Posterior variance from ten noisier readings

Question

You estimate an unknown quantity \mu with a Normal prior of variance 5. You take 10 independent readings, each with known noise variance 50. Both the prior and the readings are Normal. What is the variance of the posterior distribution for \mu?

Accepted Answer

Intuition first. Turn uncertainty into precision, "one over variance," since precisions add. Each single reading is noisier than the prior, but ten of them together still add real information, so the posterior variance ends up below the prior's. 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 = 1/5 = 0.2. Data precision = 10/50 = 0.2. Total precision: \frac{1}{5} + \frac{10}{50} = \frac{1}{5} + \f…

Posterior variance from ten noisier readings

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