Estimating the parameter $\beta$ The lifetime of computer monitors has a exponential distribution where the expected value can be written as:
$\mu(s) = \frac{\beta}{s}$
Where $s$ is how bright the monitor is and both $s$ and $\beta$ are >0
How do I estimate the maximum-likelihood of parameter $\beta$?
I know that the log-likelihood for a exponential distribution is:
$l(\lambda) = n(\ln(\lambda) - \lambda\bar{t})$
And taking the derivative and equals it to zero yields:
$\frac{dl(\lambda)}{d\lambda} = n(\frac{1}{\lambda} - \bar{t}) = 0$
$\bar{t} = \frac{1}{\lambda}$
Here is where I get stuck.
EDIT 1
I have data for both the lifetime of the computer monitors and their brightness.

$
L(\beta) = P(\mathrm{Data} \mid \beta) = \prod_i f(x_i)
$ and $\lambda = \frac{1}{\mu} = \frac{s}{\beta}$
Log is increasing so maximizing log is same as maximizing likelihood:
$\ell(\beta) = \sum \log f(x_i).$
$
\log f(x_i) = \log \lambda e^{-\lambda x_i} = \log s - \log \beta  - \frac{s}{\beta}x_i$
and

 \frac{d}{d \beta} \log f = -\frac{1}{\beta} - \frac{s}{\beta^2}x_i.

So,

 \frac{d}{d \beta}\ell(\beta) = -\frac{n}{\beta} - \frac{s}{\beta^2}\sum x_i = 0

gives

 \hat{\beta} = \frac{s}{n}\sum x_i = s \bar{x}.

Note: the second derivative of $\ell$ is positive since $x_i, s, \beta>0$ so  LL is concave.
