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For $p>1$, the zero-interest curve $C(K)=(1+K^p{)}^{1/p}-K$ has positive second derivative $f_p$. Its mass and first moment are both one, and $\int (u-K{)}^{+}{f}_p(u)du=C(K)$. It prices an integrable payoff by $\int g(u)f_p(u)du$ when this law is equivalent to the physical terminal stock law, or on a canonical model with this pricing law. Finite-strike consistency alone is insufficient for that equivalence.

A power put has financial payoff $(K-S{)}^n$ below the strike price $K$ and zero at or above it, where $n$ is a positive integer. For $n=1$ this is the ordinary put option. For every $n\ge 1$ the payoff is decreasing and is a convex function of the stock price. A European option permits exercise only at expiry; a perpetual power put option permits exercise at any finite time without a fixed expiry.

Put-call parity fixes the price difference between a European call option and a European put option on the same underlying stock, with the same strike price $K$ and expiry $T$. Assume no dividends or other payments from the stock before or at expiry, a known constant continuously compounded interest rate $r$, and a frictionless market. Borrowing and lending are available at that same interest rate; short selling of the stock, European call option, European put option, and zero-coupon bond is permitted. Settlement is certain, and the buy-and-hold portfolios below may be held until $T$ without interim funding restrictions or forced liquidation. These assumptions exclude transaction costs, taxes, borrowing fees, and execution or settlement risks from the arbitrage argument.

At expiry, the two positive parts cancel their corners:Thus a European call option and a short European put option together have the same terminal financial payoff as one stock share with a debt of $K$. A particularly readable claim replication uses two portfolios with nonnegative terminal amounts: portfolio A holds one European call option and a risk-free zero-coupon bond paying $K$ at $T$; portfolio B holds one European put option and one stock share. The zero-coupon bond costs $K{e}^{-rT}$ today, since the discount factor is $e^{-rT}$. There are no intermediate cash flows. For every terminal stock price,For the strike price $K=100$, even the following very different outcomes leave the two portfolios equal:

The law of one price forces $A_0=B_0$ in this setting: otherwise buy the cheaper portfolio and sell the more expensive portfolio using short selling. Their terminal obligations cancel in every scenario, while the initial price difference can grow in the bank account. Hence absence of arbitrage givesWith a constant per-period interest rate $i$ over $N$ discrete periods, the same reasoning gives $C_0-P_0=S_0-K(1+i{)}^{-N}$. This is a different compounding convention; $i$ is not the continuously compounded interest rate $r$.

For an illustrative stock price $S_0=100$, strike price $K=100$, expiry one year away, interest rate $r=0.05$ per year, and given European call option price $C_0=10$, put-call parity requiresBoth portfolios therefore cost approximately $105.12294245$. The inputs specify no stock price dynamics: $C_0=10$ is supplied for illustration, not calculated from the Black-Scholes model. Put-call parity determines the European put option price conditional on that European call option price; it does not determine both prices by itself. At zero interest rate, the same $S_0=K=100$ and $C_0=10$ give the exact result $P_0=10$.

To compute both directions of arbitrage, keep all the preceding inputs except the quoted European put option price. If it is $6$, portfolio B costs $106$ while portfolio A costs $105.12294245$. Buy A, sell B, and place the initial surplus $6-(10+100e^{-0.05}-100)\approx 0.87705755$ in the bank account. At expiry, A and the obligations from selling B cancel exactly, leaving $96e^{0.05}-100\approx 0.92202525$ in every scenario. If the European put option instead costs $4$, buy B for $104$ and sell A for $105.12294245$; the initial surplus is approximately $1.12294245$, growing to $100-94e^{0.05}\approx 1.18051694$. Both constructions start with zero external capital. They are model calculations under the stated frictionless market assumptions.

The beauty of put-call parity is that uncertain individual financial payoffs produce an exact relation between today's prices. No forecast of the terminal stock price and no probability distribution are needed. It provides a consistency check and lets claim replication exchange one contract for a combination of others. Dividends introduce extra cash flows, and American put options can be exercised early, so the displayed European equality cannot simply be transferred to those contracts. K. Foster's ^[ref] explain the matching portfolios; Jun Pan's ^[ref] discuss put-call parity and its dividend adjustment.

An original modern research source is Hans R. Stoll's ^[ref], The Journal of Finance 24(5), 801-824 (December 1969). Stoll studied the relation between contract prices and tested it against market data; this was a relation between prices rather than a formula fixing each price individually. Robert C. Merton's ^[ref], 28(1), 183-184, highlighted the importance of excluding premature exercise for an exact arbitrage equality. This distinction explains why the exercise assumption belongs in the statement of put-call parity.

For the normalized Black-Scholes call function, completing the square gives $F(v,m)=\Phi (d_1)-m\Phi (d_2)$, with $d_1=(-\log m+v/2)/\sqrt{v}$ and $d_2=d_1-\sqrt{v}$. The identities $d_1(v,1/m)=-d_2(v,m)$ and $\Phi (-x)=1-\Phi (x)$ implyFor a stock with risk-neutral drift $r$, maturity difference $\tau$, and positive strike $K$, put-call parity then givesThe positive sign of $r\tau$ in the reciprocal argument is forced by taking the reciprocal of $K{e}^{-r\tau }/S_t$. The identity includes $v=0$ by continuity and requires $m>0$.

For a European put option with deterministic discount factor $D$ and an integrable terminal stock price, $P(K)=D{\mathbb{E}}_{\mathbb{Q}}[(K-S_T{)}^{+}]$. If $\mathbb{Q}(S_T=K)=0$, bounded difference quotients and the dominated convergence theorem give the displayed derivative, the price of a digital put option. At an atom, the left derivative is $D\mathbb{Q}(S_T<K)$ and the right derivative is $D\mathbb{Q}(S_T\le K)$, so the ordinary derivative need not exist.

A risk-neutral measure $Q$ is a probability measure equivalent to the physical probability measure $P$, meaning they agree on which events have zero probability. For assets without intermediate payments, prices expressed in the chosen numéraire must be martingales under $Q$. For a positive bank account $B$ and a stock without dividends, this requires $S/B$ to be a martingale; with dividends, use discounted dividend gains. In an unrestricted finite-state one-period financial market, existence of such a measure is equivalent to absence of arbitrage, by the fundamental theorem of asset pricing.

Physical probability $p=P(\text{up})$ describes the modeled frequency or belief about an up move. The risk-neutral probability $q=Q(\text{up})$ is constrained by traded prices. In the worked binomial market, $q=5/8$ even if $p=1/2$ or $p=4/5$. Calling $Q$ risk-neutral does not assume that investors have no risk aversion; it changes the weights used for risk-neutral pricing.

The martingale condition for the discounted stock is $qu{S}_0+(1-q)d{S}_0=R{S}_0$. Therefore the up risk-neutral probability isWith down factor $d=1+a$, up factor $u=1+b$ and $R=1+r$, this is equivalently $(r-a)/(b-a)$. For $d=0.8$, $u=1.2$, $R=1.05$, it gives $q=0.25/0.4=0.625$. Under the risk-neutral measure, the terminal stock price has expected value $(5/8)120+(3/8)80=105$, whose discounted value is 100. The physical probability $p$ need only satisfy $0<p<1$; it does not enter this calculation.

For a non-dividend-paying stock in the Black-Scholes model, use density $d{Q}^S/dQ=e^{-\rho T}{S}_T/S_0$. Under $Q^S$, $W_t^S=W_t^Q-\sigma t$ is Brownian motion, and $\log S_t=\log S_0+(\rho +{\sigma }^2/2)t+\sigma W_t^S$. Thus $e^{-\rho T}{\mathbb{E}}_Q[S_{T}H]=S_0{\mathbb{E}}_{Q^S}H$. This change of numeraire is particularly useful for payments delivered in stock units.

Fix $T>0$, let $X$ be Brownian motion, and let $\tau =\inf \{u:X_u=-1\}$. Set $q(t)=t/(T-t)$ for $t<T$ andThe hitting time is finite almost surely, so each price path reaches one before $T$ and stays there. Stopping additionally on hitting $n+2$ makes the stock price a bounded martingale; these stopping times increase to $T$. Hence $S$ is a positive local martingale, and the original probability measure is an equivalent local martingale measure, excluding admissible arbitrage. Nevertheless $S_T=B_T$ while $S_0=2\ne 1=B_0$. The zero-cost short-stock/long-two-cash portfolio has terminal gain one but wealth $2-S_t$, which has no deterministic lower bound. It is not admissible, resolving the apparent contradiction.

If $Z$ is nondegenerate, $g$ is strictly increasing and the relevant products are integrable, an independent copy $Z^{\prime }$ gives $2\mathrm{Cov}(g(Z),Z)=\mathbb{E}[(g(Z)-g(Z^{\prime }))(Z-Z^{\prime })]$. The integrand is positive exactly when $Z\ne Z^{\prime }$, so the covariance is strictly positive. The same argument applies within each conditional law and yields a positive terminal stock hedge for an attainable increasing payoff.

For a non-dividend-paying stock, buy $1/F_t^T$ shares and short a forward contract for the same number of shares at its fair delivery price $F_t^T>0$. The combined payoff is one at $T$, and the initial cost is $S_t/F_t^T$. This replicates a unit zero-coupon bond without initially assuming an independently traded bank account. A single maturity quote determines that discount factor; it does not determine the entire future short-rate process.

An up-and-in claim activates when the stock reaches an upper barrier $b>S_0$ during its lifetime. After activation its terminal payoff is the prescribed function $f(S_T)$; without activation it pays zero. Under continuous stock paths the barrier event is described by the displayed running maximum.

A digital put option extinguished if the stock reaches an upper barrier $B$. Under the dividend-free Black-Scholes model, conditional on not having hit the barrier before the current time, let $0<S<B$, $\tau =T-t>0$, $m=\min (K,B)$, $\gamma =1-2r/{\sigma }^2$, andThen its price isThe Black-Scholes reflection symmetry proves the pricing equation, and the two terms cancel at $S=B$. At expiry the reflected argument is above $B\ge m$, so its payoff is zero, while $S<m$ is equivalent to $S<K$ inside the alive region. Thus the correct free strike is capped at the barrier. For $K\ge B$, the value is the discounted upper-barrier survival probability, independent of $K$, as for a barrier digital put. For $K>B$ its terminal data have a corner discontinuity at the absorbing barrier. Set the price to zero after a hit; the reflected expression on the other side is not a valid contract value.

An up-and-out power claim pays a power of the terminal stock price provided an upper barrier has never been reached. In the Black-Scholes model, set $A=\log (c/S_0)/\sigma$ and $m_p=(\rho +(p-\frac{1}{2}){\sigma }^2)/\sigma$, where $\sigma >0$, $c>S_0$ and $T>0$. The risk-neutral pricing value isHere $\Phi$ is the standard normal distribution function. Weighting the Brownian motion endpoint by $e^{p\sigma W_T-p^2{\sigma }^{2}T/2}$ changes the logarithmic drift to $m_p$; the finite-horizon maximum of Brownian motion with drift then supplies the survival factor.