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Tightening a normal posterior

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An unknown mean has a normal prior with variance $8$ . Each observation adds independent normal noise of variance $16$ , known in advance. How many observations bring the posterior variance down to exactly $1$ ? Give a whole number.

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Hint

Work in precisions, which add.

Solution

The prior precision is $\frac{1}{8}$ , and each observation adds $\frac{1}{16}$ .
The target precision is $1$ , so $\frac{1}{8} + \frac{n}{16} = 1$ .
That gives $(1 - \frac{1}{8}) \times 16 = 14$ observations.

Answer: $14$

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