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Given a model where $ x_i | \mu \sim \mathcal{N} ( \mu, \sigma^2 ) $ where $ \mu \sim \mathcal{N} ( \mu_0, \sigma_0^2 ) $, is there a closed form formula for the PDF of $ x_i $? Namely, what's $ p (x_i) $?

I know the solution by Bayes, but I wonder if there is a closed form solution. My intuition is a Normal distribution with updated mean and variance according to the prior.

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2 Answers 2

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One way to model this would be by a sum of 2 variables:

$$ {x}_{i} = {y}_{i} + {z}_{i}, \quad {y}_{i} \sim \mathcal{N} \left( 0, {\sigma}_{2}^{2} \right), \; {z}_{i} \sim \mathcal{N} \left( {\mu}_{0}, {\sigma}_{0}^{2} \right) $$

Since $ {z}_{i} \perp {y}_{i} $ then the variance of $ {x}_{i} $ is the sum of variances.
Hence $ {x}_{i} \sim \mathcal{N} \left( {\mu}_{0}, {\sigma}^{2} + {\sigma}_{0}^{2} \right) $.

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  • $\begingroup$ In the first normal variance, the subindex 2 seems superfluous? $\endgroup$ Commented Dec 31, 2023 at 17:43
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Do it with moment-generating functions and iterated expectations. We have:

$$ \begin{align*} M_X(t) & = \mathbb{E}[\exp(tX)] = \mathbb{E}[\mathbb{E}(\exp(tX)|\mu)] \\ & = \mathbb{E}[\exp(\mu t + \sigma^2 t^2 / 2 )] = (\sigma^2 t^2 / 2) \, \mathbb{E}[\exp(\mu t)] \\ & =(\sigma^2 t^2 / 2)\, \exp(\mu_0 t + \sigma_0^2 t^2 / 2 ) \\ & = \exp\{\mu_0 t + (\sigma^2+\sigma_0^2) t^2 / 2 \}, \end{align*} $$

which is the MGF of a $\mathcal{N}(\mu_0, \sigma^2 +\sigma_0^2)$.

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  • $\begingroup$ This is nice! I had a different way in mind. +1. $\endgroup$
    – Royi
    Commented Apr 22, 2022 at 16:10
  • $\begingroup$ @Royi well it is essentially the proof that you can add Gaussian variables and get another Gaussian variable $\endgroup$
    – qwr
    Commented Apr 22, 2022 at 21:05

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