# Bayesian posterior of censored exponential distribution

Suppose $y|\theta \sim$ Exponential$(\theta)$, and the prior distribution of $\theta$ is Gamma$(α,β)$.

• (a) Suppose we observe that $y \geq y'$, but do not observe the exact value of $y$. What is the posterior distribution, $p(\theta|y≥y')$, as a function of $\alpha$ and $\beta$? What are the posterior mean and variance of $\theta$?

• (b) Suppose that we are now told that y is exactly 100. Now what are the posterior mean and variance of θ?

It seems clear that this distribution will exploit the memoryless property of the prior exponential distribution. However, I am not exactly clear on how to prove this; in particular I am not sure how to properly incorporate the information that $y \geq y'$ into the likelihood function in order to derive the posterior distribution for the censored data. (Note: assume $y'$ is a known constant.)

This question is a generalized version of a question from Gelman, Bayesian Data Analysis, 3rd. ed., ch. 2 exercise 20.

• Do I understand correctly that $y'$ is a (known) constant, so that the aswer to (a) would depend on $\alpha,\beta, y'$? Have you tried a "brute force" approach of writing the posterior as proportional to prior * likelihood? – Juho Kokkala Sep 25 '17 at 15:38
• Yes, you are correct; question updated to reflect this. I have indeed tried the "brute force" approach but the solution is not obvious from this; in particular, as I ask, the question that remains to me is how the truncation affects the posterior (otherwise, the distribution seems to be the same; a truncated exponential distribution is just another exponential). – jpgard Sep 25 '17 at 20:19
• (NeilG seems to have given a full answer, but in case you did not look at that yet and want a hint) By "brute force" I meant: don't try to reason about "truncation", just plug in $p(\theta \mid \textrm{Data}) \propto p(\textrm{Data} \mid \theta)\,p(\theta)$ – Juho Kokkala Sep 26 '17 at 6:04
• Just a note to future readers, I found the example (without truncation) useful here, in the slide titled "the exponential-gamma system" halweb.uc3m.es/esp/Personal/personas/mwiper/docencia/English/… – jpgard Sep 26 '17 at 15:49

## 1 Answer

1. Calculate the likelihood, which is the probability than an exponential distribution with rate $\theta$ has a sample greater than $y'$: $L(\theta \mid y>y') = e^{-y'{\theta}}$. This is an exponential distribution with rate $y'$, i.e., a gamma with shape 1 and rate $y'$. The posterior is therefore $\alpha$ and $\beta + y'$.

2. This is just the Bayesian update of $\alpha$ and $\beta$, which gives a gamma distribution with parameters $\alpha+1$ and $\beta + 100$.

• Thanks for your answer. Can you clarify how to actually show what you state in (1) -- i.e., how the likelihood and prior generate that posterior? – jpgard Sep 23 '17 at 13:37
• @jpgard I don't see how; it seems immediate to me. – Neil G Sep 23 '17 at 13:38
• The question asks "how to prove this"; your answer just gives the result. – jpgard Sep 23 '17 at 15:10
• @jpgard write out the density $f(x) \propto e^{-\frac{x}{\theta}}$, after truncation, it is the same density over a different support. – Neil G Sep 23 '17 at 15:34
• Are you referring to the likelihood, or the posterior in the above comment? It is definitely clear from the problem that the likelihood is truncated at y', but the question is how that affects the posterior distribution and its mean/variance in both cases, which is still not clear to me from this answer. – jpgard Sep 25 '17 at 0:44