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
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Consider the pdf
$f(x;\theta,\sigma) = \frac{0.9}{\sigma}\phi\left(\frac{x-\theta}{\sigma}\right) + 0.1\ \phi(x-\theta)$, where $-\infty < \theta < \infty$ and $\sigma > 0$ are unknown parameters and $\phi$ denotes the pdf of $N(0,1)$. Let $X_1, X_2, \cdots, X_n$ be a random sample from this probability distribution. Then which of the following is (are) correct?Suppose $X$ has density $f(x|\theta)$ where $\theta$ is $0$ or $1$. Also,
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