Deriving the European Call Formula

From the Heat Equation to a Call Price

In the Black–Scholes model, the call payoff at maturity is \((S_T-K)^+\), where \(K>0\) is the strike and \(S_T\) is the stock price at maturity. Earlier results in the Black–Scholes development give a pricing PDE and a transformation of that PDE to the heat equation. Here we solve the transformed terminal-value problem and evaluate its integrals. The main point is that the two terms in the payoff produce two Gaussian tail probabilities, with different shifts.

Use a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\). Let \(r\) be a constant interest rate, \(\sigma>0\) a constant volatility, \(T\) the maturity, and \(0\le t<T\) the current time. Write \(\tau_{\mathrm{cal}}=T-t\) for the time remaining. The calculation below solves the pricing PDE; it is not an expectation under the physical measure \(\mathbb{P}\). The risk-neutral measure \(\mathbb{Q}\) is not needed for this PDE derivation. The integrals evaluated below are ordinary Gaussian integrals, not Itô integrals.

For clarity, use \(\theta=\sigma^2\tau_{\mathrm{cal}}/2\) as the heat-equation time and \(x=\log(S/K)\) as the log-price coordinate. Set \(k=2r/\sigma^2\), and divide the call value by \(K\), writing \(c(x,\theta)=C(t,S)/K\). The transformed equation and initial payoff are

The exponential substitution \(c(x,\theta)=e^{(1-k)x/2-(k+1)^2\theta/4}u(x,\theta)\) removes the first-derivative and zero-order terms, giving \(u_\theta=u_{xx}\). Its initial function is \(g(y)=e^{(k+1)y/2}-e^{(k-1)y/2}\) for \(y>0\), and \(g(y)=0\) for \(y\le0\). Thus the heat-kernel representation uses only \(y>0\).

The Gaussian Integral Behind the Formula

For a real number \(a\) and \(\theta>0\), the heat kernel applied to the exponential \(e^{ay}\) on the positive half-line gives the integral below. Completing the square identifies it as a normal tail with a shifted argument.

Proof. Expand the exponent and complete the square: \(-\frac{(x-y)^2}{4\theta}+ay=-\frac{(y-x-2a\theta)^2}{4\theta}+ax+a^2\theta\). After taking out the factor \(e^{ax+a^2\theta}\), set \(q=(y-x-2a\theta)/\sqrt{2\theta}\). The lower endpoint \(y=0\) becomes \(q=-(x+2a\theta)/\sqrt{2\theta}\), and the upper endpoint remains \(+\infty\). Since \(dy=\sqrt{2\theta}\,dq\), the remaining integral is \((2\pi)^{-1/2}\int_{-(x+2a\theta)/\sqrt{2\theta}}^\infty e^{-q^2/2}\,dq\). Substituting \(v=-q\) and using the evenness of \(e^{-q^2/2}\) gives \(N((x+2a\theta)/\sqrt{2\theta})\), as required. \(\square\)

The heat-kernel solution for the initial function \(g\) is the difference of two integrals in this lemma. The heat kernel is the standard solution operator for \(u_\theta=u_{xx}\): its Gaussian kernel integrates to one, satisfies the heat equation for positive \(\theta\), and converges to the initial function at its continuity points as \(\theta\) decreases to zero. Applying the lemma evaluates that solution. The following result completes the transformation back to the call price.

Proof. The initial function for \(u\) is the difference of \(e^{a_1y}\) and \(e^{a_2y}\) on \(y>0\), where \(a_1=(k+1)/2\) and \(a_2=(k-1)/2\). By the lemma, its heat-kernel solution is \(u(x,\theta)=e^{a_1x+a_1^2\theta}N((x+2a_1\theta)/\sqrt{2\theta})-e^{a_2x+a_2^2\theta}N((x+2a_2\theta)/\sqrt{2\theta})\). Multiplying by \(e^{(1-k)x/2-(k+1)^2\theta/4}\) recovers \(c\). In the first term, the combined coefficient of \(x\) is \(1\), and the combined coefficient of \(\theta\) is \(0\). In the second term, both combined coefficients are \(0\) and \(-k\), respectively. Therefore \(c=e^xN((x+(k+1)\theta)/\sqrt{2\theta})-e^{-k\theta}N((x+(k-1)\theta)/\sqrt{2\theta})\). Now \(K e^x=S\), \(k\theta=r\tau_{\mathrm{cal}}\), and \(\sqrt{2\theta}=\sigma\sqrt{\tau_{\mathrm{cal}}}\). The first normal argument is \((x+(r+\sigma^2/2)\tau_{\mathrm{cal}})/(\sigma\sqrt{\tau_{\mathrm{cal}}})=d_1\); the second is \(d_2\). Multiplying by \(K\) proves the formula. Finally, subtracting the two arguments gives \(d_1-d_2=\sigma\sqrt{\tau_{\mathrm{cal}}}\), which checks their stated relation. \(\square\)

Worked Applications

Expiry, Interpretation, and a Common Mistake

At expiry the call price must equal its payoff, not the formula with a zero denominator substituted directly into \(d_1\) and \(d_2\). To check the limit as \(\tau_{\mathrm{cal}}\) decreases to zero, if \(S>K\), both normal arguments tend to \(+\infty\), so the formula tends to \(S-K\). If \(S<K\), they tend to \(-\infty\), so it tends to zero. If \(S=K\), both arguments tend to zero: the difference of the two CDF terms tends to zero, while the discount factor tends to one, so the call value tends to zero. In all three cases the limit is \((S-K)^+\).

A common mistake is to use \(d_2\) in both terms, or to omit the discount factor on the strike term. For the first worked example, replacing \(d_1\) by \(d_2\) in both CDF terms would give \(100[N(-0.1)-N(-0.1)]=0\), rather than the correct positive value. The two different arguments arise because the payoff contains both \(e^y\) and \(1\) on the positive half-line: their exponential factors shift the Gaussian tails by different amounts. The factor \(e^{-r\tau_{\mathrm{cal}}}\) also follows directly from the transformed solution, where the second term carries \(e^{-k\theta}=e^{-r\tau_{\mathrm{cal}}}\).

The formula is a solution of the Black–Scholes pricing PDE with the call payoff as its terminal condition. Its derivation here uses the PDE and the heat kernel, rather than calculating an expectation under \(\mathbb{P}\). Under the Black–Scholes pricing setup, this distinction matters: \(\mathbb{P}\) is the physical probability measure, whereas \(\mathbb{Q}\), when used for risk-neutral pricing, is a different measure. The PDE calculation alone does not assert that the physical expected payoff equals the price.