# Log-likelihood of a exponential distribution

I have an exercise that I don't quite understand: The life of 100 lamps has been measured. Each lamp has been used with a intensity between 0 and 1, where 0 is off and 1 is the maximum intensity. It is known from experience that lamps of this type have an exponentially distributed for who long they last where the expected value can be written:

$$\mu(s) = \frac{\beta}{s},\quad s>0,$$

where β> 0 is an unknown parameter and s is the power (how bright they shine) at which the lamp is used.

The first task: Write down the log-likelihood for β given the observations.

My question is where to start? Should it be: $$Ln[\mu(\beta,s)] = ln(\beta) - ln(s)$$ or am I doing it wrong?

• Please add self-study as a tag and provide more details on your issues. Sep 9, 2020 at 14:07

$$\lambda = -\log P = -\log \prod_i p_i = -\sum_i\log p_i$$
where $$P$$ is the probability of observing the event that you observed (some lamps blew up and others did not). The big event that you observed consists of multiple small events, one for each lamp. Since lamps are independent, big event is a product of small events. Now you need to correctly plug in your probability distributions, simplify, and you should be done
• $\mu$ and $\beta$ are parameters. You must plug in the probability distributions into $p$. In your case they are exponential. Make sure that you evaluate the probability distributions at the observations that you acquired. Log-likelihood will depend both on parameters of the distributions, as well as on the exact observations on which lamps blew up and which did not Sep 9, 2020 at 13:09