Question:medium

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

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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.
Updated On: Feb 18, 2026
  • \( \frac{-1}{\sum_{i=1}^n \log(x_i)} \)
  • \( \frac{-n}{\sum_{i=1}^n \log(x_i)} - 1 \)
  • \( \sum_{i=1}^n \log(x_i) - 1 \)
  • \( \frac{-1}{\sum_{i=1}^n \log(x_i)} - 1 \)
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The Correct Option is B

Solution and Explanation

Step 1: Concept Overview:
The Maximum Likelihood Estimator (MLE) determines the parameter value that maximizes the likelihood function given the data. This involves defining the likelihood function, computing its natural logarithm (log-likelihood), differentiating with respect to the parameter, equating the derivative to zero, and solving for the parameter.

Step 2: Core Methodology:
1. Likelihood function: \( L(\beta) = \prod_{i=1}^n f(x_i; \beta) \).2. Log-likelihood function: \( l(\beta) = \ln(L(\beta)) \).3. Solve: \( \frac{d l(\beta)}{d\beta} = 0 \) for \(\beta\).

Step 3: Detailed Solution:
For a random sample \(x_1, x_2, \ldots, x_n\):1. Likelihood function: \[ L(\beta) = \prod_{i=1}^n (\beta+1)x_i^\beta = (\beta+1)^n \left(\prod_{i=1}^n x_i\right)^\beta \]2. Log-likelihood function: \[ l(\beta) = \ln(L(\beta)) = \ln\left((\beta+1)^n \left(\prod_{i=1}^n x_i\right)^\beta\right) \] \[ l(\beta) = n \ln(\beta+1) + \beta \ln\left(\prod_{i=1}^n x_i\right) \] \[ l(\beta) = n \ln(\beta+1) + \beta \sum_{i=1}^n \ln(x_i) \]3. Differentiation with respect to \(\beta\): \[ \frac{dl}{d\beta} = \frac{n}{\beta+1} + \sum_{i=1}^n \ln(x_i) \]4. Setting the derivative to zero to find 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: Result:
The MLE for \(\beta\) is \( \frac{-n}{\sum_{i=1}^n \log(x_i)} - 1 \).
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