Question
6. Let be a random sample from the Exponential distribution with pdf
(i) Obtain the maximum likelihood estimator (MLE) of .
(ii) Hence, find the maximum likelihood estimate of for the observed sample:
2.1, 1.6, 3.4, 0.9, 2.5
Answer :
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The probability density function of the exponential distribution is [ f(x; \theta) = \theta e^{-\theta x}, \quad x>0, \theta>0. ] For a random sample (X_1, X_2, \dots, X_n), the likelihood function is [ L(\theta) = \prod_{i=1}^{n} f(X_i;\theta) = \prod_{i=1}^{n} \theta e^{-\theta X_i} = \theta^n e^{-\theta \sum_{i=1}^{n} X_i}. ] The log-likelihood function is [ \ell(\theta) = \ln L(\theta) = n
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