To find the Bayesian estimator, we need to find the posterior distribution and then calculate its mean.
Given that \, p \, follows a \, \textsf{Beta}(a,b) \, distribution and \, X_1,\ldots ,X_ n\stackrel{i.i.d.}{\sim } \textsf{Ber}(p), the likelihood function is given by:
\, \displaystyle L(p;X_1,\ldots ,X_ n) = \prod _{i=1}^ n p^{X_ i}(1-p)^{1-X_ i}
The prior distribution is given by \, \textsf{Beta}(a,b).
The posterior distribution is given by:
\, \displaystyle p|X_1,\ldots ,X_ n \sim \textsf{Beta}(a+\sum _{i=1}^ n X_ i, b+n-\sum _{i=1}^ n X_ i)
The mean of the posterior distribution (Bayesian estimator) is:
\, \displaystyle \hat{p}^{\text {Bayes}} = \frac{a+\sum _{i=1}^ n X_ i}{a+b+n}
Now we need to determine if the Bayesian estimator is consistent and/or asymptotically normal.
To show consistency, we need to show that the Bayesian estimator converges to the true value of \, p \, as \, n \, increases.
To show asymptotic normality, we need to show that the Bayesian estimator follows a normal distribution as \, n \, increases, with mean equal to the true value of \, p \, and variance equal to some finite value.
Unfortunately, we do not have enough information on the values of \, a \, and \, b \, to determine if the Bayesian estimator is consistent and/or asymptotically normal. Therefore, we cannot compute the asymptotic variance. The answer is "unanswered".
On this page, you will be given a distribution and another distribution conditional on the first one. Then, you will find the posterior distribution in a Bayesian approach. You will compute the Bayesian estimator, which is defined in lecture as the mean of the posterior distribution. Then, determine if the Bayesian estimator is consistent and/or asymptotically normal.
We recall that the Gamma distribution with parameters \, q>0 \, and \, \lambda >0 \, is the continuous distribution on \, (0,\infty ) \, whose density is given by \, \displaystyle f(x)=\frac{\lambda ^ q x^{q-1}e^{-\lambda x}}{\Gamma (q)} \,, where \, \Gamma \, is the Euler Gamma function \, \Gamma (q)=\int _0^\infty t^{q-1} e^{-t} dt \,, and its mean is \, q/\lambda \,.
We also recall that the \, \textsf{Beta}(a,b) \, distribution has the density \, \displaystyle f(x) = \frac{x^{a-1}(1-x)^{b-1}}{B(a,b)} \, and expectation \, a/(a+b) \,, where \, \displaystyle B(a,b) = \frac{\Gamma (a) \Gamma (b)}{\Gamma (a+b)} \,.
\, p\sim \textsf{Beta}(a,b) \, for some \, a,b>0 \, and conditional on \, p \,, \, X_1,\ldots ,X_ n\stackrel{i.i.d.}{\sim } \textsf{Ber}(p) \,.
What is the Bayesian estimator \hat{p}^{\text {Bayes}}?
(If applicable, enter barX_n for \bar{X_ n},\, max_n for \displaystyle \text {max}_{i=1\ldots n} X_ i. )
\hat{p}^{\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(a,b,p)? If it is not asymptotically normal, type in \, 0 \,.
V(a,b,p)=\quad
1 answer
1 answer