Bayes Estimator Properties Under Different Loss Functions
The Bayes estimator is a statistical decision rule that minimizes the expected loss based on the posterior distribution of a parameter. The specific form of the Bayes estimator depends on the chosen loss function and the characteristics of the posterior distribution.
Bayes Estimator under Squared Error Loss (SEL)
For a parameter $\theta$, the squared error loss function is defined as $L(\theta, \delta) = (\theta - \delta)^2$, where $\delta$ is the estimator. The Bayes estimator under SEL is the value $\delta(\mathbf{x})$ that minimizes the posterior expectation of the loss function, $E[L(\theta, \delta(\mathbf{x})) | \mathbf{x}] = E[(\theta - \delta(\mathbf{x}))^2 | \mathbf{x}]$. This minimum occurs when the estimator equals the posterior mean: $ \delta_{SEL}(\mathbf{x}) = E[\theta | \mathbf{x}] $
In this problem, the data $X_1, \dots, X_n$ are independent and identically distributed Normal random variables with mean $\theta$ and known variance $\sigma^2$, and the prior distribution for $\theta$ is Normal with mean $\mu$ and variance $\tau^2$. A conjugate prior-Normal distribution with a Normal likelihood results in a Normal posterior distribution for $\theta$. For any Normal distribution, the mean is equal to the median.
- Statement 1 (A) is correct because the Bayes estimator under SEL is the posterior mean, which exists and is derived from the Normal posterior distribution.
- Statement 2 (B) is also correct because, for the Normal posterior distribution, the posterior mean is equal to the posterior median. Therefore, the posterior median also serves as the Bayes estimator under SEL in this specific case.
Bayes Estimator under Absolute Error Loss (AEL)
The absolute error loss function is defined as $L(\theta, \delta) = |\theta - \delta|$. The Bayes estimator under AEL minimizes the posterior expectation of this loss, $E[L(\theta, \delta(\mathbf{x})) | \mathbf{x}] = E[|\theta - \delta(\mathbf{x})| | \mathbf{x}]$. This minimum occurs when the estimator equals the posterior median: $ \delta_{AEL}(\mathbf{x}) = \text{Median}(\theta | \mathbf{x}) $
As established, the posterior distribution for $\theta$ is Normal. For a Normal distribution, the mean and the median are identical.
- Statement 3 (C) is correct because the Bayes estimator under AEL is the posterior median, which exists for the Normal posterior distribution.
- Statement 4 (D) is also correct because, given the Normal posterior, the posterior median is equal to the posterior mean. Hence, the posterior mean also serves as the Bayes estimator under AEL in this scenario.
Conclusion
All four statements correctly describe properties of the Bayes estimator in the context of a Normal posterior distribution derived from Normal data and a Normal prior, considering both squared error loss and absolute error loss functions. The equality of the mean and median for the Normal posterior makes both measures valid estimators under the respective loss functions.
Important Questions from Elementary Bayesian Inference
- Let $X|\theta \sim \text{Uniform}[0, \theta]$ and $\theta$ has an Exponential distribution with mean $\lambda$, where $\lambda > 3$ is known. If the realized value of $X$ is $2025$, then the posterior mode equals
- Suppose $X|\theta \sim \text{Binomial}(7,\theta)$, $0 < \theta < 1$, and the prior distribution of $\theta$ is $\text{Beta}(\alpha, \beta)$ where $\alpha > 0$ and $\beta > 0$ are known. Then which of the following statements MAY NOT be true?
Let $X$ be a random sample from an exponential distribution with mean $1/\lambda$. If $\lambda$ has a prior distribution with probability density function
$g(\lambda) = \begin{cases} \lambda e^{-\lambda} & ; \quad \lambda > 0 \\ 0 & ; \quad \lambda \leq 0 \end{cases}$
then the Bayes estimator of $1/\lambda$ with respect to the squared error loss function is
Let $X_1, X_2, \dots, X_7$ be a random sample from $N(\mu, \sigma^2)$ where $\mu$ and $\sigma^2$ are unknown. Consider the problem of testing $H_0: \mu = 2$ against $H_1: \mu > 2$. Suppose the observed values of $x_1, x_2, \dots, x_7$ are $1.2, 1.3, 1.7, 1.8, 2.1, 2.3, 2.7$. If we use the Uniformly Most Powerful test, which of the following is true?
Suppose $X_i \mid \theta_i \sim N(\theta_i, \sigma^2), i = 1, 2$ are independently distributed. Under the prior distribution, $\theta_1$ and $\theta_2$ are i.i.d $N(\mu, \tau^2)$, where $\sigma^2, \mu$ and $\tau^2$ are known. Then which of the following is true about the marginal distributions of $X_1$ and $X_2$?
