Question

Let X have a probability density function of the form, \( f(x;\theta) = \begin{cases} \frac{1}{\theta} e^{-x/\theta} & ; 0<x<\infty, \theta>0 \\ 0 & ; \text{otherwise} \end{cases} \) To test null hypothesis \(H_0: \theta = 2\) against the alternate hypothesis \(H_1: \theta = 1\), a random sample of size 2 is taken. For the critical region \(W_0 = \{(x_1, x_2) : 6.5 \le x_1 + x_2\}\), the power of the test is

• A. \( P(\chi^2_{(4)} \le 6.5) \)
• B. \( P(\chi^2_{(4)} \ge 6.5) \)
• C. \( P(\chi^2_{(4)} \ge 13) \)
• D. \( P(\chi^2_{(4)} \ge 2) \)

Hint:

Recognizing the relationship between common distributions is key to solving advanced problems quickly. The link between the sum of exponentials, the Gamma distribution, and the Chi-squared distribution is a frequently tested concept. Remember: \(2 \times \text{rate} \times \text{Gamma}(\text{shape, rate}) \sim \chi^2_{2 \times \text{shape}}\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The power of a test is the probability of rejecting \(H_0\) when \(H_1\) is true. The PDF is that of an exponential distribution with mean \(\theta\). We need to calculate the probability of the sample falling into the critical region \(W_0\), assuming the parameter value from \(H_1\).

Step 2: Key Formula or Approach:
1. Power = \(P(\text{Reject } H_0 | H_1 \text{ is true}) = P((X_1, X_2) \in W_0 | \theta = 1)\). 2. This is \(P(X_1 + X_2 \ge 6.5 | \theta=1)\). 3. The sum of \(n\) i.i.d. Exponential(\(\lambda\)) random variables is a Gamma(\(n, \lambda\)) variable. The given distribution has mean \(\theta\), so the rate is \(\lambda = 1/\theta\). 4. A Gamma distributed variable can be transformed into a Chi-squared variable using the relation: If \(S \sim \text{Gamma}(k, \lambda)\), then \(2\lambda S \sim \chi^2_{2k}\).

Step 3: Detailed Explanation:
Under the alternative hypothesis \(H_1: \theta = 1\), the random variables \(X_1, X_2\) are i.i.d. from an exponential distribution with mean \(\theta=1\). The rate parameter is \(\lambda = 1/\theta = 1/1 = 1\). Let \(S = X_1 + X_2\). Since \(X_1, X_2\) are i.i.d. Exp(1), their sum \(S\) follows a Gamma distribution with shape parameter \(n=2\) and rate parameter \(\lambda=1\). So, \(S \sim \text{Gamma}(2, 1)\). Now, we use the transformation to the Chi-squared distribution. If \(S \sim \text{Gamma}(k=2, \lambda=1)\), then \(2\lambda S = 2(1)S = 2S\) follows a Chi-squared distribution with \(2k = 2(2) = 4\) degrees of freedom. \[ 2(X_1 + X_2) \sim \chi^2_{(4)} \] The power of the test is \(P(X_1 + X_2 \ge 6.5)\) calculated under \(H_1\). We transform this inequality to be in terms of the Chi-squared variable: \[ X_1 + X_2 \ge 6.5 \] Multiply both sides by 2: \[ 2(X_1 + X_2) \ge 2(6.5) \] \[ 2(X_1 + X_2) \ge 13 \] Since \(2(X_1+X_2)\) is a \(\chi^2_{(4)}\) variable, the probability is: \[ \text{Power} = P(\chi^2_{(4)} \ge 13) \]
Step 4: Final Answer:
The power of the test is \( P(\chi^2_{(4)} \ge 13) \).