Data's weight against a Normal prior's pretend readings

Question

You estimate a true value with a Normal prior whose precision equals that of 4 pretend readings (each reading having the same noise as your real measurements). You then take 6 real readings. The posterior mean is a weighted average of the prior mean and the sample average.

What fraction of the posterior-mean estimate is contributed by the real readings (rather than the prior)?

Accepted Answer

Intuition first. For a Normal prior meeting Normal data of the same per-reading noise, the prior acts like a stack of pretend readings. Here it is worth 4 pretend readings. Your real data adds 6 more. The posterior mean is a precision-weighted tug-of-war, and with equal per-reading noise the pull is just each side's share of the reading count. The rule. Pretend readings = n0 and real readings = n, so: 	ext{data weight} = \frac{n}{n0 + n}. Plug in. Here n0 = 4 and n = 6: 	ext{data weight} = \frac{6}{4 + 6} = \frac{6}{10} = 0.6. What the number means. Sixty percent of the estimate comes from …

Data's weight against a Normal prior's pretend readings

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