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?

  • $\begingroup$ $\theta = P(X_{n+1} > 100) = e^{-100\lambda}$ and $\hat{\lambda} = n/\sum_{i=1}^n X_i$. Hence $\hat\theta = exp\{-100n / \sum_{i=1}^n X_i\}$ $\endgroup$
    – Chia
    Jan 28, 2023 at 1:33
  • $\begingroup$ 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?) $\endgroup$ Jan 28, 2023 at 12:21

1 Answer 1


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.

  • $\begingroup$ I see now my mistake, thank you $\endgroup$ Jan 29, 2023 at 15:53

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .