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

$$f(x\mid \theta) = \theta e^{−\theta x},\quad \theta > 0, x > 0$$

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

I don't understand this topic very well, how can I derive the likelihood function for $\theta$?



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 $\theta$.

  • $\begingroup$ 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 $\endgroup$ – cheeseman123 Nov 8 '12 at 2:21
  • $\begingroup$ @cheeseman123 It means the distribution when the sample size $n$ goes to infinity. $\endgroup$ – Patrick Li Nov 8 '12 at 3:54

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