Exponential distribution with maximum Here the question
Hello everyone, I'm been studying for an exam and I'm stuck on this exercise. I know how to solve  the second part (finding Mle once I have the distribution) however I don't understand how to rewrite the parameter in Point 1.
What I got to so far:
Since we need to have $ \theta=P(X_{n+1} > 100)$, and $X_{i} \  \ldots \ X_{n} \sim exp(\lambda)$, we have that also $X_{n+1}$ is distributed as an exponential.
Here I think I need to rewrite the CDF of the exponential in terms of $\theta$, you know $1- P (X_i < 100)$ and then use it to find the solution to point 1... I don't know. I'm pretty lost. Can someone help me with both these points?
 A: Here I'm going to propose my own solution.
Part 1: In the problem $\theta$ is defined as the probability that next year's annual maximum will exceed 100mm.
$$
\theta = P(X_{n+1} > 100) = \int_{100}^\infty \lambda e^{-\lambda x} dx = e^{-100\lambda}
$$
I'm not sure what you mean for $e^{-100\lambda}-e^{-100x}$
Part 2: we firstly find the MLE for $\lambda$. The likelihood function is
$$
\begin{align}
L(\lambda) &= f(X_1) f(X_2) \cdots f(X_n) \\
&= \lambda^n e^{-\lambda(X_1 + \cdots + X_n)}
\end{align}
$$
Log likelyhood is
$$
\log L(\lambda) = n\log\lambda - \lambda (X_1 + \cdots + X_n)
$$
Taking the derivate with respect to $\lambda$ and setting it to zero yield that the MLE of $\lambda$ is
$$
\hat\lambda = \frac n{X_1 + \cdots +X_n}
$$
Hence the MLE of $\theta$ is
$$
\hat\theta = e^{-\frac{100n}{X_1 + \cdots X_n}}
$$
To get the asymptotic distribution,
$$ 
E\frac{\sum_{i=1}^n X_i}{n} = \frac1\lambda \\
Var\frac{\sum_{i=1}^n X_i}{n} = \frac1{n\lambda^2}
$$
Applying the Continuous mapping theorem derives that
$$
\hat\theta \sim N(e^{-100\lambda}, e^{-100n\lambda^2})
$$
when $n$ is sufficiently large.
