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 $X_2$?
$X_1$ and $X_2$ are i.i.d $N(\mu, \tau^2 + \sigma^2)$.
To solve this question, we need to determine the marginal distribution of $X_1$ and $X_2$ given the provided information. Let's break down the problem step-by-step:
Step 1: Understanding the Distribution Relationships
- We know that $X_i \mid \theta_i \sim N(\theta_i, \sigma^2)$ for $i = 1, 2$. This means that given $\theta_i$, $X_i$ follows a normal distribution with mean $\theta_i$ and variance $\sigma^2$.
- The prior distribution given is $\theta_i \sim N(\mu, \tau^2)$, where $\theta_1$ and $\theta_2$ are independent and identically distributed (i.i.d).
Step 2: Deriving the Marginal Distribution of $X_i$
- According to the problem, $\theta_1$ and $\theta_2$ are drawn from $N(\mu, \tau^2)$. Therefore, the overall distribution of $X_i$ can be obtained by integrating over all possible values of $\theta_i$.
- This means we need to find the marginal distribution by considering both the distribution of $\theta_i$ and the conditional distribution of $X_i$ given $\theta_i$.
Step 3: Calculation using properties of Normal Distributions
- Since $X_i \mid \theta_i \sim N(\theta_i, \sigma^2)$ and $\theta_i \sim N(\mu, \tau^2)$, the marginal distribution of $X_i$ is obtained as follows:
- When you integrate out $\theta_i$, $X_i$ follows a normal distribution with mean $\mu$ and variance $\tau^2 + \sigma^2$.
- This is because the variance of a normal distribution $N(a, b^2)$ when summed with another normal distribution $N(c, d^2)$ has a combined variance of $b^2 + d^2$.
Step 4: Conclusion
- The marginal distributions of $X_1$ and $X_2$ are therefore $N(\mu, \tau^2 + \sigma^2)$.
- Since $\theta_i$ were independent and identically distributed, and given that $X_i$ are independent based on $\theta_i$, $X_1$ and $X_2$ are indeed i.i.d.
Correct Option: $X_1$ and $X_2$ are i.i.d $N(\mu, \tau^2 + \sigma^2)$.
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?
Let $(X,Y)$ have the joint discrete distribution such that $X \mid Y=y \sim \text{Binomial} \quad (y, 0.5)$ and $Y \sim \text{Poisson}(\lambda), \lambda > 0$, where $\lambda$ is an unknown parameter. Let $T = T(X,Y)$ be any unbiased estimator of $\lambda$. Then
