Past exam of the statistics course of the University of Oxford / 2015 / 3 / b / i

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- Solution i

Solution

Take

σ, τ > 0

to be the positive standard deviations. Multiply the normal likelihood function by the normal prior density. Terms independent of

θ

can be absorbed into the normalizing constant:

π (θ ∣ x) \propto exp [- \frac{1}{2 σ ^{2}} \sum_{i} (x_{i} - θ)^{2} - \frac{θ ^{2}}{2 τ ^{2}}] .

The identity

\sum_{i} (x_{i} - θ)^{2} = \sum_{i} (x_{i} - \overline{x})^{2} + n (θ - \overline{x})^{2}

makes the quadratic term explicit. Completing the square gives

π (θ ∣ x) \propto exp [- \frac{( θ - μ _{n} ) ^{2}}{2 v _{n}}],

where

v_{n} = (\frac{n}{σ ^{2}} + \frac{1}{τ ^{2}})^{- 1} = \frac{σ ^{2} τ ^{2}}{σ ^{2} + n τ ^{2}}, μ_{n} = \frac{n τ ^{2}}{σ ^{2} + n τ ^{2}} \overline{x} .

Thus

θ ∣ x \sim N (μ_{n}, v_{n})

. This is normal-normal conjugacy: prior and data precisions add, and the posterior mean shrinks the sample mean towards the prior mean zero.

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