Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2012 / S9 / 3

Financial data show approximate, scale-dependent regularities rather than an exact universal scaling law. Heavy return tails can resemble a power law over an intermediate range, but the fitted exponent, cutoff and range depend on instrument, period and sampling. A straight segment on a log-log plot alone is not enough to establish a power law.

For independent finite-variance increments, variance grows linearly with horizon and typical fluctuations scale as $\Delta t^{1/2}$. Central-limit aggregation can make longer-horizon returns more Gaussian even when short-horizon returns are heavy-tailed. Real data can depart from a single self-similar distribution because of volatility clustering, changing regimes, asymmetric responses, intraday seasonality and market microstructure. Apparent different scaling of higher moments may reflect intermittent volatility, but moments estimated from rare tails can also be unstable or may not exist in an idealized model.

Useful tests compare rescaled distributions, structure functions and volatility correlations across several scales, with uncertainty estimates and checks for nonstationarity. Scaling is a valuable approximate modelling guide over a documented range, not proof that one exponent describes every asset and horizon or that historical tail behaviour fixes future extreme losses.

After three multiplicative steps, $k$ upward moves and $3-k$ downward moves giveFor $k=0,1,2,3$, respectively, the prices are $\boxed{72.9,89.1,108.9,133.1}$. Since the steps are independent, the endpoint probabilities are binomial:In the displayed ascending-price order they are $(1-p{)}^3$, $3p(1-p{)}^2$, $3p^2(1-p)$, and $p^3$. Different paths with the same count of up moves recombine because multiplication by $1.1$ and $0.9$ commutes.

The one-step expected price multiplier under the stated physical probability is $1.1p+0.9(1-p)=0.9+0.2p$. Independence lets the expectation of the product factorize, givingThe same expression follows by summing the four binomial endpoint probabilities times their prices. This is a physical-measure expectation; option replication below uses a separately determined risk-neutral probability.

Let $q$ be the risk-neutral up probability and $R=1.04$ the one-step risk-free gross return. Discounted asset price must be a martingale, soThe no-arbitrage condition $0.9<R<1.1$ is satisfied, so $q$ lies strictly between zero and one. This pricing probability need not equal the actual up frequency $p$.

At strike $K=108$, the European put payoff is $(108-S_3{)}^{+}$. In the ascending-price order $72.9,89.1,108.9,133.1$, its values areRisk-neutral expectation discounted for three steps givesThe physical probability $p$ drops out because the complete two-state tree admits a replicating hedge. The strike is $108$, so the $108.9$ endpoint is out of the money; replacing it by a positive payoff would materially change the price.

At each time-2 node, replicate one positive put payoff using $\Delta$ shares and a current cash deposit $b$. Its two terminal values must satisfy $\Delta S_u+Rb=P_u$ and $\Delta S_d+Rb=P_d$, with $R=1.04$. ThusFor the three nodes the hedge isAt $S_2=121$, both child puts expire worthless. At $99$, the child payoffs are $0$ and $18.9$; at $81$, they are $18.9$ and $35.1$. Substituting the hedge holdings into both child states verifies those payoffs exactly.

The writer is short the option and therefore holds this positive replicating portfolio to offset the liability. At the lower nodes that means shorting shares and depositing the proceeds together with the option value, rather than buying positive-delta stock. Rebalancing the replicating holdings along the tree is self-financing in the ideal model; the combined short option plus hedge has zero terminal liability in either child state.