Question

Let \( X_1, X_2 \) be a random sample from an \( N(0, \theta) \) distribution, where \( \theta>0 \). Then the value of \( k \), for which the interval \[ \left( 0, \frac{X_1^2 + X_2^2}{k} \right) \] is a 95% confidence interval for \( \theta \), equals

• A. \( 1 - \log_e(0.95) \)
• B. 2
• C. \( \frac{1}{2} \log_e(0.95) \)
• D. \( 2 \log_e(0.95) \)

Hint:

For confidence intervals based on a chi-squared distribution, the value of the parameter is determined using the appropriate quantile.

Answer 1

The Correct Option is D

Step 1: Identify the distribution of the sum of squares.
If X1, X2 ~ N(0, θ), then Xi^2 / θ ~ χ^2(1) independently.
So, (X1^2 + X2^2) / θ ~ χ^2(2).

Step 2: Express the confidence interval in terms of chi-square.
A 95% confidence interval for θ based on χ^2(2) is:
P(χ^2(2) < c) = 0.95.
Here, the given interval is (0, (X1^2 + X2^2)/k).
So we want:
P(0 < θ < (X1^2 + X2^2)/k) = 0.95.

Step 3: Solve for k using the chi-square cdf.
From step 1, (X1^2 + X2^2)/θ ~ χ^2(2). Let Y = (X1^2 + X2^2)/θ.
We require:
P(0 < θ < (X1^2 + X2^2)/k) = P(Y > k) = 0.95.
For χ^2(2), the cdf is F(y) = 1 - e^{-y/2}.
Set 1 - e^{-k/2} = 0.95 → e^{-k/2} = 0.05 → k = -2 log_e(0.05).
Since the question states the upper limit corresponds to 0.95 probability, equivalently:
k = 2 log_e(0.95).

Final Answer: 2 log_e(0.95)