Question

Let \( X_1, \dots, X_5 \) be a random sample from \( N(\theta, 6) \), where \( \theta \in \mathbb{R} \), and let \( c(\theta) \) be the Cramer-Rao lower bound for the variances of unbiased estimators of \( \theta \) based on the above sample. Then \( 15 \cdot \inf_{\theta \in \mathbb{R}} c(\theta) \) equals _________ (answer in integer).

Hint:

For the Cramer-Rao lower bound, remember that it depends on the Fisher information, which can often be computed directly for common distributions like the normal distribution.

Answer 1

Step 1: Fisher Information for a single observation
For a normal distribution \( \mathcal{N}(\theta, \sigma^2) \), the Fisher Information is:
\[ I(\theta) = \frac{1}{\sigma^2} \] Given \( \sigma^2 = 6 \), we have: \[ I_1(\theta) = \frac{1}{6} \]

Step 2: Fisher Information for the full sample
Since the sample size is 5: \[ I_5(\theta) = 5 \cdot \frac{1}{6} = \frac{5}{6} \]

Step 3: Cramér-Rao Lower Bound
The CRLB is the reciprocal of the Fisher Information: \[ c(\theta) = \frac{1}{I_5(\theta)} = \frac{6}{5} \] This value is constant and does not depend on \( \theta \), so: \[ \inf_{\theta \in \mathbb{R}} c(\theta) = \frac{6}{5} \]

Step 4: Final Computation
\[ 15 \cdot \inf_{\theta \in \mathbb{R}} c(\theta) = 15 \cdot \frac{6}{5} = 18 \]

Final Answer: 18