The usual exception I have come across regarding non-existence of conjugate prior outside the exponential family is the uniform distribution on $(0,\theta)$ (i.e. $U(0,\theta)$) where $\theta$ has a Pareto prior. Pareto distribution also acts as a conjugate prior in the $U(-\theta,\theta)$ family but this is basically the same example. Other common examples outside exponential family where the support depends on the unknown parameter are the shifted exponential distribution with shift $\theta$ and the Pareto distribution with scale $\theta$. Turns out they also allow conjugate priors with a sufficiently 'nice' distribution as I found out browsing some textbooks, but there was no motivation for how they came up with the priors.

For real $\theta$, suppose $\text{Exp}(\theta,1)$ denotes the shifted exponential density $$f(x)=e^{-(x-\theta)}\mathbf1_{[\theta,\infty)}(x)$$

And for positive $\alpha,\theta$, let $\text{Pareto}(\alpha,\theta)$ be the density $$f(x)=\frac{\alpha \theta^{\alpha}}{x^{\alpha+1}}\mathbf1_{[\theta,\infty)}(x)$$

These are related to the uniform distribution as follows:

$$X \sim \text{Pareto}(1,\theta)\implies \frac1X \sim U\left(0,\frac1{\theta}\right)$$

$$X \sim \text{Exp}(\theta,1) \implies e^{-X} \sim U\left(0,e^{-\theta}\right)$$

Using the Pareto prior for the uniform distribution, I considered $\frac1{\theta}\sim \text{Pareto}(\alpha,a)$ for the Pareto data and $e^{-\theta}\sim \text{Pareto}(\alpha,a)$ for the exponential data.

Now one can easily show that the prior for $\theta$ in the Pareto data has pdf (taking $\beta=\frac1a$) $$\pi(\theta)=\frac{\alpha}{\beta^\alpha}\theta^{\alpha-1}\mathbf1_{[0,\beta]}(\theta) \tag{1}$$

And for the exponential data, the prior has pdf (taking $\beta=-\ln a$)

$$\pi(\theta)=\alpha e^{\alpha(\theta-\beta)}\mathbf1_{(-\infty,\beta]}(\theta) \tag{2}$$

I verified that the distributions in $(1)$ and $(2)$ are indeed conjugate priors for $\theta$ in the $\text{Pareto}(1,\theta)$ and $\text{Exp}(\theta,1)$ distributions respectively.

Is this how the derivation of a conjugate prior works out given that I already have one for a related distribution? Is it always the case that if $g(\theta)$ has a conjugate prior in a given data $X\sim F_{g(\theta)}$, then $\theta$ also has a conjugate prior in the same data $X\sim F_{\theta}$? I guess this does not really make the priors in $(1)$ and $(2)$ distinct from the Pareto prior in $U(0,\theta)$.

The fact that conjugate priors can exist outside exponential family is apparently not surprising since one can construct a conjugate prior whenever a sufficient statistic of fixed dimension exists for the parametric family in question. Indeed the examples above show that not being a member of exponential family does not in itself make the distributions ineligible for a conjugate prior.

But I am not sure what exactly 'fixed dimension' means here. Is a sufficient statistic of fixed dimension essentially referring to a non-trivial sufficient statistic? Consider other distributions outside exponential family like $\text{Laplace}(\theta,1)$ or $\text{Cauchy}(\theta,1)$ with unknown location $\theta$. Suppose a sample of size $n$ is drawn from them. Am I correct in saying that because they do not allow non-trivial sufficient statistics, $\theta$ is guaranteed to not have any conjugate prior? Does this make sense when $n=1$?

  • $\begingroup$ I'm fairly sure "fixed dimension" means you can make $n$ observations and summarise them in a statistic that's a vector of a fixed size, and it remains a sufficient statistic regardless of $n$. For an example of a sufficient statistic that isn't of constant dimension, just take the sample values themselves. They form a sufficient statistic (though not a useful one) but its dimension is proportional to $n$. $\endgroup$
    – N. Virgo
    Commented Jul 29, 2020 at 15:06
  • $\begingroup$ The conjugate prior does not depend on the parameterisation of the data, ie, it is the same when you consider $x$, $1/x$, $\exp(x)$, and so on. $\endgroup$
    – Xi'an
    Commented Jul 29, 2020 at 19:56
  • $\begingroup$ "Does this make sense when $n=1$?" - I think not, as the fixed dimension and conjugate families of distributions relates to all values of $n$, not just a single one. The case of $n=1$ is just moving from the prior to the posterior distribution: you would need to show that any prior from a particular family led to a posterior from the same family with the same dimension of parameters to be able to extend this to all $n$ and so have a conjugate family. $\endgroup$
    – Henry
    Commented Nov 17, 2021 at 14:01

1 Answer 1


The non-existence of conjugate priors outside exponential families is related to the Fisher-Darmois-Piman-Koopman lemma. Which states that, for parameterised families with fixed support (hence excluding the Uniform counterexamples), there cannot exist a sufficient statistic $S_n$ of fixed dimension whatever the sample size $n$ is. Here is a version of the Lemma due to H. Jeffreys (1939) [and reproduced from Oban (2009)):

Fisher-Darmois-Pitman-Koopman Lemma

Let the random quantities $X_1,X_2,...$ be conditionally i.i.d. given the value of some random quantity $\theta$, and assume that the conditional distribution $P_X(X_i|\theta)$ is dominated by a measure ν. Let $p(\cdot|θ)$ be the corresponding conditional density.

Assume further that the support of $f_{X|θ}$ is independent of the value of θ:$$∀θ_1,θ_2∈Ω_θ:\ \text{supp} p(.|θ_1) = \text{supp} p(.|θ_2)\quad ν-a.e.$$ Then if there is a sufficient statistic $S_n: Ω^n_x\mapsto Ω_s$ for each sample size $n\ge n_0$, and if $Ω_s$ has finite dimension,$P_X(\cdot|Θ)$ is an exponential family model.

Indeed, if there exists a conjugate family with a fixed and finite number $p$ of hyper-parameters, the posterior update of these hyper-parameters is sufficient (since Bayesian and classical sufficiencies are equivalent for dominated models).

  • $\begingroup$ Sorry but I don't understand how this answers my questions. $\endgroup$ Commented Jul 29, 2020 at 18:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.