Question

A random variable X has a distribution with density function
\( f(x;\beta) = \begin{cases} (\beta+1)x^\beta; & 0<x<1; \beta>-1 \\ 0; & \text{otherwise} \end{cases} \) Based on 'n' observations on X, Maximum Likelihood Estimator (MLE) of \(\beta\) is

• A. \( \frac{-1}{\sum_{i=1}^n \log(x_i)} \)
• B. \( \frac{-n}{\sum_{i=1}^n \log(x_i)} - 1 \)
• C. \( \sum_{i=1}^n \log(x_i) - 1 \)
• D. \( \frac{-1}{\sum_{i=1}^n \log(x_i)} - 1 \)

Hint:

The process of finding an MLE is standardized: Likelihood \(\rightarrow\) Log-Likelihood \(\rightarrow\) Differentiate \(\rightarrow\) Set to Zero \(\rightarrow\) Solve. Using the log-likelihood function simplifies the process by converting products to sums, which are easier to differentiate.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The Maximum Likelihood Estimator (MLE) of a parameter is the value of the parameter that maximizes the likelihood function for a given sample of data. The standard procedure is to write the likelihood function, take its natural logarithm (log-likelihood), differentiate with respect to the parameter, set the derivative to zero, and solve for the parameter.

Step 2: Key Formula or Approach:
1. Write the likelihood function: \( L(\beta) = \prod_{i=1}^n f(x_i; \beta) \). 2. Write the log-likelihood function: \( l(\beta) = \ln(L(\beta)) \). 3. Solve the equation: \( \frac{d l(\beta)}{d\beta} = 0 \) for \(\beta\).

Step 3: Detailed Explanation:
Given a random sample \(x_1, x_2, \ldots, x_n\): 1. The likelihood function is: \[ L(\beta) = \prod_{i=1}^n (\beta+1)x_i^\beta = (\beta+1)^n \left(\prod_{i=1}^n x_i\right)^\beta \] 2. The log-likelihood function is: \[ l(\beta) = \ln(L(\beta)) = \ln\left((\beta+1)^n \left(\prod_{i=1}^n x_i\right)^\beta\right) \] \[ l(\beta) = n \ln(\beta+1) + \beta \ln\left(\prod_{i=1}^n x_i\right) \] \[ l(\beta) = n \ln(\beta+1) + \beta \sum_{i=1}^n \ln(x_i) \] 3. Differentiate with respect to \(\beta\): \[ \frac{dl}{d\beta} = \frac{n}{\beta+1} + \sum_{i=1}^n \ln(x_i) \] 4. Set the derivative to zero and solve for the MLE, \( \hat{\beta} \): \[ \frac{n}{\hat{\beta}+1} + \sum_{i=1}^n \ln(x_i) = 0 \] \[ \frac{n}{\hat{\beta}+1} = - \sum_{i=1}^n \ln(x_i) \] \[ \hat{\beta}+1 = \frac{n}{- \sum_{i=1}^n \ln(x_i)} = \frac{-n}{\sum_{i=1}^n \ln(x_i)} \] \[ \hat{\beta} = \frac{-n}{\sum_{i=1}^n \ln(x_i)} - 1 \]
Step 4: Final Answer:
The MLE of \(\beta\) is \( \frac{-n}{\sum_{i=1}^n \log(x_i)} - 1 \).