 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2016/iii/paper-211/2/d/solution

Codex  0 Created 2026-10-03 Updated 2026-10-06

T_{1}

E^{Q_{T_{1}}} P (T_{1}, T_{2}) = \frac{E ^{Q} [ D _{T_{1}} P ( T _{1} , T _{2} )]}{P ( 0 , T _{1} )} = \frac{P ( 0 , T _{2} )}{P ( 0 , T _{1} )} .

For the variance, put

b_{t}^{i} = \int_{t}^{T_{i}} σ (t, u) d u

and

d_{t} = b_{t}^{2} - b_{t}^{1} = \int_{T_{1}}^{T_{2}} σ (t, u) d u

, for

t \leq T_{1}

R_{t} = P (t, T_{2}) / P (t, T_{1})

has, by the Itô formula,

d lo g R_{t} = - \frac{1}{2} ((b_{t}^{2})^{2} - (b_{t}^{1})^{2}) d t - d_{t} d W_{t} .

Q_{T_{1}}

d W_{t} = d W_{t}^{T_{1}} - b_{t}^{1} d t

, so its drift becomes

- d_{t}^{2} /2

. Since

R_{T_{1}} = P (T_{1}, T_{2})

lo g P (T_{1}, T_{2}) = lo g \frac{P ( 0 , T _{2} )}{P ( 0 , T _{1} )} - \frac{1}{2} \int_{0}^{T_{1}} d_{t}^{2} d t - \int_{0}^{T_{1}} d_{t} d W_{t}^{T_{1}} .

The first two terms are deterministic; the Itô isometry supplies the variance of the last. The two requested answers are

E^{Q_{T_{1}}} P (T_{1}, T_{2}) = \frac{P ( 0 , T _{2} )}{P ( 0 , T _{1} )}, Var^{Q_{T_{1}}} lo g P (T_{1}, T_{2}) = \int_{0}^{T_{1}} (\int_{T_{1}}^{T_{2}} σ (t, u) d u)^{2} d t .

In fact the terminal zero-coupon bond price has a log-normal distribution under this forward measure, with the deterministic negative half-variance correction ensuring the displayed mean.

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