Posterior variance from one equally noisy reading

Question

You estimate an unknown quantity \mu with a Normal prior of variance 9. You take a single reading with known noise variance 9. Both the prior and the reading are Normal. What is the variance of the posterior distribution for \mu?

Accepted Answer

Intuition first. Flip uncertainty into precision, "one over variance," because precisions add under Bayesian updating. The prior and the one reading here carry the same precision, so combining them should roughly halve the variance. The rule. With prior variance 	au0^2 and one reading of variance \sigma^2: 	ext{posterior precision} = \frac{1}{	au0^2} + \frac{1}{\sigma^2}, \qquad 	ext{posterior variance} = \frac{1}{	ext{posterior precision}}. Plug in. Prior precision = 1/9. Reading precision = 1/9. Total precision: \frac{1}{9} + \frac{1}{9} = \frac{2}{9}. Posterior variance is the recipro…

Posterior variance from one equally noisy reading

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