Part BCSIR NET December 2024the-prior-adds-one-to-the-exponent

The prior adds one to the exponent

Let $X_{1}, X_{2}, \dots, X_{n}$ be a random sample from distribution Poisson $(θ), θ > 0$ . Let $π (θ) = e^{(} - θ), θ > 0$ , be the prior distribution of $θ$ . Under the squared error loss function, which of the following is the Bayes estimator of $e^{(} - 2 θ)$ ?

A. $((n + 1) / (n + 3))^{(} \sum (i = 1.. n) X_{i} + 1)$ ✓
B. $((n + 1) / (n + 3)) e^{(} - (2/ n) \sum (i = 1.. n) X_{i})$
C. $((n + 1) / (n + 3))^{(} \sum (i = 1.. n) X_{i})$
D. $((n + 1) / (n + 3)) e^{(} - (2/ n) (\sum (i = 1.. n) X_{i} + 1))$

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.

See pricing

50 are analysed free — try those first.

Related counterexample: The MLE is unbiased

The prior adds one to the exponent

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