Asymptotic MLE Distribution With Two Random Samples 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.

It's possible to solve this problem via "brute force" from just grinding through the calculus of the problem.  To facilitate this analysis, let $\bar{x}_n \equiv \sum_{i=1}^n x_i/n$ and let $\bar{y}_m \equiv \sum_{i=1}^m y_i/m$ denote the sample means of the two groups.  We can then write the log-likelihood function as:
$$\begin{align}
\ell_{\mathbf{x}, \mathbf{y}}(\lambda)
&= \sum_{i=1}^n \log \text{Exp}(x_i|\lambda) + \sum_{i=1}^m \log \text{Bern}(y_i|e^{-\lambda}) \\[6pt]
&= \text{const} + n \log (\lambda) - n \bar{x}_n \lambda + m(1-\bar{y}_m) \log (1-e^{-\lambda}) - m \bar{y}_m \lambda \\[14pt]
&= \text{const} - (n \bar{x}_n + m \bar{y}_m) \lambda + n \log (\lambda) + m(1-\bar{y}_m) \log (1-e^{-\lambda}). \\[6pt]
\end{align}$$
We then have:
$$\begin{align}
\frac{d \ell_{\mathbf{x}, \mathbf{y}}}{d \lambda}(\lambda)
&= - (n \bar{x}_n + m \bar{y}_m) + \frac{n}{\lambda} + m(1-\bar{y}_m) \cdot \frac{e^{-\lambda}}{1-e^{-\lambda}}, \\[6pt]
\frac{d^2 \ell_{\mathbf{x}, \mathbf{y}}}{d \lambda^2}(\lambda)
&= - \frac{n}{\lambda^2} - m(1-\bar{y}_m) \cdot \frac{e^{-\lambda}}{(1-e^{-\lambda})^2}. \\[6pt]
\end{align}$$
Since the second derivative is (almost surely) negative the log-likelihood function is strictly concave, so the MLE occurs at the unique critical point.  The Fisher information is:
$$\begin{align}
I(\lambda)
&= \mathbb{E} \Bigg[ - \frac{d^2 \ell_{\mathbf{X}, \mathbf{Y}}}{d \lambda^2}(\lambda) \Bigg| \lambda \Bigg] \\[6pt]
&= \mathbb{E} \Bigg[ \frac{n}{\lambda^2} + m(1-\bar{Y}_m) \cdot \frac{e^{-\lambda}}{(1-e^{-\lambda})^2} \Bigg| \lambda \Bigg] \\[6pt]
&= \frac{n}{\lambda^2} + m(1-e^{-\lambda}) \cdot \frac{e^{-\lambda}}{(1-e^{-\lambda})^2} \\[6pt]
&= \frac{n}{\lambda^2} + m \cdot \frac{e^{-\lambda}}{1-e^{-\lambda}}, \\[6pt]
\end{align}$$
so its inverse is:
$$\begin{align}
\frac{1}{I(\lambda)} 
= \frac{1}{n/\lambda^2 + m e^{-\lambda}/(1-e^{-\lambda})} 
= \frac{\lambda^2 (1-e^{-\lambda})}{n (1-e^{-\lambda}) + m \lambda^2 e^{-\lambda}}.
\end{align}$$
Since the setup of your problem obeys the regularity conditions for asymptotic normality of the MLE, as $\min(n,m) \rightarrow \infty$ we have the asymptotic distribution:
$$\hat{\lambda}_{n,m} \overset{\text{Approx}}{\sim} \text{N} \Bigg( \lambda, \frac{\lambda^2 (1-e^{-\lambda})}{n (1-e^{-\lambda}) + m \lambda^2 e^{-\lambda}} \Bigg).$$
As $\min(n,m) \rightarrow \infty$ the variance converges to zero, so the MLE is a consistent estimator of $\lambda$.
