I am looking for advice to make sure I am understanding an example. If I am not , I am looking for advice on where I can fix things or how to think about it in a different way.
Suppose we have a random variable $T$ that denotes the time to death of a fly in a jar. $T$ is distributed as an exponential random variable with mean $\frac{\mu}{s}$ where $s$ is the moisture level in the jar. Say we test this $n$ independent times at moisture levels $s_{1},..s_{n}$ and we observe times of death $t_{1},...t_{n}$.
I am looking for an MLE of $\mu$ and an estimate of its asymptotic variance.
What I tried:
I noted that $T_{i} ~ \exp(s_{i}/ \mu)$ write $$G_{i}=\frac{s_{i}}{\mu}$$ and derived that the log likelhood would be
$$l(\mu;t)= \sum_{i=1}^{n} log(G_{i})-\sum_{i=1}^{n} t_{i}G_{i}$$
Taking derivatives and setting equal to zero I came to the conclusion that
mle was $$\mu = \frac{ \sum_{i=1}^{n} t_{i}s_{i}}{n}$$
But I am not sure if I took the correct approach. Does anyone have any comments or can confirm? Does anyone get a different answer?
For the asymptotic variance, assuming the above makes sense I would use Fisher criterion ie $I^{-1} (\mu)$ I will update depending on what I hear about the first part.
Thanks all
Taking it from where you stopped, I'm sure this is correct \begin{equation} l(\mu;t)= \sum_{i=1}^{n} log(G_{i})-\sum_{i=1}^{n} t_{i}G_{i} \end{equation} Using $G_i = \frac{s_i}{\mu}$ we get \begin{equation} l(\mu;t)=\sum_{i=1}^{n} s_i- \sum_{i=1}^{n} log(\mu)-\frac{1}{\mu}\sum_{i=1}^{n} t_{i}s_i \end{equation} Deriving w.r.t $\mu$, we get \begin{equation} \frac{\partial}{\partial \mu} l(\mu;t)=-\frac{n}{\mu}+\frac{1}{\mu^2}\sum_{i=1}^{n} t_{i}s_i = 0 \end{equation} which gives you the answer that you've arrived to. Well done.
And yes knowing that the MLE variance is $I^{-1}(\mu)$, you can get your variance. I think you're on the right track.