Could anyone check that the alternative hypothesis is valid? I wanted to prove that the "Mahalanobis distance ($\mathbf{T_i} = \mathbf{(x_i - \bar{x})^T \Sigma^{-1}(x_i - \bar{x})}$)" is a Log Likelihood Ratio Test statistics for the following hypothesis.

For validating the following hypothesis, (all notations are vector notation)

\begin{cases} H_0 : \mathbf{x_i} \sim N(\mathbf{x_i} | \mathbf{\mu, \Sigma}) \\ H_1 : \mathbf{x_i} \sim N(\mathbf{x_i} | \mathbf{\mu + \delta_i, \Sigma}),\;\; \mathbf{\delta_i := (x_i - \bar{x})} \end{cases}

I used the Log Likelihood Ratio Test (LRT) as followed.

\begin{split} \lambda_i &= \log \left( \frac{ N(\mathbf{x_i} | \mathbf{\mu + \delta_i, \Sigma} ) }{N(\mathbf{x_i} | \mathbf{\mu, \Sigma} )} \right) \\ &= -\frac{1}{2} \left( (\mathbf{x_i - \mu - \delta_i})^T\mathbf{\Sigma}^{-1}(\mathbf{x_i - \mu - \delta_i}) - (\mathbf{x_i - \mu})^T \mathbf{\Sigma}^{-1}(\mathbf{x_i - \mu})\right) \\ &= -\frac{1}{2} \left( \mathbf{\delta_i}^T \mathbf{\Sigma}^{-1}\mathbf{\delta_i} - 2(\mathbf{x_i - \bar{x}})^T\mathbf{\Sigma}^{-1}\mathbf{\delta_i} \right)\\ & \approx \frac{1}{2} (\mathbf{x-\bar{x}})^T \mathbf{\Sigma}^{-1}(\mathbf{x-\bar{x}}) \end{split}

One thing that I cannot be certain is that the alternative hypothesis ($H_1$) can have such a form (dependent to $\mathbf{x_i}$). Is the above hypothesis testing formulation valid?

  • 5
    $\begingroup$ $H_1$ doesn't even make sense, since it is attempting to define the distribution of a variable in terms of itself! Could you perhaps state in words what you are hoping $H_1$ should be expressing? $\endgroup$ – whuber Sep 1 '15 at 15:03

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