Understanding the Problem
We are given four independent and identically distributed (i.i.d.) random variables $X_1, X_2, X_3, X_4$, each following a normal distribution with mean $\mu$ and variance 2. The prior distribution for the mean $\mu$ is also normal, with mean 0 and variance $\frac{1}{2}$. We need to determine the correct statement regarding the posterior distribution of $\mu$ after observing the data.
Conjugacy Check
The likelihood function is based on a normal distribution with known variance ($\sigma^2=2$). The prior distribution for the mean ($\mu$) is a normal distribution ($N(0, \frac{1}{2})$). A normal prior is a conjugate prior for the mean parameter of a normal distribution with known variance. Therefore, the posterior distribution will also be a normal distribution.
- Option 1 states the prior is not conjugate. This is false because the normal prior is conjugate to the normal likelihood (with known variance).
Calculating Posterior Parameters
Let the sample data be $X_1, X_2, X_3, X_4$. The sample size is $n=4$. The parameters are:
- Likelihood variance: $\sigma^2 = 2$
- Prior mean: $\mu_0 = 0$
- Prior variance: $\sigma_0^2 = \frac{1}{2}$
- Sample mean: $\bar{X} = \frac{1}{n}\sum_{i=1}^4 X_i = \frac{\sum_{i=1}^4 X_i}{4}$
The posterior distribution for $\mu$ is normal, $N(\mu_{post}, \sigma_{post}^2)$. The formulas for the posterior mean and variance are:
$ \mu_{post} = \frac{\sigma^2 \mu_0 + n \sigma_0^2 \bar{X}}{\sigma^2 + n \sigma_0^2} $ $ \sigma_{post}^2 = \frac{\sigma^2 \sigma_0^2}{\sigma^2 + n \sigma_0^2} $Calculation of Posterior Mean/Mode:
$ \mu_{post} = \frac{(2)(0) + (4)(\frac{1}{2})(\bar{X})}{2 + (4)(\frac{1}{2})} = \frac{0 + 2\bar{X}}{2 + 2} = \frac{2\bar{X}}{4} = \frac{1}{2}\bar{X} $Substituting $\bar{X} = \frac{\sum_{i=1}^4 X_i}{4}$:
$ \mu_{post} = \frac{1}{2} \left( \frac{\sum_{i=1}^4 X_i}{4} \right) = \frac{\sum_{i=1}^4 X_i}{8} $Since the posterior distribution is normal, its mean, median, and mode are equal. Therefore, the posterior mode is $\frac{\sum_{i=1}^4 X_i}{8}$.
Calculation of Posterior Variance:
$ \sigma_{post}^2 = \frac{(2)(\frac{1}{2})}{2 + (4)(\frac{1}{2})} = \frac{1}{2 + 2} = \frac{1}{4} $The posterior variance is $\frac{1}{4}$.
Evaluating the Options
- Option 2: Posterior mode of $\mu$ given $X_1, X_2, X_3$ and $X_4$ is $\frac{\sum_{i=1}^4 X_i}{8}$. Our calculation shows the posterior mode is $\frac{\sum_{i=1}^4 X_i}{8}$. This statement is true.
- Option 3: Posterior median of $\mu$ given $X_1, X_2, X_3$ and $X_4$ is $\frac{\sum_{i=1}^4 X_i}{4}$. The posterior median equals the posterior mean, which is $\frac{\sum_{i=1}^4 X_i}{8}$, not $\frac{\sum_{i=1}^4 X_i}{4}$. This statement is false.
- Option 4: Posterior variance of $\mu$ given $X_1, X_2, X_3$ and $X_4$ is $\left(\frac{\sum_{i=1}^4 X_i}{4}\right)^2$. The posterior variance is $\frac{1}{4}$, not $\left(\frac{\sum_{i=1}^4 X_i}{4}\right)^2$. This statement is false.
Based on the calculations, Option 2 is the only true statement.
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 $X_1, X_2, . . ., X_n$ 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?
- 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
