Part CCSIR NET June 2025prior-precision-not-prior-variance-is-what-adds-to-n

Prior precision not prior variance is what adds to n

Let $X_{1}, X_{2}, \dots, X_{n}$ be a random sample from $N(\theta, 1), \theta \in \mathbb{R}$ . If $\theta$ ̂ is the Bayes estimator of $\theta$ with respect to some prior $\pi(\theta)$ and loss function $L(\theta, d)$ . Then, which of the following statements are true?

A. $\theta$ ̂ $= \sum_{i}$ ₌ $_{1}^{n} X_{i}/(n + \tau^{2})$ , if the prior is $N(0, 1/\tau^{2}), \tau^{2}$ known and $L(\theta, d) = (\theta - d)^{2}$ ✓ Correct
B. $\theta$ ̂ $= \sum_{i}$ ₌ $_{1}^{n} X_{i}/(n + \tau^{2})$ , if the prior is $N(0, \tau^{2}), \tau^{2}$ known and $L(\theta, d) =$ | $\theta - d$ |
C. $\theta$ ̂ $= \sum_{i}$ ₌ $_{1}^{n} X_{i}/(n + 1/\tau^{2})$ , if the prior is $N(0, 1/\tau^{2}), \tau^{2}$ known and $L(\theta, d) =$ | $\theta - d$ |
D. $\theta$ ̂ $= \sum_{i}$ ₌ $_{1}^{n} X_{i}/n$ , if the prior is the Jeffreys prior and $L(\theta, d) = (\theta - d)^{2}$ ✓ Correct

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward, and why this one tests execution slip.

See pricing

50 are analysed free — try those first.

The trap it tests

Execution slip

The idea was right. The computation was not.

Drill statements like this

Related counterexample: Pairwise independent events are mutually independent

Prior precision not prior variance is what adds to n

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