it's concave in \(\pi\) as long as the q's are non-negative (which is true).
is it also concave in \(\mu_k\) and \(\Sigma_k\), or jointly convex in the three variable types?
for that you have to plug in the definition of the normal distribution, which will make the log kill the e^, and you will have a quadratic left there. you see that it will not necessarily concave/convex in \(\mu_k\), but that it will depend if the quadratic defined by \(\Sigma_k\) is curved up or down or a mix of that. in other words it depends on if \(\Sigma_k^{-1}\) is positive definite or not.
concave
Hey, is this function concave? if yes, can someone please explain why?
Thanks in advance,
1
it's concave in \(\pi\) as long as the q's are non-negative (which is true).
is it also concave in \(\mu_k\) and \(\Sigma_k\), or jointly convex in the three variable types?
for that you have to plug in the definition of the normal distribution, which will make the log kill the e^, and you will have a quadratic left there. you see that it will not necessarily concave/convex in \(\mu_k\), but that it will depend if the quadratic defined by \(\Sigma_k\) is curved up or down or a mix of that. in other words it depends on if \(\Sigma_k^{-1}\) is positive definite or not.
so in general the answer would be no
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