Exercise 4.13 - Gaussian posterior credible interval

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Assume the prior distribution for an 1d normal distribution:

p (μ) = 𝒩 (μ | μ_{0}, σ_{0}^{2} = 9) .

And the likelihood is:

p (x) = 𝒩 (x | μ, σ^{2} = 4) .

Having observed $n$ variables, we want that the probability measure of $μ$ ’s posterior distribution is no less than 0.95 within an interval no longer than 1. The posterior for $μ$ is:

\begin{align} p (μ | D) \propto p (μ) \cdot p (D | μ) = & 𝒩 (μ | μ_{0}, σ_{0}^{2}) \cdot \prod_{i = 1}^{n} 𝒩 (x_{n} | μ, σ^{2}) \\ \propto & \exp {- \frac{1}{2 σ_{0}^{2}} {(μ - μ_{0})}^{2}} \cdot \prod_{i = 1}^{n} \exp {- \frac{1}{2 σ^{2}} {(x_{i} - μ)}^{2}} \\ = & \exp {(- \frac{1}{2 σ_{0}^{2}} - \frac{n}{2 σ^{2}}) μ^{2} + . . .}, \end{align}

where we have dropped the terms inrevelent with $μ$ . The posterior variance of $μ$ is determined by the coefficient of $μ^{2}$ in the exponential of the posterior distribution:

σ_{post}^{2} = \frac{σ_{0}^{2} σ^{2}}{σ^{2} + n σ_{0}^{2}} .

Since 0.95 of the probability mass for a normal distribution lies within $- 1.96 σ$ and $1.96 σ$ , we have:

n \geq 611 .

solour_lfq

2021-03-24 13:42

Exercise 4.13 - Gaussian posterior credible interval

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