 Solution

ID: past-exam-of-the-probability-statistics-and-options-ii-course-of-the-university-of-oxford/2008/p/3/c/solution

Write

q (x) = a + b x

x^{2} - q

1, x

gives

(lo g 3 2 24) (b a) = (26/3 4) .

The entries are obtained by integrating

1/ x, 1, x, x^{2}

over the given interval. Subtracting the second equation divided by two from the first gives

(lo g 3 - 1) a = - 1/3

. Thus

q (x) = - \frac{1}{3 ( lo g 3 - 1 )} + (\frac{13}{6} + \frac{1}{6 ( lo g 3 - 1 )}) x .

The Gram matrix is positive definite, with determinant

4 (lo g 3 - 1) > 0

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