Say I have this situation with an exponential distribution and it's gamma conjugates:

$y\mid\lambda \sim exp(\lambda)$

$\lambda \sim gamma(\theta,\beta)$

$\lambda \mid y,\theta,\beta \sim gamma(\theta + 1, \beta + y)$

A trial shows that $y>x$, (where $x$ is just a constant) and we'd like to update $\lambda$. Am I correct to think that the posterior density would be given by the following equation?

$p(\lambda\mid \theta,\beta,y>x)= \int_{y=x}^\infty 1- gamcdf(x|\theta+1,\beta+y)dy$

Is there are better way to do this?

  • 1
    $\begingroup$ That does not look correct to me. Why don't you start by writing the likelihood and substitute into Bayes' theorem. $\endgroup$
    – Ben
    Jul 28, 2019 at 10:34
  • $\begingroup$ Hey @Xi'an would really appreciate if you could explain what you mean. $\endgroup$
    – emir
    Jul 29, 2019 at 12:21
  • $\begingroup$ Hey @Xi'an also, x in the above example is just a constant. It doesn't have a prior. Would really appreciate if you could share how you'd tackle this. $\endgroup$
    – emir
    Jul 29, 2019 at 12:52
  • $\begingroup$ Oh really ? @Xi'an so in my case it would just be : $\propto gamma(\lambda \mid \theta,\beta)(1-expcdf(x \mid \lambda)$. Is that correct? $\endgroup$
    – emir
    Jul 29, 2019 at 12:55
  • $\begingroup$ @Xi'an it's quite an interesting answer because it basically means when there is an inequality a gamma conjugate prior does not imply a gamma posterior ... that's basically what you're saying. $\endgroup$
    – emir
    Jul 29, 2019 at 12:58

1 Answer 1


When $\lambda\sim\mathcal G(\theta,\beta)$ and it is known that $Y>c$ for $Y\sim\mathcal E(\lambda)$, this amouns to observing a Bernoulli random variable$$Z\sim\mathcal{Be}(\mathbb P_\lambda(Y>c))$$to be equal to one. The posterior is therefore $$\pi(\lambda) \propto \lambda^{\theta-1}e^{-\beta\lambda}\mathbb P_\lambda(Y>c)=\lambda^{\theta-1}e^{-\beta\lambda}e^{-\lambda c}=\lambda^{\theta-1}e^{-(\beta+c)\lambda}$$


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