Assume that $X_1, ..., X_n, X_{n+1}| \theta \sim \text{iid } Exp(\theta) $ and the posterior is $\theta \sim Gamma(\alpha, \lambda)$. Task is to compute the posterior $\theta|x_{1:n}$.

$\pi(\theta|x_{1:n}) \propto \pi(x_{1:n}|\theta) \pi(\theta)$

First I look for $\pi(x_{1:n}|\theta)$

$P_{\theta}(\min(X_1, X_2, ..., X_n)<t) = 1 - P(\min(X_1, X_2, ..., X_n)>t) = $ $1 - (P(X_1>t))^n = 1 - (e^{-\theta t} -1)^n$

So $\pi(x_{1:n}|\theta) = \frac{d}{dt} (1 - (e^{-\theta t} -1)^n)= n t (e^{-\theta t} -1 )^{n-1} e^{-\theta t}$

$\pi(\theta|x_{1:n}) \propto (e^{-\theta t} -1 )^{n-1} e^{-\theta t}\theta^{\alpha-1} e^{-\lambda\theta}$.

I have no idea what kind of distribution is it. Am I doing sth wrong?

  • $\begingroup$ Does $x_{1:n}$ refer to the first $n$ data points of the sample or to the smallest observation? Typically I would interpret it as referring to the sample, but you seem to be interpreting it as referring to the smallest observation. $\endgroup$
    – jbowman
    Jan 23 '20 at 17:15
  • $\begingroup$ I see. I though it refers to the smallest observation. If it refers to the sample, the task is much easier :) $\endgroup$ Jan 23 '20 at 17:20
  • $\begingroup$ Better check! But this problem would typically be structured to use the sample rather than the smallest datapoint. If it is the smallest data point, I'll observe that your derivation of the dist'n of the smallest data point isn't correct - you should get an Exponential distribution with parameter $n\theta$ (or something similar, depending on how you parameterize the exponential.) $\endgroup$
    – jbowman
    Jan 23 '20 at 17:28
  • $\begingroup$ Ok, I will. Thanks. Just looks like the posterior makes no sens if we assume the minimal value. $\endgroup$ Jan 23 '20 at 17:30

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