Question
Let $X_1, X_2, . . ., X_n$ be a random sample from $N(\theta, 1)$, $\theta \in R$. If $\hat{\theta}$ is the Bayes estimator of $\theta$ with respect to some prior $\pi(\theta)$ and loss function $L(\theta, d)$. Then, which of the following statements are true?
To solve this problem, we need to determine which of the provided statements are correct concerning the Bayes estimator of the parameter \(\theta\) from a normal distribution sample \(N(\theta, 1)\). Let's analyze each statement one by one.
- Statement 1: \(\hat{\theta} = \frac{\sum_{i=1}^n X_i}{n + \tau^2}\), if the prior is \(N(0, \frac{1}{\tau^2})\), \(\tau^2\) known and \(L(\theta, d) = (\theta – d)^2\)
- This statement refers to the Bayes estimator under the squared error loss function. When we have a normal prior \(N(0, \frac{1}{\tau^2})\) and a sample from \(N(\theta, 1)\), the Bayes estimator is given by a weighted average of the sample mean and the prior mean. The formula is indeed correct because it considers the sample information and the prior information balanced by their variances.
- Thus, Statement 1 is correct.
- Statement 2: \(\hat{\theta} = \frac{\sum_{i=1}^n X_i}{n + \tau^2}\), if the prior is \(N(0, \tau^2)\), \(\tau^2\) known and \(L(\theta, d) = |\theta – d|\)
- This statement changes the loss function to an absolute deviation. However, the given estimator does not correspond to the minimizer of the expected absolute error loss in Bayesian analysis.
- Therefore, Statement 2 is incorrect.
- Statement 3: \(\hat{\theta} = \frac{\sum_{i=1}^n X_i}{\,n + \frac{1}{\tau^{2}}}\), if the prior is \(N(0, \frac{1}{\tau^2})\), \(\tau^2\) known and \(L(\theta, d) = |\theta – d|\)
- Similar to Statement 2, this also involves an absolute error loss function, which is not compatible with the stated estimator for this prior configuration. The correct form for an absolute error loss differs conceptually from the squared error loss.
- Thus, Statement 3 is also incorrect.
- Statement 4: \(\hat{\theta} =\frac{\sum_{i=1}^n X_i}{n}\), if the prior is the Jeffreys prior and \(L(\theta, d) = (\theta – d)^2\)
- Jeffreys prior is often used to represent non-informative priors. In this case, the Bayes estimator under the squared error loss function corresponds to the sample mean itself when the variance is known.
- Hence, Statement 4 is indeed accurate under these conditions.
Based on the above analysis, the correct statements are:
- \(\hat{\theta} = \frac{\sum_{i=1}^n X_i}{n + \tau^2}\), if the prior is \(N(0, \frac{1}{\tau^2})\), \(\tau^2\) known and \(L(\theta, d) = (\theta – d)^2\)
- \(\hat{\theta} = \frac{\sum_{i=1}^n X_i}{n}\), if the prior is the Jeffreys prior and \(L(\theta, d) = (\theta – d)^2\)
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Important Questions from Elementary Bayesian Inference
- Suppose the distribution of $X$ given $\theta$ is normal with mean $\theta$ and variance $15$. Further, let the prior (improper) distribution of $\theta$ be proportional to $1, \ -\infty<\theta<\infty$. If the observed value of $X$ is $13$, then which of the following statements is true?
- 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
- $X_1, X_2, \cdots, X_n$ are independent and identically distributed $N(\theta, 1)$ random variables, where $\theta$ takes only integer values i.e.
$\theta \in \{\cdots, -2, -1, 0, 1, 2, \cdots\}$.
Which of the following is the maximum likelihood estimator of $\theta$? - Suppose the probability mass function of a random variable X under the parameter $\theta = \theta_0$ and $\theta = \theta_1 (\ne \theta_0)$ are given by
x 0 1 2 3 $p_{\theta_0}(x)$ 0.01 0.04 0.5 0.45 $p_{\theta_1}(x)$ 0.02 0.08 0.4 0.5
Define a test $\phi$ such that $\phi(x) = 1$ if $x = 0, 1$, and $0$ if $x = 2, 3$.
For testing $H_0: \theta = \theta_0$ against $H_1: \theta = \theta_1$, the test $\phi$ is - $\theta$ is the probability of obtaining a head in the toss of a coin. The coin is tossed three times and we record
$Y = 1$ if all the three tosses result in heads
$Y = 2$ if all the three tosses result in tails
$Y = 3$ otherwise
If the prior density of $\theta$ is Beta $(\alpha, \beta)$, and $\hat{\theta}_i$ is the posterior mean of $\theta$ given $Y = i$, for $i = 1, 2$, then
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