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If $\det(\Lambda_0) \to 0$, what does $$ \exp\left(-\frac{1}{2}\text{trace}\left(\Lambda_0 \Sigma^{-1}\right)\right)\det\left(\Lambda_0\right)^{-1/2} $$ approach?

I was trying to answer the following question, and I couldn't quite finish it off. It asks for a derivation of why the Jeffreys prior for a multivariate normal distribution is proportional to $\det(\Sigma)^{-(d+1)/2}$, where $d$ is the size/length of an observation.

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  • $\begingroup$ Have you tried simple examples? For instance, work in two dimensions, set $\Sigma$ to the identity, and let $\Lambda_0$ be diagonal with entries $(a,b).$ Consider the cases (1) $(a,b)=(e,1/\sqrt{e})$ as $e\to 0$ and (2) $(a,b)=(e,e)$ as $e\to 0.$ (If you like, make both limits occur through positive values of $e$ only). Clearly $\det\Lambda_0\to 0$ in each case, but what are the two limits? $\endgroup$ – whuber Aug 22 '18 at 11:34
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    $\begingroup$ @whuber from the right, $0$ and $\infty$. well that's weird because I remember this happens in the univariate case $\endgroup$ – Taylor Aug 22 '18 at 18:50
  • $\begingroup$ I'm not conversant with the original question, but my initial impression was that you might want to focus on exactly how the determinant of $\Lambda_0$ grows small. (That was the intuition that led to the two examples I provided.) $\endgroup$ – whuber Aug 22 '18 at 20:50

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