Question

If, \(X \sim N(\theta, 1)\) and in order to test \(H_0: \theta=1\) against the alternate \(H_1: \theta=2\) a random sample \((x_1, x_2)\) of size 2 is taken. Then, the best critical region (B.C.R.) is given by (where \(Z_\alpha = 1.64\))

• A. \( W = \{(x_1, x_2): x_1 + x_2 \ge 4.32\} \)
• B. \( W = \{(x_1, x_2): x_1 + x_2 \ge 1.64\} \)
• C. \( W = \{(x_1, x_2): x_1 + x_2 \ge 2\} \)
• D. \( W = \{(x_1, x_2): x_1 + x_2 \ge 3.96\} \)

Hint:

The Neyman-Pearson Lemma is the fundamental tool for finding the most powerful test for simple hypotheses. The process always involves setting up the likelihood ratio, simplifying it to a condition on a sufficient statistic (like the sum or mean), and then using the size of the test \(\alpha\) to find the exact critical value.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
To find the Best Critical Region (BCR) for testing a simple null hypothesis against a simple alternative, we use the Neyman-Pearson Lemma. The lemma states that the BCR is based on the likelihood ratio.

Step 2: Key Formula or Approach:
The Neyman-Pearson Lemma states that the BCR of size \(\alpha\) is the region W such that for some \(k>0\): \[ W = \left\{ \mathbf{x} : \frac{L(\theta_1 | \mathbf{x})}{L(\theta_0 | \mathbf{x})} \ge k \right\} \] where \(L\) is the likelihood function. We then find \(k\) such that \(P(\mathbf{X} \in W | \theta_0) = \alpha\).

Step 3: Detailed Explanation:
Here, \(\theta_0 = 1\) and \(\theta_1 = 2\). The PDF is \(f(x;\theta) = \frac{1}{\sqrt{2\pi}}e^{-(x-\theta)^2/2}\). The likelihood function for a sample of size 2 is \(L(\theta) = f(x_1;\theta)f(x_2;\theta)\). The likelihood ratio is: \[ \frac{L(2)}{L(1)} = \frac{e^{-(x_1-2)^2/2}e^{-(x_2-2)^2/2}}{e^{-(x_1-1)^2/2}e^{-(x_2-1)^2/2}} = \exp\left[-\frac{1}{2}\left(\sum(x_i-2)^2 - \sum(x_i-1)^2\right)\right] \] Let's simplify the exponent term: \[ \sum(x_i-2)^2 - \sum(x_i-1)^2 = \sum(x_i^2-4x_i+4) - \sum(x_i^2-2x_i+1) = \sum(-2x_i+3) = -2\sum x_i + 2(3) = -2(x_1+x_2)+6 \] The ratio test \(\frac{L(2)}{L(1)} \ge k\) becomes: \[ \exp\left[-\frac{1}{2}(-2(x_1+x_2)+6)\right] \ge k \] \[ \exp(x_1+x_2-3) \ge k \] Taking the natural log of both sides: \[ x_1+x_2-3 \ge \ln(k) \implies x_1+x_2 \ge 3+\ln(k) \] The BCR is of the form \(x_1+x_2 \ge k'\) for some constant \(k'\). To find \(k'\), we use the size of the test, \(\alpha\). \(P(Z>1.64) = \alpha\), which corresponds to \(\alpha \approx 0.05\). \[ \alpha = P(\text{Reject } H_0 | H_0 \text{ is true}) = P(X_1+X_2 \ge k' | \theta=1) \] Under \(H_0: \theta=1\), \(X_1, X_2\) are i.i.d. \(N(1, 1)\). Let \(S = X_1+X_2\). The distribution of S is Normal with: - Mean: \(E(S) = E(X_1) + E(X_2) = 1+1=2\) - Variance: \(\text{Var}(S) = \text{Var}(X_1) + \text{Var}(X_2) = 1+1=2\) So, under \(H_0\), \(S \sim N(2, 2)\). We standardize the probability statement: \[ P\left( \frac{S - E(S)}{\sqrt{\text{Var}(S)}} \ge \frac{k' - 2}{\sqrt{2}} \right) = \alpha \] \[ P\left( Z \ge \frac{k' - 2}{\sqrt{2}} \right) = \alpha \] We are given that \(P(Z \ge 1.64) = \alpha\). Therefore: \[ \frac{k' - 2}{\sqrt{2}} = 1.64 \] \[ k' - 2 = 1.64\sqrt{2} \approx 1.64 \times 1.4142 = 2.3193 \] \[ k' = 2 + 2.3193 = 4.3193 \] Rounding to two decimal places, \(k' = 4.32\).
Step 4: Final Answer:
The best critical region is \( W = \{(x_1, x_2): x_1 + x_2 \ge 4.32\} \).