Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 39 / 1 / b

Solution

Put

n = T - t

A_{0} = 0, A_{j} = \sum_{k = 0}^{j - 1} β^{k} = {(1 - β^{j}) / (1 - β), j, β \neq = 1, β = 1.

t

gives

r_{t + k} = β^{k} r_{t} + \sum_{j = 1}^{k} β^{k - j} ξ_{t + j}

. Consequently

\sum_{k = 1}^{n} r_{t + k} = β A_{n} r_{t} + \sum_{j = 1}^{n} A_{n - j + 1} ξ_{t + j}, \frac{B _{t}}{B _{T}} = exp (- \sum_{k = 1}^{n} r_{t + k}) .

F_{t}

K

yields

P (t, T) = e^{- β A_{n} r_{t}} \prod_{j = 1}^{n} E [e^{- A_{n - j + 1} ξ_{1}}] = exp (- β A_{n} r_{t} + \sum_{j = 1}^{n} K (- A_{j})) .

Q (t, T) = - β A_{T - t}, R (t, T) = j = 1 \sum T - t K (- A_{j}) .

They vanish at maturity, giving

P (T, T) = 1

. Everywhere finiteness of

K

ensures that each zero-coupon bond price is positive and finite, including when

β

is zero, negative, or one.

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