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We know that there are two versions of exponential

$\big(1\big)$ $exp(\lambda)$ with pdf $$f(x;\lambda)=\lambda e^{-\lambda x}$$ ,$\lambda>0,x>0$ and $\mathbb{E}[x]=\frac{1}{\lambda}$ for this version we know that a conjugate prior is $gamma(a,b)$.

$\big(2\big)$ $exp(b)$ with pdf

$$f(x;b)=\frac{1}{b}e^{\frac{-x}{b}},$$ $, \mathbb{E}[x]=b$.For version $\big(2\big)$ which is the conjugate prior ?

In the following link they mention only the conjugate prior for $\lambda$ version

https://en.wikipedia.org/wiki/Conjugate_prior

I started wondering because I was asked to simulate exponential data with unknown mean parameter let's say $\theta$ and then fit a non informative prior distribution for $\theta$.

At first I was confused and I did bayesian inference for $\theta$ following the $\big(1\big)$ exponential distribution $exp(\theta)$, but then I clarified what the question was asking and now I'm stuck.

Any help or idead would be great.

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  • $\begingroup$ You can switch from one parameterisation to the other by a change of variable $b=1/\lambda$, hence apply the change of variable to the Gamma conjugate prior as well. $\endgroup$ – Xi'an Jun 4 '18 at 6:47
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The conjugate prior for the second case is the inverse-Gamma distribution:

$$p(b;\alpha,\theta) = {\theta^{\alpha} \over \Gamma(\alpha)}b^{-\alpha-1}e^{-\theta/b}$$

where $b$ is the random variable as per your notation. It is related to the regular Gamma distribution in the obvious way (given the motivating problem) - if $\lambda$ is distributed according to a Gamma distribution, $1/\lambda$ is distributed according to an inverse-Gamma distribution with the same parameters.

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