Question

Let \(X_1,X_2,\ldots,X_n\ (n\geq2)\) be a random sample from the probability density function \(f(x)\). Consider the following hypotheses:
\[ H_0: f(x)=\frac{1}{\sqrt{2\pi}}\,e^{-\frac{x^2}{2}};\quad -\infty<x<\infty \]
\[ H_1: f(x)=\frac{1}{2}\,e^{-|x|};\quad -\infty<x<\infty. \]
For testing \(H_0\) against \(H_1\), let \(R\) denote the critical region based on the likelihood ratio test having level \(0.05\). Then, for some constant \(c\), the region \(R\) is

• A. \(\{(x_1,x_2,\ldots,x_n):\sum_{i=1}^n(x_i-1)^2>c\}\)
• B. \(\{(x_1,x_2,\ldots,x_n):\sum_{i=1}^n(x_i-1)^2<c\}\)
• C. \(\{(x_1,x_2,\ldots,x_n):\sum_{i=1}^n(|x_i|-1)^2>c\}\)
• D. \(\{(x_1,x_2,\ldots,x_n):\sum_{i=1}^n(|x_i|-1)^2<c\}\)

Hint:

Take the log likelihood ratio and complete the square in \(|x_i|\); remember \(x_i^2=|x_i|^2\), so the exponent naturally becomes a function of \(|x_i|-1\).

Answer 1

The Correct Option is C

Step 1: Write the likelihood ratio.
\[ \Lambda=\frac{L_1}{L_0}=\frac{\left(\frac{1}{2}\right)^n\exp\left(-\sum_{i=1}^n|x_i|\right)}{\left(\frac{1}{\sqrt{2\pi}}\right)^n\exp\left(-\sum_{i=1}^n \frac{x_i^2}{2}\right)}. \]

Step 2: Take logs.
\[ \ln\Lambda = n\ln\!\left(\frac{\sqrt{2\pi}}{2}\right) + \sum_{i=1}^n\left(\frac{x_i^2}{2}-|x_i|\right). \] Reject for large \(\Lambda\), so reject when \(\sum(x_i^2/2-|x_i|)>c_1\).

Step 3: Complete the square in \(|x_i|\).
\[ \frac{x_i^2}{2}-|x_i| = \frac{1}{2}\left[(|x_i|-1)^2-1\right]. \]

Step 4: Simplify the region.
\[ \sum_{i=1}^n(|x_i|-1)^2 > c. \]

Step 5: Rule out other options.
Options using \((x_i-1)^2\) don't match since \(x_i\) can be negative; the reversed inequality is wrong since large \(\sum(|x_i|-1)^2\) favors \(H_1\).

Final Answer:
\[ \boxed{R=\left\{(x_1,\ldots,x_n):\sum_{i=1}^n(|x_i|-1)^2>c\right\}} \]