Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X_1, X_2, \dotsc , X_n$ be a random sample from a distribution having probability density function (pdf)

$f(x|θ) = θ e^{−θx},\quad θ > 0, x > 0$.

Derive the likelihood function for $θ$, maximum likelihood estimator (MLE) of $θ$ and its asymptotic distribution.

I don't understand this topic very well, how can I derive the likelihood function for θ

share|cite|improve this question
up vote 1 down vote accepted

Hint:

Likelihood function for $\theta$ is just the probability density function of your sample, but regarded as a function of the parameter $\theta$ instead of the observations. $L(\theta|\mathbf{x})=f(\mathbf{x}|\theta)$.

In this case, since $X_i$ are independent, the probability density function of the sample is the product of each individual pdf. Then you maximize this function with respect to $\theta$ to get the MLE estimate of $\theta$.

share|cite|improve this answer
    
That's brilliant, I think that worked out well. If you don't mind me asking, what does it mean by its asymptotic distribution. All i know is that it is somehow a follow on from the MLE – cheeseman123 Nov 8 '12 at 2:21
    
@cheeseman123 It means the distribution when the sample size $n$ goes to infinity. – Patrick Li Nov 8 '12 at 3:54

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.