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As we know, when the Poisson rate $\lambda$ is large, we can approximate the Poisson distribution as a normal distribution with equal mean and variance.

Is there a conjugate family for this normal distribution with equal mean and variance? Neither gamma nor inverse-gamma work for that.

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    $\begingroup$ Why not use the conjugate prior for the Poisson distribution? $\endgroup$ – Christoph Hanck Jul 24 '15 at 19:02
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Let $X_1, X_2, \dotsc, X_n$ be iid from the distribution $\text{N}(\theta,\sigma^2=\theta)$, that is, a normal distribution with equal expectation and variance ($\theta>0$). There is a conjugate distribution for this model, which is known as generalized inverse gaussian distribution, see https://en.wikipedia.org/wiki/Generalized_inverse_Gaussian_distribution#Conjugate_prior_for_Gaussian

The likelihood from the normal model above can be written (after dropping constants) as ( * ) $$ L(\theta) \propto \theta^{-n/2} e^{-(n/2)\theta} e^{-\frac{\sum_i x_i^2/2}{\theta}}, \qquad \theta>0 $$ so we need a prior distribution with three factors that can absorb the three factors from the likelihood. We propose then $$ \pi(\theta) =K^{-1} \theta^{-a-1} e^{-b/\theta} e^{-c\theta} $$ with $a\in\mathbb{R}, b>0, c>0$ and $$K={\frac {{c}^{a/2-1}{b}^{a/2}{c}^{-a/2} \left( -{b}^{-a/2}a{{\it K}_{a} \left(2\,\sqrt {cb}\right)}+{b}^{1/2-a/2}\sqrt {c}{{\it K}_{a+1}\left( 2\,\sqrt {cb}\right)} \right) }{{{\it K}_{a}\left(2\,\sqrt {cb}\right) }}} $$, $K$ is the modified Bessel function of the second kind.

Then we can find the posterior as $$ \pi(\theta | x) \propto \theta^{-(n/2+a)-1} e^{-(n/2+c)\theta} e^{-(b+\sum_i x_i^2/2)/\theta} $$

( * ) I think this model is an interesting example. Note from the likelihood factorization above that $T=\sum_i X_i^2$ is a sufficient statistic. In the question the model was developed as an approximation for a poisson model, but in the poisson model the sufficient statistic is $T^* = \sum_i X_i$. So, however good the approximation is, the two models lead to different sufficient statistics!

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    $\begingroup$ The question asks for equal mean and variance while this answer seems to be about equal mean and standard deviation $\endgroup$ – Juho Kokkala May 12 '18 at 11:16
  • $\begingroup$ @Juho Kokkala: I added a correct answer to the correct question ... $\endgroup$ – kjetil b halvorsen May 12 '18 at 13:37
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    $\begingroup$ The conjugate prior of this question (as derived in the new question) seems to be known as the generalized inverse Gaussian en.wikipedia.org/wiki/… (The "old answer" part feels irrelevant, I'd delete it or move it to another question) $\endgroup$ – Juho Kokkala May 12 '18 at 14:01
  • $\begingroup$ @Juho Kokkala: Thanks for helping! I will move the old answer to a new question. $\endgroup$ – kjetil b halvorsen May 12 '18 at 16:03
  • $\begingroup$ I was going to mention that this distribution looked like a generalisation of the inverse Gaussian distribution. The generalisation being in the power $a$. $\endgroup$ – Xi'an May 13 '18 at 10:19

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