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Let $y_i$ have simplex distribution with density $$f(y;\mu,\sigma^2)=[2\pi\sigma^2\{y(1-y)\}^3]^{-1/2}\exp\{-\frac{1}{2\sigma}2d(y,\mu)\}$$ where $\mu\in(0,1)$ , $\sigma^2>0$ is a unit deviance function and $d(y,\mu)=\frac{(y-\mu)^2}{y(1-y)\mu^2(1-\mu)^2}$.

I need to find the likelihood of each observation, because I need to simulate data from this distribution (Bayesian simulation). Suppose I take the logarithm of the density function, as follows $$-\log(f(y,\mu,\sigma^2))\propto\frac{1}{2}\log[\sigma^2[\{y(1-y)\}^3]+\frac{1}{2\sigma^2}d(y,\mu)$$,

is this the likelihood of each observation, or is there some other method of finding the maximum likelihood?

Carl
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  • Your modified question is answered at https://stats.stackexchange.com/questions/2641/what-is-the-difference-between-likelihood-and-probability?s=1|0.0000. How you state your question, though, is puzzling, because since $d(y,\mu)$ depends on the observations $y$ it does not seem to be any kind of "parameter." You should watch out for the disappearing factor of $2$ in the exponential, as well. – whuber Apr 14 '17 at 20:34
  • @whuber the parameter in $d(y,\mu)$ is $\mu$. I can disappear with both $\frac{1}{2}$ in $-log(f(y,\mu,\sigma^2))$? –  Apr 14 '17 at 21:09
  • @whuber I made a mistake while typing, $\sigma^2$ is the dispersion parameter. The way I had written it seemed to be the $d(y,\mu)$. –  Apr 14 '17 at 21:11
  • @whuber, AFAICT the edits addressed your comment. Do you think it's clear enough now? Do you think it's a duplicate? – gung - Reinstate Monica Apr 16 '17 at 15:00
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    @gung I have not voted to reopen it because I cannot figure out what is being asked: (1) if it's asking for the likelihood, that's explicitly given in the question; (2) maybe it's about simulation, but that is unclear; (3) maybe it's about maximizing the likelihood, but again that's unclear. – whuber Apr 16 '17 at 19:44

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