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According to a book, a distribution belongs to the exponential family if it can be written in the form of

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I wrote the Bernoulli distribution as $\exp\Big(y \log\,[{\mu}/{(1-\mu)}] + \log\,(1-\mu)\Big)$. In this case $a(y)=y$, $b(\theta)= \log\,[{\mu}/{(1-\mu)}], c(\theta)=\log(1-\mu)$, but don't know what $d(y)$ is. Any idea what it is and why?

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  • $\begingroup$ If you have the entire pdf without using $d$, then won't it be the case that you just have $d(y)=0$? $\endgroup$
    – Glen_b
    Commented Nov 27, 2014 at 23:39

1 Answer 1

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$${}{}{}{}{}d(y)=0{}{}{}{}{}$$

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  • $\begingroup$ That is what I thought, but since this is the first case I have come across in which $d(y)=0$, I wasn't sure. $\endgroup$
    – Günal
    Commented Nov 28, 2014 at 12:55
  • $\begingroup$ Let me suggest to reread the definition in your book, this seems the simplest way to be sure. $\endgroup$
    – Did
    Commented Nov 28, 2014 at 14:15

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