Question

Consider the probability density function \( f(x;\theta) = \begin{cases} \frac{2x}{5\theta} & ; 0 \le x \le \theta \\ \frac{2(5-x)}{5(5-\theta)} & ; \theta \le x \le 5 \end{cases} \) For a sample of size 3, let the observations are, \( x_1 = 1, x_2 = 4, x_3 = 2 \). Then, the value of likelihood function at \( \theta=2 \) is

• A. \( \frac{4}{125} \)
• B. \( \frac{1}{125} \)
• C. \( \frac{8}{125} \)
• D. \( \frac{4}{375} \)

Hint:

When dealing with a piecewise PDF, be very careful to use the correct formula for each data point based on its value relative to the parameter(s). A common mistake is using the same piece of the function for all observations.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The likelihood function, \(L(\theta)\), for a given set of observations \(x_1, x_2, \dots, x_n\) is the product of the probability density function (PDF) evaluated at each of these points. We need to calculate this value for the specific parameter value \( \theta = 2 \).

Step 2: Key Formula or Approach:
The likelihood function is given by: \[ L(\theta | x_1, x_2, \dots, x_n) = \prod_{i=1}^n f(x_i; \theta) \] We need to calculate \( L(\theta=2 | x_1=1, x_2=4, x_3=2) \).

Step 3: Detailed Explanation:
First, we evaluate the PDF \(f(x; \theta)\) at \( \theta = 2 \) for each observation. The PDF at \(\theta=2\) is: \( f(x; 2) = \begin{cases} \frac{2x}{5(2)} = \frac{x}{5} & ; 0 \le x \le 2
\frac{2(5-x)}{5(5-2)} = \frac{2(5-x)}{15} & ; 2 \le x \le 5 \end{cases} \) Now, we find the value of the PDF for each observation: - For \( x_1 = 1 \): Since \( 0 \le 1 \le 2 \), we use the first case. \[ f(1; 2) = \frac{1}{5} \] - For \( x_2 = 4 \): Since \( 2 \le 4 \le 5 \), we use the second case. \[ f(4; 2) = \frac{2(5-4)}{15} = \frac{2(1)}{15} = \frac{2}{15} \] - For \( x_3 = 2 \): The value \(x=2\) is the boundary point and is included in both intervals. The function is continuous at \(x=\theta\), so either formula gives the same result. Let's use the first one. \[ f(2; 2) = \frac{2}{5} \] Now, we compute the likelihood by multiplying these values: \[ L(2) = f(1; 2) \times f(4; 2) \times f(2; 2) \] \[ L(2) = \frac{1}{5} \times \frac{2}{15} \times \frac{2}{5} \] \[ L(2) = \frac{1 \times 2 \times 2}{5 \times 15 \times 5} = \frac{4}{375} \]
Step 4: Final Answer:
The value of the likelihood function at \( \theta=2 \) is \( \frac{4}{375} \).