0,I_p\right), \end{align*} and \begin{align*} \left\Vert\Sigma_\overline{x}^{-\frac{1}{2}}\left(\overline{x}-\mu\right)\right\Vert^2\sim\chi_p^2, \end{align*} after removing dimensions where $\overline{x}_i-\widehat{\mu}_i=0$ under $H_0$ (and $H_1$) after the linear transformation of $\Theta$ by $\Sigma_{\overline{x}}^{-\frac{1}{2}}$, and where $p=k-\dim\left(\Theta_0\right)$. Now, if $\Sigma_\overline{x}^{-\frac{1}{2}}\left(\overline{x}-\mu\right)\sim{}N\left(0,I_p\right)$, then \begin{align*} \widehat{\Sigma}_\overline{x}\sim\frac{1}{n-1}W_p\left(n,\Sigma_\overline{x}\right) \end{align*} and \begin{align*} \left\Vert\widehat{\Sigma}_\overline{x}^{-\frac{1}{2}}\left(\overline{x}-\mu\right)\right\Vert^2\sim\frac{p\left(n-1\right)}{n-p}F_{p,n-p} \end{align*} follows.

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    $\begingroup$ How do you estimate the parameters for a null hypothesis? If it's part of the mean vector, wouldn't you just use $\bar{x}_i$ for the estimate of the $i^{th}$ component of the mean vector $\mu_i$, effectively removing it from the test altogether? $\endgroup$ – jbowman Jul 15 '18 at 23:46
  • $\begingroup$ Thanks for the reply. I see what you are saying. Yes, in a simple example you would just use the MLE. Unfortunately my example is more complicated than that. Suppose $\mu=\left(\mu_1,\mu_2,\mu_3\right)$, where $\mu_3=\mu_1\mu_2$. Then $\mu$ cannot be estimated simply by the MLEs due to the extra equality constraint and you would need to used a constrained optimization technique. Nonetheless, I am pretty sure in this situation you would still have $p=\dim\left(\Theta_1\right)-\dim\left(\Theta_0\right)=3-2=1$. $\endgroup$ – Jonathan Mitchell Jul 16 '18 at 0:05
  • $\begingroup$ You're comment was helpful. I think I've worked it out now, but I'm not 100% sure. Thanks. $\endgroup$ – Jonathan Mitchell Jul 16 '18 at 4:51

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