# 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?

• $\theta = P(X_{n+1} > 100) = e^{-100\lambda}$ and $\hat{\lambda} = n/\sum_{i=1}^n X_i$. Hence $\hat\theta = exp\{-100n / \sum_{i=1}^n X_i\}$
– Chia
Jan 28, 2023 at 1:33
• Hello thanks for the comment! That's what I thought too, but when doing the CDF doesn't it come out as $e^{-100\lambda} - e^{-\lambda x?$ Also, I've calculated the MLE thanks to the invariance property (as you suggested), but what does the pdf and asymptotic variance of MLE (I think I need the inverse of the fisher information matrix, but what does the log likelihood look like?) Jan 28, 2023 at 12:21

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.