Question

Let \(X_{1}, X_{2}, \ldots, X_{n}\) represent a random sample from each of the distributions having the following pdfs: (a) \(f(x ; \theta)=\theta x^{\theta-1}, 0

Short answer

The MLE for \(\theta\) for part (a) is \(\frac{n}{\sum_{i=1}^n \ln x_i}\), and for part (b) it is \(\min(x_i)\).

Step-by-step solution

- Compute Log-likelihood function for Part (a)

The first step is to compute the logarithm of the likelihood function, the log-likelihood. Given the pdf \(f(x ; \theta)=\theta x^{\theta-1}\), the joint likelihood of n observations is \(\prod_{i=1}^{n} \theta x_{i}^{\theta-1}\). Taking the natural log gives \(\ln L(\theta)=n \ln \theta+(\theta-1) \sum_{i=1}^{n} \ln x_{i}\).

- Find MLE for \(\theta\) for Part (a)

To find the MLE, take the derivative of the log-likelihood function with respect to \(\theta\), set that equal to zero and solve for \(\theta\). Differentiating and setting equal to zero, \(\frac{n}{\hat{\theta}}+\sum_{i=1}^{n} \ln x_{i}=0\), you find \(\hat{\theta}_{\text{MLE}} = \frac{n}{\sum_{i=1}^n \ln x_i}\).

- Compute Log-likelihood function for Part (b)

Similarly, for part (b) with pdf \(f(x ; \theta)=e^{-(x-\theta)}\), the joint likelihood of n observations is given by \(\prod_{i=1}^{n} e^{-(x_i - \theta)}\). Taking the natural log gives \(\ln L(\theta)= \sum_{i=1}^{n} - (x_i - \theta)\).

- Find MLE for \(\theta\) for Part (b)

As in Step 2, the MLE is found by taking the derivative of the log-likelihood function with respect to \(\theta\), setting it equal to zero, and solving for \(\theta\). Differentiating and setting equal to zero, \(\sum_{i=1}^n - (x_i - \hat{\theta})=0\), you find \(\hat{\theta}_{\text{MLE}} = \min(x_i)\). Note that the minimum value of \(x_i\) is used as the estimator because \(\theta\) is less than or equal to \(x\), as per the given distribution.