From Risk-Neutral Payoff to Price
In the Black–Scholes model, a European claim pays a specified function of the stock price at maturity. Its value before maturity can be expressed as the conditional expectation of its discounted payoff under the risk-neutral measure. The conditional expectation matters: at time \(t\), the observed stock price and other information in \(\mathcal{F}_t\) are known, while the future payoff is not.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\), and let \(\mathbb{Q}\) be the risk-neutral measure from the change of measure in the Black–Scholes model. Under \(\mathbb{Q}\), write \(B_t^{\mathbb{Q}}\) for the Brownian motion driving the stock. Assume a constant interest rate \(r\), volatility \(\sigma>0\), no dividends, and \(S_0>0\). The stock dynamics under \(\mathbb{Q}\) are $$ dS_t=rS_t\,dt+\sigma S_t\,dB_t^{\mathbb{Q}}. $$ Here \(rS_t\) is the drift coefficient and \(\sigma S_t\) is the diffusion coefficient. For every stochastic integral below, the integrand is progressively measurable and belongs to \(\mathcal{L}^2\) on the stated finite time interval under the measure in use; that is, its expected time integral of the square is finite.
Fix a strike \(K>0\) and maturity \(T\). For \(t<T\), set \(\tau=T-t\). A call pays \((S_T-K)^+=\max(S_T-K,0)\), and a put pays \((K-S_T)^+\). The risk-neutral pricing expressions are $$ C_t=e^{-r\tau}\mathbb{E}^{\mathbb{Q}}\!\left[(S_T-K)^+\mid\mathcal{F}_t\right], \qquad P_t=e^{-r\tau}\mathbb{E}^{\mathbb{Q}}\!\left[(K-S_T)^+\mid\mathcal{F}_t\right]. $$ At \(t=T\), these expressions are the payoffs themselves. We derive the formulas for \(t<T\), then verify their connection to the PDE and hedge.
Evaluating the Conditional Expectations
The explicit solution of the risk-neutral stock equation gives $$ S_T=S_t\exp\!\left(\left(r-\frac{\sigma^2}{2}\right)\tau+\sigma\sqrt{\tau}\,Z\right), $$ where, conditionally on \(\mathcal{F}_t\), \(Z=(B_T^{\mathbb{Q}}-B_t^{\mathbb{Q}})/\sqrt{\tau}\) has the standard normal distribution. Define $$ d_1=\frac{\ln(S_t/K)+(r+\sigma^2/2)\tau}{\sigma\sqrt{\tau}}, \qquad d_2=d_1-\sigma\sqrt{\tau}, $$ and let \(\Phi\) and \(\phi\) denote the standard normal distribution function and density.
Proof. Let \(a=\sigma\sqrt{\tau}>0\). The event \(S_T>K\) is the event \(Z>-d_2\), since taking logarithms of the stock solution gives the inequality \(Z>-[\ln(S_t/K)+(r-\sigma^2/2)\tau]/a=-d_2\). For any real \(x\), completing the square in the normal density yields $$ \mathbb{E}^{\mathbb{Q}}\!\left[e^{aZ}\mathbf{1}_{\{Z>x\}}\mid\mathcal{F}_t\right] =\int_x^\infty e^{az}\phi(z)\,dz =e^{a^2/2}\Phi(a-x). $$ Therefore the discounted call expectation is $$ e^{-r\tau}\left( S_t e^{(r-\sigma^2/2)\tau}e^{a^2/2}\Phi(a+d_2) -K\Phi(d_2)\right). $$ Since \(a+d_2=d_1\) and \(-r\tau+(r-\sigma^2/2)\tau+a^2/2=0\), this is \(S_t\Phi(d_1)-Ke^{-r\tau}\Phi(d_2)\).
For the put, the event \(S_T<K\) is \(Z<-d_2\). Completing the square on this lower tail gives $$ \mathbb{E}^{\mathbb{Q}}\!\left[e^{aZ}\mathbf{1}_{\{Z<-d_2\}}\mid\mathcal{F}_t\right] =e^{a^2/2}\Phi(-d_2-a)=e^{a^2/2}\Phi(-d_1). $$ Thus its discounted expectation is \(Ke^{-r\tau}\Phi(-d_2)-S_t\Phi(-d_1)\), as claimed. The boundary event \(S_T=K\) has conditional probability zero for \(\tau>0\), because the conditional law is lognormal with positive variance. Both conditional expectations are finite: the call payoff is bounded by \(S_T\), whose conditional expectation is \(S_te^{r\tau}<\infty\), and the put payoff is bounded by \(K\). \(\square\)
Worked Example: A Call and Put at the Money
At time \(t\), suppose \(S_t=100\), \(K=100\), \(r=0\), \(\sigma=\sqrt{2}\), and \(\tau=\tfrac12\). Then \(\sigma\sqrt{\tau}=\sqrt{2}\sqrt{\tfrac12}=1\), and \(\ln(S_t/K)=\ln(1)=0\). Hence \(d_1=\tfrac12\) and \(d_2=-\tfrac12\). Substitution gives $$ C_t=100\Phi\!\left(\tfrac12\right)-100\Phi\!\left(-\tfrac12\right), \qquad P_t=100\Phi\!\left(\tfrac12\right)-100\Phi\!\left(-\tfrac12\right). $$ The two values agree because \(r=0\) and the strike and spot coincide. Check: subtracting the put expression from the call expression gives \(100[\Phi(\tfrac12)-\Phi(-\tfrac12)-\Phi(\tfrac12)+\Phi(-\tfrac12)]=0\), consistent with the zero difference between spot and discounted strike.
The PDE and the Price Formula
The Black–Scholes PDE for a European claim value \(u(t,s)\) is $$ u_t+rsu_s+\frac12\sigma^2s^2u_{ss}-ru=0, \qquad 0\le t<T,\ s>0, $$ with terminal condition \(u(T,s)=(s-K)^+\) for a call. The expectation formula agrees with this PDE solution: the call formula satisfies the equation and terminal condition. To verify the equation, write \(C(t,s)=s\Phi(d_1)-Ke^{-r\tau}\Phi(d_2)\), now treating \(d_1,d_2\) as functions of \((t,s)\).
Using \(d_{1,s}=d_{2,s}=1/(s\sigma\sqrt{\tau})\) and the identity \(s\phi(d_1)=Ke^{-r\tau}\phi(d_2)\), differentiation and cancellation give \(C_s=\Phi(d_1)\) and \(C_{ss}=\phi(d_1)/(s\sigma\sqrt{\tau})\).
Since \(\tau=T-t\), direct differentiation gives \(C_t=-\sigma s\phi(d_1)/(2\sqrt{\tau})-rKe^{-r\tau}\Phi(d_2)\).
The diffusion term is \(\tfrac12\sigma^2s^2C_{ss}=\sigma s\phi(d_1)/(2\sqrt{\tau})\). Its sum with the first term of \(C_t\) is zero. The remaining terms cancel because \(rsC_s- rC= rKe^{-r\tau}\Phi(d_2)\), which cancels the remaining term of \(C_t\).
As \(t\) increases to \(T\), the lognormal stock converges to \(s\) in the terminal condition, and the formula tends to \((s-K)^+\). At \(s=K\), both terms involving \(\Phi\) tend to their common half-probability contribution, and their difference tends to zero, as required. This verifies the PDE and terminal payoff directly.
The same link can be seen from Itô’s formula. Under \(\mathbb{Q}\), apply it to the function \(f(t,s)=e^{-rt}C(t,s)\) along \(S_t\). Its derivatives are \(f_t=e^{-rt}(C_t-rC)\), \(f_s=e^{-rt}C_s\), and \(f_{ss}=e^{-rt}C_{ss}\). The explicit second-order correction is \(\tfrac12 f_{ss}\sigma^2S_t^2\,dt\), because \(dB_t^{\mathbb{Q}}dB_t^{\mathbb{Q}}=dt\), \(dB_t^{\mathbb{Q}}dt=0\), and \(dt\,dt=0\). Thus $$ d(e^{-rt}C(t,S_t)) =e^{-rt}\left(C_t+rS_tC_s+\frac12\sigma^2S_t^2C_{ss}-rC\right)dt +e^{-rt}\sigma S_tC_s\,dB_t^{\mathbb{Q}}. $$ The PDE makes the drift zero. The remaining stochastic integral has integrand \(e^{-rt}\sigma S_tC_s\), which belongs to \(\mathcal{L}^2\) on \([0,T]\): \(0\le C_s=\Phi(d_1)\le1\), and \(\mathbb{E}^{\mathbb{Q}}[S_t^2]=S_0^2e^{(2r+\sigma^2)t}\) is finite and integrable over a finite interval. By the Itô isometry the integral is a true martingale, not merely a local martingale. This connects the PDE value to the risk-neutral conditional expectation.
Worked Example: Checking the Call Delta and Hedge Integrability
For the call formula, the differentiation above gives \(C_s=\Phi(d_1)\), so the replicating strategy holds \(\Phi(d_1)\) shares. In particular, its stock holding lies between zero and one. For a concrete integrability check, take \(S_0=100\), \(r=0\), \(\sigma=1\), and \(T=1\). Then $$ \mathbb{E}^{\mathbb{Q}}\!\left[\int_0^1 e^{-2rt}\sigma^2S_t^2C_s(t,S_t)^2\,dt\right] \le \int_0^1 10000e^t\,dt=10000(e-1)<\infty. $$ The inequality uses \(C_s^2\le1\) and \(\mathbb{E}^{\mathbb{Q}}[S_t^2]=10000e^t\). Therefore the stochastic integral in the discounted hedge wealth is square-integrable, and its expectation conditional on \(\mathcal{F}_t\) has the martingale property. Check: at \(t=0\), the second moment is \(10000\); the displayed bound integrates that moment growth over the whole unit interval rather than assuming the integrand is deterministic.
Why the Martingale Qualification Matters
It is not valid to infer a conditional-expectation price from the statement that every discounted self-financing wealth process is a martingale. In general, discounted wealth from a self-financing strategy is a local martingale. Under a common admissibility condition that discounted wealth is bounded below, it is a supermartingale, but need not be a true martingale. A local martingale need not have constant expectation, so the conditional-expectation conclusion requires additional integrability or uniform-integrability conditions.
For the call and put considered here, the required condition is established for the particular replicating hedges. The discounted terminal payoff is integrable: the call payoff is bounded by \(S_T\), with \(\mathbb{E}^{\mathbb{Q}}[e^{-rT}S_T]=S_0\), and the put payoff is bounded by \(K\). Consequently, the process $$ M_t=\mathbb{E}^{\mathbb{Q}}\!\left[e^{-rT}(S_T-K)^+\mid\mathcal{F}_t\right] $$ is an integrable true martingale by the tower property of conditional expectation. The put analogue has the same property. The explicit formulas show \(M_t=e^{-rt}C_t\) and the corresponding put identity. Their delta hedges have square-integrable stochastic integrands, as in the example, so the discounted hedge wealth is also a true martingale and matches the conditional-expectation value. This argument is specific to these claims and hedges; it does not promote every admissible discounted wealth process to a true martingale.
Worked Example: A Common Call-Pricing Error
A tempting but incorrect calculation is to replace the call payoff by \(S_T-K\) inside the expectation. That gives \(e^{-r\tau}\mathbb{E}^{\mathbb{Q}}[S_T-K\mid\mathcal{F}_t]=S_t-Ke^{-r\tau}\), which ignores the event on which the call pays zero. For example, with \(S_t=K=100\), \(r=0\), \(\sigma=2\), and \(\tau=1\), this incorrect answer is \(100-100=0\). The correct parameters are \(d_1=1\) and \(d_2=-1\), so the call value is \(100[\Phi(1)-\Phi(-1)]\), which is positive because \(\Phi(1)>\Phi(-1)\). Check: the correct formula retains only the upper-tail stock contribution \(S_T>K\) and subtracts the strike payment on that same event; the untruncated calculation incorrectly includes negative values of \(S_T-K\) when \(S_T<K\).
For both calls and puts, the conditional expectation is under the risk-neutral measure \(\mathbb{Q}\), not the physical measure \(\mathbb{P}\). Under \(\mathbb{Q}\), the stock drift is \(r\); under \(\mathbb{P}\), its drift generally reflects a different expected return. The discount factor \(e^{-r\tau}\), the risk-neutral stock distribution, and the PDE’s discount term \(-ru\) are three parts of the same pricing relation.
Check Your Understanding
Use the conditional normal variable and the stated risk-neutral dynamics in your reasoning.
- For \(t<T\), show that \(S_T>K\) is equivalent to \(Z>-d_2\).
- Why does the call expectation require the upper-tail truncated exponential moment rather than just \(\mathbb{E}^{\mathbb{Q}}[S_T\mid\mathcal{F}_t]\)?
- Verify from the formulas that \(C_t-P_t=S_t-Ke^{-r\tau}\).
- State the integrability condition that makes the discounted call hedge stochastic integral a true martingale.
- Explain why the local-martingale property of a general discounted wealth process alone does not justify a conditional-expectation pricing formula.