Bayesian weight estimate from repeated readings

Question

You estimate the true weight \mu of an object in grams. Your prior is Normal with mean 500 and variance 100 (standard deviation 10). You place it on a scale 3 times and get readings averaging 530 grams, where each reading has known noise variance 300. Both the prior and the reading noise are Normal. Using Bayesian updating, what is the posterior mean estimate of the true weight \mu?

Accepted Answer

Intuition first. Normal prior plus Normal data gives a Normal answer (conjugate). The updated mean is a precision-weighted tug-of-war. With several readings, the data's precision grows: n readings each of variance \sigma^2 behave like one average of variance \sigma^2/n, so their combined precision is n/\sigma^2. The rule. With prior mean \mu0, prior variance 	au0^2, and n readings averaging \bar x with per-reading variance \sigma^2: 	ext{posterior mean} = \frac{\dfrac{\mu0}{	au0^2} + \dfrac{n\,\bar x}{\sigma^2}}{\dfrac{1}{	au0^2} + \dfrac{n}{\sigma^2}}, a precision-weighted average of the…

Bayesian weight estimate from repeated readings

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