Past exam of the mathematical models of financial derivatives course of the University of Oxford / 2021 / 2

A self-financing replicating portfolio for a terminal payoff $F(S_T)$ has the same terminal value on every relevant path, with all earlier rebalancing funded internally. In a complete market its initial cost is unique under no-arbitrage: otherwise buy the cheaper payoff and sell the more expensive identical payoff.

In the Black-Scholes model, the portfolio holds $V_s$ shares and bond value $V-S{V}_s$, where $V$ solves the Black-Scholes equation with terminal data $F$. Itô's formula verifies the self-financing condition. Equivalently,This relates claim replication, the Black-Scholes equation, and risk-neutral valuation.

Buy $1/F_0^T$ shares and take a short forward contract for $1/F_0^T$ shares at delivery price $F_0^T$. The forward has zero initial value, and the combined terminal payoff isTherefore the synthetic zero-coupon bond has initial priceIts implied continuously compounded maturity yield is $r_T=T^{-1}\log (F_0^T/S_0)>0$.

To replicate a constant-rate risk-free asset throughout the interval, the market must support dynamically coherent discounting, not just this one terminal payment. A sufficient condition is that for all $0\le t\le u\le T$ the fair forward prices satisfy $F_t^u=S_t{e}^{r(u-t)}$ for the same deterministic $r>0$, with the required stock/forward positions tradable and no intervening payments. Then $P(t,u)=S_t/F_t^u=e^{-r(u-t)}$, and a suitably scaled fixed-maturity zero-coupon bond grows deterministically as $e^{rt}$. A single initial forward quote does not establish these intermediate-time conditions; even a flat initial yield curve need not imply deterministic future rates.

Under the risk-neutral measure, writeComplete the square in the normal density after multiplying by the power of $S_T/s$. The truncated lognormal moment gives, for $K>0$,All real powers have finite moments in this lognormal distribution. If $K=0$, the indicator is identically one and $V_p=M_p$. The function $V_p$ is an undiscounted conditional moment; the option prices below additionally include the discount factor.

If $p>0$ and $K>0$, the payoff is positive exactly when $S_T>k:=K^{1/p}$. Splitting it into two truncated moments givesFor $p=0$, the payoff is the constant $(1-K{)}^{+}$, so its price is $e^{-r\tau }(1-K{)}^{+}$. For $K=0$ and $p>0$, its price is $e^{-r\tau }{s}^p{M}_p(\tau )$. These boundary cases avoid taking $K^{1/p}$ when $p=0$.

The payoff is $(S_0{S}_T-S_T^2)1_{\{S_T<S_0\}}$. Subtract the upper-tail truncated lognormal moments from the full moments in part c. With current spot $s=S_t$, the price iswhere $M_1=e^{r\tau }$ and $M_2=e^{(2r+{\sigma }^2)\tau }$. The threshold is the original $S_0$, not the current spot $s$. Since the conditional lognormal distribution has no atoms, the strict endpoint convention does not affect the price for $t<T$.

The logarithm of the geometric average isBy stochastic Fubini theorem, ${\int }_0^T{W}_{u}du={\int }_0^T(T-v)d{W}_v$, a centered Gaussian random variable with variance $T^3/3$. ThereforeThe Geometric Asian option is consequently a call on this lognormal random variable. Completing the square as in part c givesThe averaging changes both the log-mean and log-variance; replacing $S_T$ by an ordinary stock price at a shorter maturity without these adjustments would give the wrong formula.