The continuous Bernoulli is a distribution I recently discovered. What the maximum likelihood estimate of the distribution's parameter? I'm struggling with the normalizing constant.
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4$\begingroup$ Funny; that distribution has been known since at least the 80s. But they have an ICML 2020 paper calling it "novel" in the title. $\endgroup$– Arya McCarthyCommented Apr 9, 2021 at 0:03
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$\begingroup$ Do you have a link to the original work? $\endgroup$– Rylan SchaefferCommented Apr 9, 2021 at 15:46
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$\begingroup$ @AryaMcCarthy I came back to your comment to find the proper paper(s) to cite. Can you help me? $\endgroup$– Rylan SchaefferCommented Aug 22, 2021 at 15:47
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1 Answer
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While the continuous Bernoulli is defined as$$p(x)=C(\lambda)\lambda^x(1-\lambda)^{1-x}\qquad x\in(0,1)\tag{1}$$on its Wikipedia page, rewriting it as $$p()\propto \lambda^x(1-\lambda)^{-x} = \{\lambda/(1-\lambda)\}^x=\exp\{\eta x\}$$ shows that it can be reparameterised as an exponential family with natural parameter$$\eta=\log\{\lambda/(1-\lambda)\}.$$The normalising constant is then $$\int_0^1 \exp\{\eta x\}\,\text dx=\eta^{-1}[e^\eta-1]$$ Given a sample $x_1,\ldots,x_n$ from (1), the log-likelihood is $$\sum_{i=1}^n \eta x_i - n \log\{e^\eta-1\}+n\log\eta$$ whose maximum in $\eta$ satisfies $$\frac{-1}{\hat\eta}+\frac{e^{\hat \eta}}{e^{\hat \eta}-1} = \bar x_n$$ which does not afford a closed-form solution.
