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Consider $\log f(x) = -0.5\log (2 \pi \theta) - 0.5 \frac{(x - \theta)^2}{\theta}$ and $$ \frac{\partial}{\partial\theta} \log f(x) \propto -\frac{1}{\theta}+\frac{x^2}{\theta^2} -1 $$ Thus, $$ \frac{ …

$L(\phi) = \prod p(x_j, y_j; \phi)$ So $\log L = \ell(\phi) = \sum \log p(x_j, y_j; \phi)$ Since $p(x\mid y)$ does not depend on $\phi$, $p(x, y) = p(x \mid y)p(y) \propto p(y)$ so $\frac{d}{d\phi} \l …

Perhaps not relevant in this case but note that if the E-step is estimated, with Monte Carlo methods or another approximation, it is possible for the likelihood to decrease. A thought that might be r …

In the plots below the result from the code is plotted. The first plot show loglikelihood, the second shows the x-estimate and the third shows the error compared to x. I used logmvnpdf found here: htt …

First point: Since $d(\theta, \theta_0)<\delta$ it should be possible to select $\theta = \theta_0$ which gives $\sup=0>-C\cdot0^\alpha$. Without being specific for this problem, if $x^*=\arg\max a(x) …

$ L(\beta) = P(\mathrm{Data} \mid \beta) = \prod_i f(x_i) $ and $\lambda = \frac{1}{\mu} = \frac{s}{\beta}$ Log is increasing so maximizing log is same as maximizing likelihood: $\ell(\beta) = \sum \l …