 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-3/29k/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

Put

d_{\pm} = \frac{l o g ( S _{0} / K ) + ( r \pm σ ^{2} /2 ) T}{σ T} .

The call payoff is

(S_{T} - K)^{+}

, and it is positive when

y \geq \frac{l o g ( K / S _{0} ) - ( r - σ ^{2} /2 ) T}{σ T} = - d_{-} .

ϕ

C_{0} = e^{- r T} \int_{- d_{-}}^{\infty} [S_{0} e^{σ T y + (r - σ^{2} /2) T} - K] ϕ (y) d y .

e^{σ T y} ϕ (y) = e^{σ^{2} T /2} ϕ (y - σ T),

so the two tail integrals are

Φ (d_{+})

and

Φ (d_{-})

, where

Φ

C_{0} = S_{0} Φ (d_{+}) - K e^{- r T} Φ (d_{-}) .

The payoff identity

(S_{T} - K)^{+} - (K - S_{T})^{+} = S_{T} - K

replicates the call-minus-put position by one stock and borrowing the present value of

K

C_{0} - P_{0} = S_{0} - K e^{- r T},

and therefore

P_{0} = K e^{- r T} Φ (- d_{-}) - S_{0} Φ (- d_{+}) .

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