Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 35 / 3 / b

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- Solution b

Solution

Let

q_{0} = n_{0}

,

q_{1} = n_{0} / c^{2}

and

y = \overline{y}

. These are the two prior precision parameters. Completing the square in

n (β - y)^{2} + q_{i} β^{2}

gives the Normal-normal conjugacy update

β ∣ y, H_{i} \sim N (\frac{n}{n + q _{i}} y, \frac{1}{n + q _{i}}), i = 0, 1.

Thus the posterior mean is a precision-weighted average of the observed mean and the prior center zero. The posterior variance is the reciprocal of the total precision.

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