 Solution

ID: past-exam-of-the-probability-statistics-and-options-i-course-of-the-university-of-oxford/2008/s/1/d/solution

For the shifted normal proposal,

w (x) = e^{- μx + μ^{2} /2}

and the estimator is

\overset{p}{^}_{N} = \frac{1}{N} \sum_{i} 1_{{X_{i} > z}} e^{- μ X_{i} + μ^{2} /2}, X_{i} \sim N (μ, 1) .

Let

\overset{ˉ}{Φ} = 1 - Φ

I (μ) = \int_{z}^{\infty} \frac{p ( x ) ^{2}}{q _{μ} ( x )} d x = e^{μ^{2}} \overset{ˉ}{Φ} (z + μ) .

[I (μ) - \overset{ˉ}{Φ} (z)^{2}] / N

, so only

I

needs minimization. It tends to infinity at both ends of the real

μ

axis and has a finite minimizer. Differentiation at a minimizer yields

2 μ \overset{ˉ}{Φ} (z + μ) = φ (z + μ) .

For a standard normal variable, integration of

x φ (x) = - φ^{'} (x)

shows

E_{p} [X ∣ X > a] = φ (a) / \overset{ˉ}{Φ} (a)

. Consequently

2 μ = E_{p} [X ∣ X > z + μ] .

The shift is chosen to reduce the second moment of the weighted rare-event indicator, rather than merely to increase the proposal's tail probability.

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