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Let X be a normal random variable with mean $\mu$ and variance $\sigma^2$. Let us further assume that $\mu \sim \mathcal N(\mu_p, \sigma_p^2)$ and its prior distribution is $\pi(\mu)$. The variables $\mu_p$ and $\sigma_p^2$ are assumed to be either known or estimated. Find the marginal distribution of the given random variable. I am struggling with this question, any help would be greatly appreciated. Even just a hint on how to start this. Thanks in advance.

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HINT:

The marginal distribution of $X$ is given by :

$m(x)=\int \pi(x|\mu)\pi(\mu) \mathrm{d}\mu\,, $ where $\pi(\mu)$ is the prior distribution and $\pi(x|\mu)$ is the sampling distribution.

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    This answer stops short of actually answering the question that was asked, and fails to exploit some simple well-known properties of the normal distribution.2011-07-17
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The conditional distribution of $X$ given $\mu$ is $N(\mu,\sigma^2)$. So the conditional distribution of $X-\mu$ given $\mu$ is $N(0,\sigma^2)$, and no $\mu$ appears in that last expression: $N(0,\sigma^2)$. Hence $X-\mu$ is independent of $\mu$! So $X$ is the sum of two independent random variables: $X-\mu$ which is distributed as $N(0,\sigma^2)$, and $\mu$, which is distributed as $N(\mu_p,\sigma_p)$. Hence $X \sim N(\mu_p,\sigma^2+\sigma_p^2)$. That's the marginal distribution of $X$.