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

The prior adds one to the exponent

Let X1,X2,…,XnX_{1}, X_{2}, \dots, X_{n} be a random sample from distribution Poisson(θ),θ>0(\theta), \theta > 0. Let π(θ)=e(−θ),θ>0\pi(\theta) = e^(-\theta), \theta > 0, be the prior distribution of θ\theta. Under the squared error loss function, which of the following is the Bayes estimator of e(−2θ)e^(-2\theta)?

  1. A.((n+1)/(n+3))(∑(i=1..n)Xi+1)((n+1)/(n+3))^(\sum(i=1..n)X_{i} + 1)✓ Correct
  2. B.((n+1)/(n+3))e(−(2/n)∑(i=1..n)Xi)((n+1)/(n+3))e^(-(2/n)\sum(i=1..n)X_{i})
  3. C.((n+1)/(n+3))(∑(i=1..n)Xi)((n+1)/(n+3))^(\sum(i=1..n)X_{i})
  4. D.((n+1)/(n+3))e(−(2/n)(∑(i=1..n)Xi+1))((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, 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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