Question

6. Let equation be a random sample from the Exponential distribution with pdf

equation

(i) Obtain the maximum likelihood estimator (MLE) of equation.

(ii) Hence, find the maximum likelihood estimate of equation for the observed sample: 

2.1, 1.6, 3.4, 0.9, 2.5

19 Jan 2026
Answer :
Word Count : 160
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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