I've encountered a sample problem in my statistics notes which I don't understand.
Suppose $X_1,...,X_{100}$ are r.v.'s which are Poisson-distributed with parameter $\frac{\mu}{100}$. We're looking for the MLE for $\mu$. However, we can only observe $Y_1,...,Y_{100}$ which are Bernoulli distributed with value $0$ if $X_i=0$ and $1$ otherwise (this is due to technical constraints). We find that the MLE for $p=e^{{-\frac{\mu}{100}}}$ is the sample proportion $\frac{1}{100} \sum_{i=1}^{100} Y_i$. Based on this, the notes say we can derive that
$-100\log(\frac{1}{100} \sum_{i=1}^{100} Y_i)$
is the MLE for $\mu$
Does anyone have any idea how we arrive at this result? It's been driving me crazy.
In order to get the same result of your solution, we must amend a typo in the text
It must be the converse:
They are Bernoulli distributed with value $1$ if $X_i=0$ and $0$ otherwise
Now it works because now your rv's become $B\Big(e^{-\frac{\mu}{100}}\Big)$
The MLE of $g(\mu)=e^{-\frac{\mu}{100}}$ is obviously
$$\widehat{g(\mu)}=\overline{Y}_{100}$$
For invariance property we can write
$$g(\hat{\mu})=\overline{Y}_{100}$$
$$e^{-\frac{\hat{\mu}}{100}}=\overline{Y}_{100}$$
now solve in $\hat{\mu}$ and get your result
If you use exatly the text you posted, the result is not very different...try, FYK