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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

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If this is homework, please add the homework tag and indicate what effort you have already put forward. Do you understand the relationship of the likelihood function to the pdf? – Jonathan Christensen Nov 8 '12 at 1:14
I've read this statement - 'The interpretation of the likelihood function L(θ) is that L(θ) is the probability (or p.d.f.) of obtaining the observed data if θ were the true value of the parameter' though I do not understand what it means. – cheeseman123 Nov 8 '12 at 1:28

1 Answer 1

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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 $\theta$.

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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

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