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 $N (θ, 1), θ \in R$ . If $θ$ ̂ is the Bayes estimator of $θ$ with respect to some prior $π (θ)$ and loss function $L (θ, d)$ . Then, which of the following statements are true?

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

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

More on this topic