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Part BCSIR NET June 2024conjugate-prior-posterior-mean

Conjugate prior posterior mean

Let X1,…,X10X_{1}, \dots, X_{10} be a random sample from a distribution with the probability density function f(x|θ)=θx(θ−1)\theta) = \theta{}x^(\theta-1) if 0 < x < 1, and 0 otherwise, where θ>0\theta > 0 is an unknown parameter. The prior distribution of θ\theta is given by π(θ)=θe(−θ)\pi(\theta) = \theta{}e^(-\theta) if θ>0\theta > 0, and 0 otherwise. The Bayes estimator of θ\theta under squared error loss is

  1. A.12/(1−∑i12 / (1 - \sum_{i}₌110ln⁡Xi)_{1}^{10} \ln X_{i})✓ Correct
  2. B.11/(2−∑i11 / (2 - \sum_{i}₌110ln⁡Xi)_{1}^{10} \ln X_{i})
  3. C.(3+∑i(3 + \sum_{i}₌110ln⁡Xi)/13_{1}^{10} \ln X_{i}) / 13
  4. D.(2+∑i(2 + \sum_{i}₌110ln⁡Xi)/11_{1}^{10} \ln X_{i}) / 11

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: The MLE is unbiased

More on this topic

The chapter behind this: Maximum likelihood and the method of moments — free to read

From Estimation › MLE and method of moments

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