Step 1: Concept Overview:
The likelihood function, \(L(\theta)\), for observations \(x_1, x_2, \dots, x_n\) is the product of the PDF at each point. We aim to find this value when \( \theta = 2 \).
Step 2: Formula:
The likelihood function is:
\[ L(\theta | x_1, x_2, \dots, x_n) = \prod_{i=1}^n f(x_i; \theta) \]Calculate \( L(\theta=2 | x_1=1, x_2=4, x_3=2) \).
Step 3: Calculation Details:
Evaluate the PDF \(f(x; \theta)\) at \( \theta = 2 \) for each observation.The PDF at \(\theta=2\) is:\( f(x; 2) = \begin{cases} \frac{2x}{5(2)} = \frac{x}{5} & ; 0 \le x \le 2
\frac{2(5-x)}{5(5-2)} = \frac{2(5-x)}{15} & ; 2 \le x \le 5 \end{cases} \)PDF values for each observation:- For \( x_1 = 1 \): Since \( 0 \le 1 \le 2 \):
\[ f(1; 2) = \frac{1}{5} \]- For \( x_2 = 4 \): Since \( 2 \le 4 \le 5 \):
\[ f(4; 2) = \frac{2(5-4)}{15} = \frac{2(1)}{15} = \frac{2}{15} \]- For \( x_3 = 2 \): Since \(x=2\) is the boundary point, using either formula will yield the same result; using the first case:
\[ f(2; 2) = \frac{2}{5} \]Compute the likelihood:\[ L(2) = f(1; 2) \times f(4; 2) \times f(2; 2) \]\[ L(2) = \frac{1}{5} \times \frac{2}{15} \times \frac{2}{5} \]\[ L(2) = \frac{1 \times 2 \times 2}{5 \times 15 \times 5} = \frac{4}{375} \]
Step 4: Result:
The likelihood function at \( \theta=2 \) is \( \frac{4}{375} \).