Pricing a Put by Its Payoff
A European put with strike \(K\) pays \((K-S_T)^+=\max(K-S_T,0)\) at maturity \(T\). Its payoff is positive precisely when \(S_T<K\); at \(S_T=K\), the payoff is zero. Under the Black–Scholes assumptions, the put price can be found by taking the risk-neutral expectation of that payoff and discounting it. The calculation reduces to two truncated normal expectations: one for the strike payment and one for the stock value in the exercise region.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\). Let \(\mathbb{Q}\) be the risk-neutral measure for the Black–Scholes market, and let \(B_t^{\mathbb{Q}}\) denote the Brownian motion under \(\mathbb{Q}\). The measure \(\mathbb{P}\) describes the physical probability model; \(\mathbb{Q}\) is the pricing measure used for the discounted payoff. We take a constant interest rate \(r\), a constant volatility \(\sigma>0\), and a positive strike \(K\). At time \(t<T\), write \(\tau=T-t>0\), and let the current stock price be \(S>0\).
Under \(\mathbb{Q}\), the stock follows geometric Brownian motion with drift \(r\) and diffusion coefficient \(\sigma S_t\). In differential notation, \(dS_t=rS_t\,dt+\sigma S_t\,dB_t^{\mathbb{Q}}\); the stochastic integral is an Itô integral. On a finite time interval its integrand is progressively measurable and square-integrable, since \(\mathbb{E}^{\mathbb{Q}}[S_s^2]=S^2e^{(2r+\sigma^2)s}\) when time is measured from the pricing date. Thus \(\mathbb{E}^{\mathbb{Q}}\int_0^\tau \sigma^2S_s^2\,ds<\infty\). The geometric Brownian motion solution gives the terminal representation \(S_T=S\exp((r-\tfrac12\sigma^2)\tau+\sigma\sqrt{\tau}\,Z)\), where \(Z\) has the standard normal distribution under \(\mathbb{Q}\). No change-of-measure argument is needed for the payoff calculation.
Deriving the European Put Formula
The risk-neutral valuation rule for this claim is to take the expectation of its maturity payoff under \(\mathbb{Q}\) and multiply by the discount factor \(e^{-r\tau}\). The positive-payoff event is strict: \(S_T<K\). In terms of \(Z\), this event becomes \(Z<-d_2\). The equality event has probability zero because \(Z\) has a continuous density, but the strict inequality is still the correct description of where the payoff is positive.
Proof. By risk-neutral valuation and the terminal stock representation, the price is $$ P=e^{-r\tau}\int_{-\infty}^{-d_2} \left(K-S e^{(r-\tfrac12\sigma^2)\tau+\sigma\sqrt{\tau}\,z}\right)\phi(z)\,dz. $$ The upper limit is \(-d_2\), because solving \(S_T<K\) for \(z\) gives \(z<[-\ln(S/K)-(r-\tfrac12\sigma^2)\tau]/(\sigma\sqrt{\tau})=-d_2\). The first integral is \(K e^{-r\tau}\Phi(-d_2)\). For the second, multiplying the stock term by the discount factor cancels \(e^{r\tau}\), leaving $$ S\int_{-\infty}^{-d_2} e^{\sigma\sqrt{\tau}\,z-\tfrac12\sigma^2\tau}\phi(z)\,dz. $$ Completing the square gives \(e^{\sigma\sqrt{\tau}\,z-\tfrac12\sigma^2\tau}\phi(z)=\phi(z-\sigma\sqrt{\tau})\). With the change of variable \(u=z-\sigma\sqrt{\tau}\), the integral is \(S\Phi(-d_2-\sigma\sqrt{\tau})=S\Phi(-d_1)\), since \(d_1=d_2+\sigma\sqrt{\tau}\). Subtracting this stock term from the strike term proves the formula. \(\square\)
Worked Example: A Put with \(S=K\)
Take \(S=K=100\), \(r=0\), \(\sigma=1\), and \(\tau=1\). Calculate the price from the formula, then recover it directly from the payoff expectation.
Since \(\ln(S/K)=\ln(1)=0\), the parameters are \(d_1=[0+(0+\tfrac12)1]/(1\cdot1)=\tfrac12\) and \(d_2=\tfrac12-1=-\tfrac12\). The formula therefore gives $$ P=100\Phi(\tfrac12)-100\Phi(-\tfrac12). $$
For a direct check, the terminal stock is \(100e^{-1/2+Z}\). The put pays a positive amount exactly when \(100e^{-1/2+Z}<100\), equivalently \(Z<\tfrac12\). Thus its expected payoff is $$ 100\int_{-\infty}^{1/2}\phi(z)\,dz -100\int_{-\infty}^{1/2}e^{z-1/2}\phi(z)\,dz. $$
The first integral is \(100\Phi(\tfrac12)\). Completing the square gives \(e^{z-1/2}\phi(z)=\phi(z-1)\); changing variables in the second integral gives \(100\Phi(-\tfrac12)\). Because \(r=0\), discounting does not change this expectation, so the direct calculation agrees with the formula. As a further check, \(\Phi(-\tfrac12)=1-\Phi(\tfrac12)\), so the price is \(100[2\Phi(\tfrac12)-1]\), which is positive because \(\Phi(\tfrac12)>\Phi(0)=\tfrac12\).
Boundary Checks: Zero Stock and Expiry
The formula is written for \(S>0\) and \(\tau>0\): its logarithm and its denominators are not defined at \(S=0\) or \(\tau=0\). Those boundaries should be checked from the payoff itself and, where appropriate, by taking limits of the formula. At zero stock, the lognormal representation with a logarithm is unavailable, but the stock remains at zero under the Black–Scholes stock dynamics. At expiry, the option is no longer valued by a nonzero-time normal formula; it is worth its payoff.
Proof. At \(S=0\), the stock is identically zero, so the maturity payoff is \(K\) and its discounted value is \(K e^{-r\tau}\). To check the positive-stock formula as \(S\downarrow0\), note that \(d_1\to-\infty\) and \(d_2\to-\infty\), because \(\ln(S/K)\to-\infty\) while all other terms in their numerators stay fixed. Hence \(\Phi(-d_2)\to1\). Also \(0\le S\Phi(-d_1)\le S\), so that term tends to zero. The formula therefore tends to \(K e^{-r\tau}\).
For expiry, first suppose \(S\ne K\) and \(S>0\). If \(S<K\), then \(\ln(S/K)<0\); as \(\tau\downarrow0\), both \(d_1\) and \(d_2\) tend to \(-\infty\). Consequently \(\Phi(-d_1)\to1\) and \(\Phi(-d_2)\to1\), and the put formula tends to \(K-S=(K-S)^+\). If \(S>K\), both parameters tend to \(+\infty\), so both normal-CDF terms in the formula tend to zero; the limit is \(0=(K-S)^+\). If \(S=K\), then \(d_1=(r+\tfrac12\sigma^2)\sqrt{\tau}/\sigma\to0\) and \(d_2=(r-\tfrac12\sigma^2)\sqrt{\tau}/\sigma\to0\). Thus the formula tends to \(K\Phi(0)-K\Phi(0)=0=(K-S)^+\). At \(S=0\) the expiry payoff is directly \(K\), consistent with \((K-S)^+\). These cases give the asserted boundary values. \(\square\)
Worked Example: Zero Stock
Let \(S=0\), \(K=120\), \(r=\tfrac1{10}\), and \(\tau=2\). The stock is zero at maturity, so the payoff is \((120-0)^+=120\). Discounting over two time units gives $$ P=e^{-(1/10)2}\cdot120=120e^{-1/5}. $$
Check: this is the zero-stock boundary value \(K e^{-r\tau}\): substituting \(K=120\), \(r=\tfrac1{10}\), and \(\tau=2\) gives \(120e^{-(1/10)2}=120e^{-1/5}\). One must not substitute \(S=0\) into \(d_1\) or \(d_2\), since \(\ln(S/K)\) is undefined there.
Worked Example: Payoff at Expiry
At expiry, take \(K=100\) and consider the three stock prices \(S=80\), \(S=100\), and \(S=125\). The payoffs are, respectively, \((100-80)^+=20\), \((100-100)^+=0\), and \((100-125)^+=0\). Thus the expiry value is 20 below the strike and zero at or above it; equality with the strike does not give a positive payoff.
Check by the positive-part definition: for \(S=80\), \(K-S=100-80=20>0\), so its positive part is 20. For \(S=100\), \(K-S=0\), so its positive part is 0. For \(S=125\), \(K-S=100-125<0\), so its positive part is 0. The limiting formula gives the same three values by the three expiry cases in the proposition.
A Common Cutoff Error
A frequent mistake is to use \(d_1\), rather than \(d_2\), to locate the event on which the payoff is positive. The exercise condition comes from comparing the terminal stock with the strike; it is the lognormal terminal-stock threshold, and therefore involves \(d_2\). The \(d_1\) term appears only after the stock part of the payoff is integrated and the normal density is shifted by completing the square.
For the parameters \(S=K=100\), \(r=0\), \(\sigma=1\), and \(\tau=1\), the correct payoff region is \(Z<\tfrac12=-d_2\). A wrong calculation would write the strike contribution as \(100\Phi(-d_1)=100\Phi(-\tfrac12)\), as though the exercise region were \(Z<-\tfrac12\). But the direct inequality \(100e^{-1/2+Z}<100\) gives \(Z<\tfrac12\), not \(Z<-\tfrac12\). The correct strike contribution is \(100\Phi(\tfrac12)=100\Phi(-d_2)\). The stock contribution is the one that evaluates to \(100\Phi(-\tfrac12)=100\Phi(-d_1)\) after completing the square.
The two CDF terms therefore have different origins: \(\Phi(-d_2)\) measures the probability of finishing below the strike under \(\mathbb{Q}\), while \(\Phi(-d_1)\) arises from the stock-weighted expectation over that same payoff region. Keeping the payoff event strict and separating these two integrals makes the formula, its boundary values, and its signs easier to check.
Check Your Understanding
Use the risk-neutral payoff calculation and boundary cases developed above.
- Starting from the terminal stock representation, derive the strict inequality in \(Z\) equivalent to \(S_T<K\).
- Why does the strike part of the put expectation involve \(\Phi(-d_2)\), while the stock part involves \(\Phi(-d_1)\)?
- For fixed \(\tau>0\), explain why \(S\Phi(-d_1)\to0\) as \(S\downarrow0\), even though \(d_1\) is undefined at \(S=0\).
- What is the expiry payoff when \(S=K\), and why can the expiry formula not be evaluated by direct substitution into \(d_1\) and \(d_2\)?
- For a stock price strictly above the strike at expiry, which limits of the normal-CDF terms make the put value zero?