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For contingent claim payoff $(S_T-K{)}_{+}^2$ in the non-dividend Black-Scholes model, differentiating the discounted expected value gives $\Delta =2e^{-r\tau }{S}^{-1}\mathbb{E}[(S_T^2-K{S}_T)1_{\{S_T\ge K\}}]$. Completing the square in the Gaussian exponent computes each truncated lognormal moment and yields the displayed formula, where $d_1=(\log (S/K)+(r+{\sigma }^2/2)\tau )/(\sigma \sqrt{\tau })$. Its terminal limit is $2(S-K{)}_{+}$.

Write the smooth claim value as $V(t,S)$. Itô's formula givesA replicating self-financing strategy with stock holding $\Delta =V_S$ and bank value $V-S{V}_S$ has gainsThe Brownian terms then agree. Equating the drift terms cancels $\mu S{V}_S$ and givesThe terminal condition enforces replication of the contingent claim payoff. Subject to the usual admissibility and growth conditions for the replicating price, this is the Black-Scholes equation. Its independence of the physical drift is a consequence of hedge cancellation, not an assumption that the drift equals the risk-free rate.

A stock holding receives both its price change and dividend, so its total gains are $dS+\delta Sdt$. The replicating portfolio therefore satisfiesItô's formula and the hedge $\Delta =(V_{\delta }{)}_S$ cancel the Brownian risk. Comparing drifts gives

To verify the claimed transformation, put $x=S{e}^{-\delta (T-t)}$ and set $U(t,S)=V_0(t,x)$. Then $x_t=\delta x$, $S{U}_S=x(V_0{)}_x$ and $S^2{U}_{SS}=x^2(V_0{)}_{xx}$. The dividend operator applied to $U$ becomesby the zero-dividend PDE. At maturity $x=S$, so the contingent claim payoff also agrees. Uniqueness in the pricing class proves

Differentiate the dividend PDE with respect to $S$. The diffusion term contributes ${\sigma }^{2}S{V}_{SS}+\frac{1}{2}{\sigma }^2{S}^2{V}_{SSS}$, and the stock-drift term contributes $(r-\delta )(V_S+S{V}_{SS})$. With $\Delta =V_S$, the coefficient of $\Delta$ is $(r-\delta )-r=-\delta$. ThereforeThe terminal condition is $\Delta (T,S)={\Psi }^{\prime }(S)$ wherever the contingent claim payoff is differentiable, with an appropriate weak or limiting interpretation at contingent claim payoff kinks. In the next part the squared positive-part contingent claim payoff is continuously differentiable, so its terminal option delta has no jump.

The contingent claim payoff derivative is ${\Psi }^{\prime }(s)=2(s-K{)}_{+}$. Differentiating the risk-neutral valuation is justified by the finite lognormal distribution, and $S_T$ is proportional to the starting stock value $s$. ThusFor a lognormal terminal stock, completing the square in its Gaussian exponent givesUse $p=2$ and $p=1$, and let $d_1=[\log (s/K)+(r+{\sigma }^2/2)\tau ]/(\sigma \sqrt{\tau })$. The result isAs $\tau \downarrow 0$ this tends to $2(s-K{)}_{+}$, including value zero at $s=K$, as required. The extra ${\sigma }^2$ in the option delta PDE's stock drift gives the same shifted Gaussian arguments. This is delta of a squared positive-part call.

Use the usual nondegenerate positive-rate case $r>0$, $\sigma >0$, $K>0$. Define the Black-Scholes equationThe American value dominates immediate exercise. Its discounted value is a supermartingale under the risk-neutral measure, so $\mathcal{L}V\le 0$, with equality where it is optimal to continue. The obstacle problem isEquivalently $\max (\mathcal{L}V,g-V)=0$.

For the put, write the exercise boundary as $S=b(t)$. On $S>b(t)$ solve $\mathcal{L}V=0$; on $0<S\le b(t)$ set $V=K-S$. Value matching and smooth pasting requireThe remaining conditions are $V(t,0)=K$, $V(t,S)\to 0$ as $S\to \infty$, and the terminal contingent claim payoff, with $b(T-)=K$ in this positive-rate setting. The boundary is an unknown to be found with the price, which is what makes this a free boundary problem.

put-call parity and nonnegativity of the European put option giveFor $r\ge 0$ this is at least $S-K$, and the call value is also nonnegative, so $C^E$ dominates the immediate call exercise contingent claim payoff. Its discounted price is a martingale. At any admissible exercise time, optional stopping theorem therefore bounds the expected discounted call option by the European call option price. Exercising at maturity attains that price. Hence American and European calls coincide for this non-dividend model with nonnegative rates.

For the strict put comparison assume $r>0$, $\sigma >0$, $S>0$ and $t<T$. Choose $u$ strictly between $t$ and $T$. A European put option at $u$ has value at most $K{e}^{-r(T-u)}$. On the eventimmediate exercise gives $K-S_u>K{e}^{-r(T-u)}\ge P^E(u,S_u)$. This event has positive probability because the lognormal stock density is positive on $(0,\infty )$. Exercise there at time $u$, and otherwise hold to maturity. This strategy strictly improves the European put option expected value, provingunder these positive-rate, pre-maturity assumptions.

The source's strict statement needs those qualifications. At maturity the prices agree, and at $r=0$ the put option is a convex function of the martingale stock and therefore is a submartingale, so waiting to maturity is optimal and the American and European puts coincide. Negative rates also invalidate the nonnegative-rate call argument. These are mathematical assumptions of the comparison, rather than changes to the printed question.

A down-and-out option with no rebate pays $(S_T-K{)}_{+}$ only if the stock has stayed above the lower barrier $B$ up to maturity; if the barrier is hit, it expires worthless. For an unknocked-out contract currently at $S>B$, apply part (a) to the ordinary European call option price:Both terms solve the PDE. At $S=B$ they cancel, enforcing the absorbing boundary. At maturity and $S>B$, the reflected stock value is $B^2/S<B\le K$, so the reflected call option is zero and the required terminal contingent claim payoff remains. Standard growth and uniqueness conditions for the killed pricing problem identify this solution as the option value.

If the contract has already hit its barrier, its value is zero regardless of a later hypothetical stock recovery. Thus the formula is conditional on survival to the current time; the barrier history is part of the contract state.

If default liquidates the firm as soon as its value crosses below $F$, equity is extinguished at that event. If default never occurs before maturity, equity receives $(V_T-F{)}_{+}$. It is therefore a no-rebate down-and-out option on firm value, with strike price and barrier both equal to $F$.

For a firm that has survived to $t$ and has $V_t>F$, part (b) givesIf default has already occurred, $E_t=0$. At maturity a surviving firm has equity $(V_T-F{)}_{+}$.

The source uses a strict crossing $V<F$. With nonzero volatility, continuity and immediate two-sided Brownian fluctuations at a first barrier hit make the pre-maturity absorbing value at $V=F$ zero as well: starting at the barrier, the process crosses below it immediately almost surely. At maturity the contingent claim payoff at equality is also zero. Thus this convention has the same pricing boundary as the usual barrier-hit convention in this nondegenerate model.

Differentiating the displayed transformation shows that it preserves the Black-Scholes equation. The first derivative coefficient fixes the exponent $1-2r/{\sigma }^2$, which also makes the zero-order coefficient correct. Subtracting this reflected image of a European call option from the call itself enforces a zero boundary at $S=B$. For strike price $K\ge B$, the reflected maturity contingent claim payoff vanishes on $S>B$, giving the no-rebate down-and-out option formula. Prices using this reflection are conditional on the barrier not having been hit already.