\, \lambda \sim \textsf{Exp}(\alpha ) \, for some \, \alpha >0 \, and conditional on \, \lambda \,, \, X_1,\ldots ,X_ n\stackrel{i.i.d.}{\sim } \textsf{Exp}(\lambda ) \,.

Question

What is the Bayesian estimator \hat{\lambda }^{\text {Bayes}}?

(If applicable, enter barX_n for \bar{X_ n},\, max_n for \displaystyle \text {max}_{i=1\ldots n} X_ i. )

\hat{\lambda }^{\text {Bayes}}=\quad
unanswered
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Determine whether the Bayesian estimator is consistent, and whether it is asymptotically normal.

Consistent and asymptotically normal

Consistent but not asymptotically normal

Asymptotically normal but not consistent

Neither consistent nor asymptotically normal
unanswered
If it is asymptotically normal, what is its asymptotic variance V(\lambda )? If it is not asymptotically normal, type in \, 0 \,. You may use the variable \lambda.

V(\lambda )=\quad

Answer 1

Since we have a conjugate prior, the exponential distribution, for the likelihood, the posterior distribution of λ is also an exponential distribution with parameter α + nλ and the Bayesian estimator of λ can be found by taking the expected value of the posterior distribution.

Thus, the Bayesian estimator \hat{λ}^{Bayes} is given by:

\hat{λ}^{Bayes} = E(λ | X_1, ..., X_n) = \frac{α + n \bar{X}_n}{n + 1}

where \bar{X}_n is the sample mean.

The Bayesian estimator is consistent, as it converges to the true value of λ as the sample size increases. However, it is not asymptotically normal, as it does not follow a normal distribution as the sample size increases.

Therefore, the correct answer is:

Consistent but not asymptotically normal

The asymptotic variance V(λ) is not applicable in this case as the Bayesian estimator is not asymptotically normal. Thus, V(λ) = 0.