For an ex-dividend stock, holding $\Delta$ shares earns both $\Delta dS$ and the dividend gain $\Delta \delta Sdt$. Freeze $\Delta =V_S$ locally in a self-financing hedge that is long the claim and short these shares. Itô's formula cancels its Brownian exposure, leaving drift $V_t+\frac{1}{2}{\sigma }^2{S}^2{V}_{SS}-\delta S{V}_S$. Its value is $V-S{V}_S$, so absence of arbitrage requires this drift to equal $r(V-S{V}_S)$. Rearrangement gives the Black-Scholes equation with continuous stock dividends:With the given parameters,The physical drift $0.1$ cancels because the hedge removes the stock-price risk; the dividend payment remains in the stock's total gain.

Let the accumulated bank account from past dividends beHolding one stock share and banking its dividends is a self-financing portfolio with wealth $X_t=S_t+A_t$ and $d{A}_t=0.05A_{t}dt+0.01S_{t}dt$. Doubling that strategy replicates the specified terminal payoff. Its price is thereforeIn particular its initial price is one hundred pounds. This illustrates banking stock dividends versus reinvesting them. The accrued-dividend term is known at time $t$ and belongs to the payoff defined from time zero; it cannot be discarded when $t>0$.

Under the physical probability, the stock has mean $\mathbb{E}{S}_t=50e^{0.1t}$. Nonnegativity permits interchanging expectation and integration in the accumulated dividends. HenceTherefore

For a call option on a power payoff, the replicating asset $V$ from part (c) has $V_T=S_T^3$, volatility $3\sigma$, and risk-neutral drift $r$. The new payoff is therefore a European call option on $V$, with strike $K$. Applying the given Black-Scholes formula givesHere $V_t=S_t^3{e}^{(2r+3{\sigma }^2)\tau }$. Equivalently $D_1=[\log (S_t^3/K)+(3r+15{\sigma }^2/2)\tau ]/(3\sigma \sqrt{\tau })$. It is the price process $V_t$, rather than the raw power $S_t^3$, that serves as the self-financing underlying for this application.

Apply Itô's formula to $V_t=A(t)S_t^3$, where $A^{\prime }(t)=-(2r+3{\sigma }^2)A(t)$. Since $d{S}_t/S_t=\mu dt+\sigma d{W}_3(t)$, the drift and diffusion areThusUnder the risk-neutral measure its drift becomes $3r-2r=r$, as required for a traded, non-dividend-paying replicating asset.

For this stock-dependent payoff, replace the stock drift by $r$ under a risk-neutral measure. Conditional lognormal moments then give, with $\tau =T-t$,This is the Power payoff in the Black-Scholes model. The delta hedge is ${\Delta }_t=\partial V/\partial S=3V_t/S_t$ shares, so the stock holding is worth $3V_t$. The rest is a bank position worth $-2V_t$, or $-2V_t/B_t$ units of the bank asset. This gives a replicating portfolio; its self-financing identity follows from Itô's formula and the Black-Scholes pricing equation, and its terminal value is $S_T^3$.

The full two-Brownian filtration contains an additional orthogonal noise direction, but this particular payoff is replicated using the traded stock and bank asset, so that extra noise does not affect its price.

Cubing the supplied geometric Brownian motion solution gives $S_t^3=s^3\exp ((3\mu -3{\sigma }^2/2)t+3\sigma W_3(t))$. The moment-generating function of a normal distribution yields $\mathbb{E}[e^{a{W}_3(t)}]=e^{a^{2}t/2}$. Therefore

Put $\sigma =\sqrt{{\sigma }_1^2+{\sigma }_2^2}>0$. The linear combination $W_3$ starts at zero and has continuous paths. An increment over an interval of length $h$ is a linear combination of independent centred normal random variables, so is centred normal with varianceIncrements on disjoint intervals are independent, since they depend on disjoint increments of the two independent Brownian motions. These are the defining properties, so $W_3$ is a standard Brownian motion.

Return to $r=0$ and $q=1/3$ from part (b). An American put option is valued by the larger of immediate exercise and discounted continuation at each node. At time one with stock price $120$, exercise gives zero, while continuation gives $(1/3)0+(2/3)9=6$. With stock price $90$, exercise gives $27$, and continuation gives $(1/3)9+(2/3)36=27$; the holder is indifferent.

At time zero, exercise gives $17$, while continuation gives $(1/3)6+(2/3)27=20$. ConsequentlyIt is optimal to wait initially and at the up node; at the down node either exercise or waiting is optimal. At this zero interest rate, the European put also costs twenty pounds.

For matching European strike and expiry with no dividends, put-call parity gives $C_0-P_0=S_0-K{e}^{-2r}$. Using the rate and put price from part (c),As a check, only the $uu$ call payoff is positive, at $27$. Its risk-neutral expectation is $27(2/3{)}^2=12$, and discounting by $1.1^2$ gives the same value.

The terminal stock prices and European put option payoffs areLet $q$ be the risk-neutral up probability. Risk-neutral pricing and the supplied discounted-price condition giveso $9q^2-27q+14=0$. Its roots are $2/3$ and $7/3$; only the first is a probability. ThereforeThe current put price is $V_0=8/1.1^2=800/121$ pounds. This interest rate is specific to part (c), and differs from the zero rate in part (b).

Under a risk-neutral measure, the expected stock multiplier is the bank multiplier. With $q=1/3$,Thus $r=0$, and the one-period effective return $e^r-1$ is also zero.

Write $R=e^r$, $u=1.2$ and $d=0.9$. In a one-step binomial options pricing model, absence of arbitrage requires $d<R<u$. If $R\le d$, borrowing to buy the stock gives a nonnegative gain in either state and a positive gain in at least one. If $R\ge u$, shorting the stock and investing the proceeds does the same. Conversely, strict inequalities give the risk-neutral probability $q=(R-d)/(u-d)\in (0,1)$, under which the discounted stock is a martingale; this excludes arbitrage at each node. Hence

Let $T$ be the first completion time of the word. A fresh gambler arrives before every throw with capital one, betting it all on the successive symbols $1,2,1,2$. Each correct bet multiplies capital by six and each wrong bet loses it, so every bet is fair. A gambler who completes the word keeps the winnings.

At time $T$, the gambler who bet on all four final throws has capital $6^4$. The gambler who bet on the final two throws has correctly matched the suffix $12$ and has capital $6^2$. There are no other survivors: the only nonempty suffixes of $1212$ that are also prefixes are $12$ and $1212$. Thus the total surviving capital is $6^4+6^2$.

To justify the expectation identity, independent blocks of four throws each have probability $6^{-4}$ of equalling the pattern, so $\mathbb{P}(T>4k)\le (1-6^{-4}{)}^k$ and $\mathbb{E}T<\infty$. Total gambling capital minus the number of arriving unit stakes is a martingale with bounded increments before $T$. Optional stopping therefore equates expected terminal capital with expected total stakes:This is the waiting time for a word in independent uniform symbols. The extra $6^2$ reflects the overlapping occurrence of the prefix $12$ at the word's end.

Put $q=1-ϵ<1$. The hypothesis implies $\mathbb{P}(\tau >n+N\mid {\mathcal{F}}_n)\le q$. Since $\{\tau >n\}\in {\mathcal{F}}_n$ and survival to $n+N$ implies survival to $n$, the tower property of conditional expectation givesApply this at $n=0,N,2N,\dots$ and use $\mathbb{P}(\tau >0)\le 1$ to obtainThis is a geometric tail bound from a uniform escape probability. The tail-sum formula for a nonnegative integer-valued variable now gives

The optional stopping theorem says that for a martingale and bounded stopping times $\sigma \le \tau$,For unbounded almost surely finite stopping times, additional hypotheses are essential. Sufficient versions include uniform integrability of the stopped martingale, or an integrable stopping time together with uniformly bounded increments. In the latter case, stop at $\tau \wedge n$ and use $|M_{\tau \wedge n}|\le |M_0|+C\tau$ and the dominated convergence theorem. This is bounded-increment optional stopping with integrable time.