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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2019

Paper 3

Put
$$
d_\pm
=\frac{\log(S_0/K)+(r\pm\sigma^2/2)T}
{\sigma\sqrt T}.
$$
The call payoff is $(S_T-K)^+$, and it is positive when
$$
y\geq
\frac{\log(K/S_0)-(r-\sigma^2/2)T}{\sigma\sqrt T}
=-d_-.
$$
Writing $\phi$ for the <standard normal density>, the given <risk-neutral pricing> formula becomes
$$
\begin{aligned}
C_0
&=e^{-rT}\int_{-d_-}^{\infty}
\left[S_0e^{\sigma\sqrt T y+(r-\sigma^2/2)T}-K\right]
\phi(y)\,dy.
\end{aligned}
$$
Completing the square gives
$$
e^{\sigma\sqrt T y}\phi(y)
=e^{\sigma^2T/2}\phi(y-\sigma\sqrt T),
$$
so the two tail integrals are $\Phi(d_+)$ and $\Phi(d_-)$, where $\Phi$ is the <standard normal distribution function>. Therefore the <Black-Scholes formula> is
$$
\boxed{C_0=S_0\Phi(d_+)-Ke^{-rT}\Phi(d_-).}
$$

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 $K$. Hence <put-call parity> gives
$$
C_0-P_0=S_0-Ke^{-rT},
$$
and therefore
$$
\boxed{
P_0=Ke^{-rT}\Phi(-d_-)-S_0\Phi(-d_+).}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 3 / 29K / a / Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 3 / 29K / a / Solution

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Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 3 / 29K / a / Solution

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