Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 48 / 2

Substitute the nonzero product $V=f(S)g(t)$ into the Black-Scholes equation with continuous stock dividends and divide by $fg$ on intervals where it is nonzero. The result isThe left side depends only on time and the right side only on spot, so both equal a constant $\lambda$. ThusThese are the separated solutions of the Black-Scholes equation. The zero solution is included separately, so division loses no relevant case.

For the unheaded parity request, the terminal difference of the power-call and power-put payoffs is $S_T^{\beta }-K^{\beta }$. Under the risk-neutral measure, $S_T=S\exp [(r-q-{\sigma }^2/2)\tau +\sigma (W_T-W_t)]$, where $\tau =T-t$. The Gaussian exponential moment gives ${\mathbb{E}}_Q[S_T^{\beta }\mid S_t=S]=S^{\beta }{e}^{{\lambda }_{\beta }\tau }$ with ${\lambda }_{\beta }=\beta (r-q)+{\sigma }^2\beta (\beta -1)/2$. Discounting the difference proves power put-call parity:For $\beta =1$ this is the familiar dividend-adjusted put-call parity, $S{e}^{-q\tau }-K{e}^{-r\tau }$.

If the value is independent of spot, both spatial derivatives vanish. The Black-Scholes equation reduces to $V_t-rV=0$, so $\boxed{V(S,t)=A{e}^{r(t-T)}}$, where $A$ is an arbitrary constant. A constant terminal payoff is therefore discounted at the interest rate $r$; its separation constant is $\lambda =0$.

For a time-independent value, $\lambda =r$. This is an Euler equation. Put $f=S^{\alpha }$ to obtainFor $\sigma >0$, let $b_0=r-q-{\sigma }^2/2$. The roots areThey are distinct and real because $r>0$, with one positive and one negative. The general stationary solution on $S>0$ is $\boxed{V(S)=A{S}^{{\alpha }_{+}}+B{S}^{{\alpha }_{-}}}$. There are no boundary restrictions in this subpart to eliminate either coefficient. In the degenerate case $\sigma =0$, the equation is first order: for $r\ne q$ its solutions are $A{S}^{r/(r-q)}$, while for $r=q$ only the zero stationary solution remains.

For the linear spatial factor, $f^{\prime \prime }=0$ and $S{f}^{\prime }=f$. Hence $\lambda =r-q$ and $g^{\prime }=qg$. Absorbing the constant $k$ into the time-factor amplitude gives $\boxed{V(S,t)=AS{e}^{q(t-T)}}$. In particular a terminal claim equal to $A{S}_T$ is worth $AS{e}^{-q(T-t)}$, as expected for an asset paying a continuous dividend yield.

For $f=k{S}^{\alpha }$, $S{f}^{\prime }=\alpha f$ and $S^2{f}^{\prime \prime }=\alpha (\alpha -1)f$. Thus $\lambda ={\lambda }_{\alpha }={\sigma }^2\alpha (\alpha -1)/2+(r-q)\alpha$, and all values with this fixed spatial factor areThis Power payoff in the Black-Scholes model calculation remains valid for any real $\alpha$ on $S>0$; the specified nonzero exponent excludes the constant-payoff subpart. It also supplies the moment term in power put-call parity above.