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
- 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 $X_2$?
