To find the Bayesian estimator, we need to calculate the posterior distribution.
The prior distribution is given by \lambda \sim \textsf{Exp}(\alpha).
The likelihood function is given by f(x_1,\ldots,x_n|\lambda) = \prod_{i=1}^{n}\textsf{Exp}(x_i\lambda) = \lambda^ne^{-\lambda\sum_{i=1}^{n}x_i}.
Using Bayes' theorem, the posterior distribution is given by:
f(\lambda|x_1,\ldots,x_n) \propto f(x_1,\ldots,x_n|\lambda)f(\lambda)
= \lambda^ne^{-\lambda\sum_{i=1}^{n}x_i}\alpha e^{-\alpha\lambda}
= \lambda^{n+1-1}e^{-\lambda(\sum_{i=1}^{n}x_i+\alpha)}.
This is the kernel of a Gamma distribution with parameters n+1 and \sum_{i=1}^{n}x_i+\alpha.
Therefore, the posterior distribution is \lambda|x_1,\ldots,x_n \sim \textsf{Gamma}(n+1, \sum_{i=1}^{n}x_i+\alpha).
The Bayesian estimator, \hat{\lambda}^{\text{Bayes}}, is the mean of the posterior distribution.
So, \hat{\lambda}^{\text{Bayes}} = (n+1)/(\sum_{i=1}^{n}x_i+\alpha).
Now, let's determine if the Bayesian estimator is consistent and/or asymptotically normal.
To determine consistency, we need to check if \hat{\lambda}^{\text{Bayes}} converges to the true value of \lambda as n approaches infinity.
Let's consider the mean of the posterior distribution:
E[\hat{\lambda}^{\text{Bayes}}] = E[(n+1)/(\sum_{i=1}^{n}x_i+\alpha)]
= (n+1)/[E[\sum_{i=1}^{n}x_i]+\alpha].
Since each X_i \sim \textsf{Exp}(\lambda), we know that E[X_i] = 1/\lambda. Therefore, E[\sum_{i=1}^{n}x_i] = n/\lambda.
Substituting this into the expression above, we get:
E[\hat{\lambda}^{\text{Bayes}}] = (n+1)/(n/\lambda+\alpha)
= (n\lambda+\alpha\lambda)/(n+\alpha)
= \lambda(\frac{n}{n+\alpha}) + \frac{\alpha}{n+\alpha).
As n approaches infinity, \frac{n}{n+\alpha} approaches 1, and \frac{\alpha}{n+\alpha} approaches 0. Therefore, E[\hat{\lambda}^{\text{Bayes}}] converges to \lambda.
Therefore, the Bayesian estimator is consistent.
To determine if the Bayesian estimator is asymptotically normal, we need to compute its asymptotic variance.
The variance of the posterior distribution can be calculated using the property of the Gamma distribution:
Var(\lambda|x_1,\ldots,x_n) = \frac{(n+1)}{(\sum_{i=1}^{n}x_i+\alpha)^2}.
Therefore, the asymptotic variance of the Bayesian estimator is \frac{1}{(\sum_{i=1}^{n}x_i+\alpha)^2}.
Since the asymptotic variance is not equal to zero, we can conclude that the Bayesian estimator is asymptotically normal.
In summary:
The Bayesian estimator \hat{\lambda}^{\text{Bayes}} = (n+1)/(\sum_{i=1}^{n}x_i+\alpha).
The Bayesian estimator is consistent and asymptotically normal.
The asymptotic variance of the Bayesian estimator V(\lambda) = \frac{1}{(\sum_{i=1}^{n}x_i+\alpha)^2}.
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)} \,.
\, \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 ) \,.
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
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
incorrect
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
1 answer
1 answer