 Solution

ID: past-exam-of-the-statistics-course-of-the-university-of-oxford/2015/3/b/ii/solution

The symmetric continuous prior distribution gives

Pr (θ < 0) = Pr (θ \geq 0) = 1/2

, so the prior odds are one. Therefore the Bayes factor equals the posterior odds. Standardizing the normal posterior distribution, put

z = \frac{μ _{n}}{v _{n}} = \frac{n τ x}{σ σ ^{2} + n τ ^{2}} .

Then

Pr (H_{0} ∣ x) = Φ (- z)

and

Pr (H_{1} ∣ x) = Φ (z)

. The endpoint

θ = 0

has zero posterior probability, so the strict/non-strict split causes no additional term. Hence

B = \frac{Φ ( - z )}{Φ ( z )} = \frac{1 - Φ ( z )}{Φ ( z )}, z = \frac{n τ x}{σ σ ^{2} + n τ ^{2}} .

In particular, positive values of the sample mean make this Bayes factor smaller, favouring the positive-mean hypothesis.

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