Posterior variance from two sharp readings

Question

You estimate an unknown quantity \mu with a Normal prior of variance 12. You take 2 independent readings, each with known noise variance 8. Both the prior and the readings are Normal. What is the variance of the posterior distribution for \mu?

Accepted Answer

Intuition first. Convert uncertainty into precision, "one over variance," since Bayesian updating adds precisions. The readings are sharper than the prior (variance 8 versus 12), and two of them together bring most of the information. The rule. With prior variance 	au0^2 and n readings each of variance \sigma^2: 	ext{posterior precision} = \frac{1}{	au0^2} + \frac{n}{\sigma^2}, \qquad 	ext{posterior variance} = \frac{1}{	ext{posterior precision}}. Plug in. Prior precision = 1/12. Data precision = 2/8 = 1/4 = 3/12. Total precision: \frac{1}{12} + \frac{2}{8} = \frac{1}{12} + \frac{3}{12} …

Posterior variance from two sharp readings

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