Question

The time between failures of a machine has an exponential distribution with parameter l. Suppose that the prior distribution for λ is exponential with mean 100 hrs. Two machines are observed, and the average time between failures is 1125 hrs.
a) Find the Bayes estimate for λ.
b) What proportion of the machines do you think will fail before 1000 hrs?

Answer 1

@ganeshie8
@surjithayer

Answer 2

@kirbykirby

Answer 3

I did a similar question on somebody else's question. I haven't had too much experience with Bayes estimation though, but I'll try it.

Also, I am not sure what parameterization of the exponential distribution you are using, but I'll assume it's of the form \(f(x;\lambda)=\lambda e^{-\lambda x}\).

Define \(L(\lambda)\) as the likelihood, \(\pi(\lambda)\) as the prior, and \(\pi(\lambda|x)\) as the posterior. Then we have:\[ \large L(\lambda)=\prod_{i=1}^{n}\lambda e^{-\lambda x_i}=\lambda^n e^{-\lambda\sum\limits_{i=1}^nx_i}\]For the prior, we have a mean of 100, thus \[\large \pi(\lambda)=\frac{1}{100}e^{-\frac{\lambda}{100}}\]Now:
\[\large \begin{align*}
\pi(\lambda|x) &\propto L(\lambda)\pi(\lambda)\\
&= \lambda^n e^{-\lambda\sum\limits_{i=1}^nx_i}\cdot \frac{1}{100}e^{-\frac{\lambda}{100}}\\
&=\frac{\lambda^n}{100}e^{-\lambda n\bar{x}-\frac{\lambda}{100}}\\
&\propto\lambda^ne^{-\left( n\bar{x}+\frac{1}{100}\right)\lambda}\\
&=\lambda^n e^{-\frac{\lambda}{1/(n\bar{x}+1/100)}}, \text{equation} (*)
\end{align*}\]
If we consider a gamma distribution, \(\text{GAM}(\alpha,\beta)\), with the following paramaterization: \[\large \frac{1}{\Gamma(\alpha)\beta^{\alpha}}x^{\alpha-1}e^{-x/\beta}\], then we observe that equation (*) is the kernel of a \(\large \text{GAM}\left( n,\frac{1}{n\bar{x}+1/100}\right)\) distribution.

Now, the Bayes estimate is the mean of the posterior, and if \(X\sim\text{GAM}(\alpha, \beta)\), then \(E(X)=\alpha\beta\).

So, Bayes estimate for \(\lambda\) is \(\large \frac{n}{n\bar{x}+1/100} \).

I hope this is correct.