I'm studiyng for an exam, and I found this problem which I can not managed to solve... I will be really grateful if someone can help me, thanks you.
Let $\left\{X_{1}, \ldots, X_{n}\right\} \sim^{iid} Exp(\lambda)$ and $\left\{Y_{1}, \ldots, Y_{m}\right\}\sim^{iid}Bern(p)$, whith $p=\exp(-\lambda)$, independent of each other.
Assume that the maximum likelihood estimator (MLE) $\widehat{\lambda}_{n, m}$ exists and $m=\alpha n$, where $\alpha$ is known. Using the Central Limit Theorem find the asymptotic distribution of the MLE where $n, m \rightarrow \infty$
Given the central limit theorem thing, I thought that the MLE is a function of the mean (then by using the delta method we have our distribution). But I don't see how the likelihood equation can be non-numerical solved, because we have the log-likelihood function
$$l(\lambda) =n\log(\lambda) -\lambda(\sum_{i=1}^{n}x_{i} +\sum_{i=1}^{m}y_{i}) +(m -\sum_{i=1}^{m}y_{i} ) \log( 1 -e^{-\lambda})$$
then
$$ \dfrac{\partial}{\partial\lambda}l(\lambda)=0:\quad \dfrac{n}{\lambda} -(\sum_{i=1}^{n}x_{i} +\sum_{i=1}^{m}y_{i}) +\dfrac{e^{-\lambda}(m -\sum_{i=1}^{m}y_{i})}{1 -e^{-\lambda}} =0$$
$$ \dfrac{n}{\lambda} -(\sum_{i=1}^{n}x_{i} +\sum_{i=1}^{m}y_{i}) +\dfrac{(m -\sum_{i=1}^{m}y_{i})}{e^{\lambda} -1} =0$$
Then, because of the "assume that the MLE exists" I thought of ussing the EMV asymptotic distribution (assuming the conditions), but I fail to see how to obtain de variance of the normal distribution because it suppose to be the fisher information for 1 observation, but we have two random samples with totally different distributon so I faced a dead end.