The Black Scholes formula and put call parity

Definition The standard normal density and the standard normal cumulative distribution function are

If $Z$ is standard normal, then $\mathbb{P}(Z \le x) = \Phi(x)$ for every real $x$. The function $\Phi$ is continuous and strictly increasing from $0$ to $1$, with $\Phi(0) = \tfrac12$, and the symmetry $\varphi(-z) = \varphi(z)$ gives $\Phi(-x) = 1 - \Phi(x)$, so $\mathbb{P}(Z > b) = \Phi(-b)$. It has no expression in elementary functions and is evaluated numerically, for example as $\Phi(x) = \tfrac12\bigl(1 + \operatorname{erf}(x/\sqrt{2})\bigr)$. Every option formula below rests on one truncated Gaussian integral. For all real numbers $a$ and $b$,

To prove it, complete the square, $az - \tfrac12 z^2 = -\tfrac12 (z - a)^2 + \tfrac12 a^2$, so that $e^{az}\varphi(z) = e^{a^2/2}\varphi(z - a)$ and

after the substitution $u = z - a$. The same steps give the lower tail $\mathbb{E}\bigl[e^{aZ}\mathbf{1}_{\{Z < b\}}\bigr] = e^{a^2/2}\,\Phi(b - a)$.

Definition In the Black-Scholes model of the next card, fix $t < T$, write $\tau = T - t$ and $S = S_t$, and assume $S > 0$, $K > 0$, $\sigma > 0$ and $\tau > 0$, with constant $r$ and $q$. Then

With the forward price $F = S e^{(r - q)\tau}$ the same numbers are

so they depend on $S$, $r$ and $q$ only through $F$. Under the pricing measure $\mathbb{Q}$ of the model the terminal price satisfies

where $Z = (W^{\mathbb{Q}}_T - W^{\mathbb{Q}}_t)/\sqrt{\tau}$ is standard normal and independent of $\mathcal{F}_t$. Solving $S_T > K$ for $Z$ gives the exercise event

The difference $d_1 - d_2 = \sigma\sqrt{\tau}$ is positive, so $d_1 > d_2$ always, and $d_1 + d_2 = 2\ln(F/K)/(\sigma\sqrt{\tau})$, so $d_1 = -d_2$ exactly when $K = F$.

Definition A European call with strike $K > 0$ and maturity $T$ pays $(S_T - K)^{+}$ at $T$, a European put pays $(K - S_T)^{+}$ at $T$, and neither can be exercised before $T$. Assume the Black-Scholes model. The market is frictionless with continuous trading, the money-market account is $B_t = e^{rt}$ with a constant continuously compounded rate $r$, and the stock pays a constant continuous dividend yield $q$ and has constant volatility $\sigma > 0$, with ex-dividend price following

under the pricing measure $\mathbb{Q}$ for the numeraire $B$. By the earlier lessons on replication and on the Feynman-Kac formula, a European claim whose payoff $g(S_T)$ grows at most polynomially in $S_T$, such as a call or a put, has the price $e^{-r\tau}\,\mathbb{E}^{\mathbb{Q}}[g(S_T) \mid \mathcal{F}_t]$. For $t < T$ and $S = S_t > 0$ the call and put prices are

For the derivation, split the call payoff on the exercise event $A$ of the previous card,

The strike term needs $\mathbb{Q}(A \mid \mathcal{F}_t) = \mathbb{Q}(Z > -d_2) = \Phi(d_2)$, since $Z$ is independent of $\mathcal{F}_t$. For the stock term write $S_T = F e^{-\sigma^2\tau/2 + \sigma\sqrt{\tau}Z}$ and apply the truncated integral with $a = \sigma\sqrt{\tau}$ and $b = -d_2$, so that $a - b = d_1$ and

Discounting gives $e^{-r\tau}F = S e^{-q\tau}$. The put uses the lower tail on $\{Z < -d_2\}$, which produces $\Phi(-d_2)$ and $\Phi(-d_2 - \sigma\sqrt{\tau}) = \Phi(-d_1)$. The stock term carries $e^{-q\tau}$ because one share delivered at $T$, without the dividends paid before $T$, costs $S e^{-q\tau}$ today, and the strike term carries $e^{-r\tau}$ because $K$ is cash paid at $T$. With deterministic time-dependent $r$, $q$ and $\sigma$ the same formulas hold after replacing $r\tau$, $q\tau$ and $\sigma^2\tau$ by $\int_t^T r(u)\,du$, $\int_t^T q(u)\,du$ and $\int_t^T \sigma(u)^2\,du$. Stochastic volatility, discrete cash dividends and American exercise are outside the formula, and averaging stochastic parameters inside it is not justified.

Definition Let a European call and a European put have the same underlying stock, strike $K$, maturity $T$ and settlement. For every value of $S_T$,

since at most one of the two terms is nonzero, and both vanish at $S_T = K$. Suppose a claim to one share at $T$, without the dividends paid before $T$, costs $S e^{-q\tau}$ at time $t$, a payment of $K$ at $T$ costs $K e^{-r\tau}$, and trading is frictionless. Then the law of one price gives

or $C - P = e^{-r\tau}(F - K)$. The argument uses only these two prices, not the lognormal law or $\sigma$, so parity holds in every arbitrage-free frictionless model with these prices for the prepaid share and the zero-coupon payment. With stochastic rates $K e^{-r\tau}$ becomes $K P(t,T)$. The Black-Scholes formulas satisfy it because $\Phi(d) + \Phi(-d) = 1$. If a quoted call and put violate it, buying the cheap side, selling the expensive side and offsetting $S_T - K$ with the share and the bond gives an arbitrage. Parity needs European exercise. For American options on a stock without dividends and with $r \ge 0$ only the inequalities

hold, and with transaction costs or different borrowing and lending rates the equality widens into a band.

Definition Under the hypotheses of the Black-Scholes formula, with $A = \{S_T > K\}$,

the probability under the money-market pricing measure that the call finishes in the money. The put finishes in the money with $\mathbb{Q}$-probability $\Phi(-d_2)$, since $\mathbb{Q}(S_T = K) = 0$, and $e^{-r\tau}\Phi(d_2)$ is the price of a claim that pays one unit of cash when $S_T > K$. The other factor is a probability under another measure. Define $\mathbb{Q}^{S}$ on $\mathcal{F}_T$ by the density $d\mathbb{Q}^{S}/d\mathbb{Q} = e^{-(r - q)T} S_T/S_0$, which is positive with $\mathbb{Q}$-mean $1$ and makes the share with dividends reinvested, worth $e^{qt}S_t$, the numeraire. Its conditional density given $\mathcal{F}_t$ is $e^{-(r - q)\tau} S_T/S_t$, so the stock-term computation of the formula card gives $\mathbb{Q}^{S}(A \mid \mathcal{F}_t) = e^{-(r - q)\tau}F\,\Phi(d_1)/S_t = \Phi(d_1)$, and

Neither number is a physical probability. If under $\mathbb{P}$ the ex-dividend price follows $dS_t = m S_t\,dt + \sigma S_t\,dW_t$, then

which equals $\Phi(d_2)$ only when $m = r - q$. The next lesson shows that $e^{-q\tau}\Phi(d_1)$ is the delta of the call.

Definition Fix $K > 0$, $S > 0$ and constants $r$ and $q$. The Black-Scholes prices have the following limits.

1. Expiry. As $\tau \downarrow 0$ with $\sigma > 0$ fixed, $C \to (S - K)^{+}$ and $P \to (K - S)^{+}$.
2. Zero volatility. As $\sigma \downarrow 0$ with $\tau > 0$ fixed, $C \to e^{-r\tau}(F - K)^{+}$ and $P \to e^{-r\tau}(K - F)^{+}$, where $e^{-r\tau}(F - K)^{+} = (S e^{-q\tau} - K e^{-r\tau})^{+}$.
3. Extreme spot prices. As $S \downarrow 0$, $C \to 0$ and $P \to K e^{-r\tau}$. As $S \to \infty$, $P \to 0$ and $C - (S e^{-q\tau} - K e^{-r\tau}) \to 0$.

In the expiry limit with $S \ne K$, the bracket in $d_1$ and $d_2$ tends to $\ln(S/K) \ne 0$ while $\sigma\sqrt{\tau} \to 0$, so both tend to $+\infty$ or both to $-\infty$ with the sign of $\ln(S/K)$, and the discount factors tend to $1$. When $S = K$, $d_{1,2} = (r - q \pm \tfrac12\sigma^2)\sqrt{\tau}/\sigma \to 0$, both probabilities tend to $\tfrac12$ and $C \to \tfrac12(S - K) = 0$. In the zero-volatility limit the forward form $d_{1,2} = (\ln(F/K) \pm \tfrac12\sigma^2\tau)/(\sigma\sqrt{\tau})$ gives the same picture with $F$ in place of $S$, applied to $C = e^{-r\tau}\bigl(F\,\Phi(d_1) - K\,\Phi(d_2)\bigr)$ with the discount factor held fixed. The formulas need $\sigma > 0$ and $\tau > 0$, and substituting $\sigma = 0$ or $\tau = 0$ into $d_1$ divides by zero. The values at the boundary come from the contract and the model. At $\tau = 0$ the option is its payoff, and with $\sigma = 0$ the stock price under $\mathbb{Q}$ grows deterministically and ends at $S_T = F$, so the value is the discounted payoff $e^{-r\tau}(F - K)^{+}$. Both agree with the limits. For $\sigma > 0$ and $\tau > 0$ the call lies strictly inside its bounds, $(S e^{-q\tau} - K e^{-r\tau})^{+} < C < S e^{-q\tau}$, because $C > 0$, $C - (S e^{-q\tau} - K e^{-r\tau}) = P > 0$ and $K e^{-r\tau}\Phi(d_2) > 0$.