QUESTION: (I translated it into English so you might find somewhat awkward..)
A company wants to get information about the lifespan of certain machines. Let's say that the lifespan of the machines follows an exponential distribution which is $p(y|\theta)=\theta e^{-\theta y}(\,y>0)$
After examining 50 machines, the average lifespan of the machines was 756 hours. Find a 95% HPD credible interval for $\theta$.(Use a non informative prior $p(\theta)=\theta^{-1}$
Hello. I'm studying Bayesian statistics with some textbook and I came across this question. I guess I'm having trouble understanding this part => 'After examining 50 machines, the average lifespan of the machines was 756 hours'. Does this mean that there are 50 samples every which follows an exponential distribution..?
If I think of it like that, the likelihood would be $p(y_{1},..,y_{50}|\theta) = \theta^{50}e^{-\theta\sum_{i=1}^{50}\large y_{i}}$ and by using a given prior we can get the posterior distribution $p(\theta|y_{1},..,y_{50}) \propto \theta^{49}e^{-\theta\sum_{i=1}^{50}\large y_{i}}$ which is Gamma(50, $\sum_{i=1}^{50}\large y_{i}$)
Now I have everything except $\sum_{i=1}^{50}\large y_{i}$. I guess I can get something from 'average lifespan was 756 hours' but isn't this average lifespan is about parameter $\theta$? How can I derive something that has to do with $\sum_{i=1}^{50}\large y_{i}$
Any help is appreciated.
Yes. All samples taken are assumed to come from an exponential distribution.
You have the sample mean of the $y_i$: $1/50 \sum_i^{50}y_i = 756$ so now you can find the summation you are looking for to compute your posterior.