Posterior variance from four readings

Question

You estimate an unknown quantity \mu with a Normal prior of variance 100. You take 4 independent readings, each with known noise variance 100. 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," because Bayesian updating adds precisions. The prior is worth one unit of this reading's precision, and four readings add four more, so total precision becomes five units and the variance shrinks fivefold. 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 = 1/100 = 0.01. Data precision = 4/100 = 0.04. Total precision: \frac{1}…

Posterior variance from four readings

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