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The likelihood function for the Poisson Distribution $P(X=k| \lambda)$, where $\lambda$ is the mean, has the property that the integral with respect to the mean integrates to unity (from $0$ to $\infty$): $$\int_0^\infty P(X=k|\lambda)\,d\lambda = \int_0^\infty \frac{\lambda^kexp(-\lambda)}{k!}\,d\lambda = \frac{\Gamma(k+1)}{k!} = \frac{k!}{k!} =1.$$ This was shown and discussed in my other post.

Question: Does anyone know of any other probability density function (or probability mass function) which integrates or sums to unity in a similar way?

Thanks.

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This is just a matter of parameterization. For almost any parametric family you can modify the parameter so that the density integrates to 1 with respect to (wrt) the Lebesgue measure defined for that modified parameter... The true question is: why do you want to integrate wrt the Lebesgue measure and not wrt a measure corresponding to reasonable prior? In your example above why did you put $d\lambda$ and not $dG(\lambda)$? Is the possibility $\lambda = 1,000,000,000,000$ as reasonable as $\lambda = 1$? It would make more sense to consider a realistic prior, specific to a given modeling task.

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EDIT: response to "how would you use parametrization of say the Normal Distribution (or any normal type distribution) so that it satisfies the integration requirement above?"

The case of normal distribution is easiest. No modification of the mean $\mu$ is necessary:

$ \int_{-\infty}^{+\infty} \frac{1}{\sqrt{2\pi\sigma^2}} \exp\{-\frac12\frac{(x - \mu)^2}{\sigma^2}\}\ \bf{d\mu} = 1. $

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  • $\begingroup$ Okay but how would you use parametrization of say the Normal Distribution (or any normal type distribution) so that it satisfies the integration requirement above? $\endgroup$
    – John Doe
    Commented Jan 22, 2018 at 10:23
  • $\begingroup$ Thanks for your response. Yes I realized that after sending the comment, since the term $(x - \mu)^2$ makes it obvious that integrating with respect to either $x$ or $\mu$ gives unity. Could you provide a simple example of the parameterization you are mentioning which allows you to modify the parameter so that the density integrates to $1$? $\endgroup$
    – John Doe
    Commented Jan 23, 2018 at 8:44
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    $\begingroup$ Great... The desired parameter modification is specific to the parametric form of the distribution your are working with (if such modification exists). It's all algebra and calculus... Note that in the gamma-distribution, for example, the rate parameter Lambda and argument X enter the density formula almost symmetrically. So we should enjoy almost the same level of simplicity as that with normal distribution. $\endgroup$
    – stans
    Commented Jan 23, 2018 at 9:19
  • $\begingroup$ That's an excellent suggestion, I will look at the gamma-distribution as well. $\endgroup$
    – John Doe
    Commented Jan 23, 2018 at 9:28
  • $\begingroup$ Do you know if something equivalent holds for the binomial distribution? $\endgroup$
    – John Doe
    Commented Jan 24, 2018 at 11:43
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As indicated in the answer by stans, the question does not make sense in that the integral depends on the parameterisation of the likelihood, which is arbitrary. And the integral also depends on the choice of the dominating measure, which again can be arbitrary.

Furthermore, outside the Bayesian framework, integrating out the parameter is not of particular interest, since the classical approach seeks the "true" parameter that led to the generation of the data. Within the Bayesian framework, the constant prior has no specific appeal, except for some special cases. Like location parameters. But in general the constant prior should not be interpreted as a "uniform prior", again because this interpretation fails to account for the dependence on the parameterisation.

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