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Slippage at a new venue

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True daily slippage across venues is normal with mean $6$ and variance $9$ . On any day a venue records its true slippage plus independent normal noise of variance $45$ . A new venue averages $10$ over $5$ days. What is the posterior mean of its true slippage? Give a number.

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Hint

Compare the noise of the average with the spread between venues.

Solution

The noise of the average is $\frac{45}{5} = 9$ , the same as the spread $9$ .
The weight on the population mean is $\frac{9}{9 + 9} = \frac{1}{2}$ .
The posterior mean is $\frac{1}{2} \times 10 + \frac{1}{2} \times 6 = 8$ .

Answer: $8$

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