CONTENTS

Solution: Black-Scholes Call Price as a Risk-Neutral Expectation

Exercise: Black-Scholes Call Price as a Risk-Neutral Expectation

Part 1: Q(ST>K)\mathbb{Q}(S_T > K)

Under Q\mathbb{Q}, ln⁡ST=ln⁡S0+(r−12σ2)T+σWTQ\ln S_T = \ln S_0 + (r - \tfrac{1}{2}\sigma^2)T + \sigma W_T^{\mathbb{Q}} with WTQ∼N(0,T)W_T^{\mathbb{Q}} \sim \mathcal{N}(0, T). So ln⁡ST\ln S_T is Gaussian with mean ln⁡S0+(r−12σ2)T\ln S_0 + (r - \tfrac{1}{2}\sigma^2)T and variance σ2T\sigma^2 T.

Q(ST>K)=Q(ln⁡ST>ln⁡K)=Q ⁣(ln⁡ST−[ln⁡S0+(r−12σ2)T]σT>ln⁡K−ln⁡S0−(r−12σ2)TσT)\mathbb{Q}(S_T > K) = \mathbb{Q}(\ln S_T > \ln K) = \mathbb{Q}\!\left(\frac{\ln S_T - [\ln S_0 + (r - \tfrac{1}{2}\sigma^2)T]}{\sigma\sqrt{T}} > \frac{\ln K - \ln S_0 - (r - \tfrac{1}{2}\sigma^2)T}{\sigma\sqrt{T}}\right)

The left side is standard normal ZZ, the right side is −d2-d_2. Hence

Q(ST>K)=Q(Z>−d2)=Φ(d2)\mathbb{Q}(S_T > K) = \mathbb{Q}(Z > -d_2) = \Phi(d_2)

using the symmetry Q(Z>−d2)=Q(Z<d2)=Φ(d2)\mathbb{Q}(Z > -d_2) = \mathbb{Q}(Z < d_2) = \Phi(d_2).

Part 2: EQ[ST1{ST>K}]\mathbb{E}^{\mathbb{Q}}[S_T \mathbf{1}_{\{S_T > K\}}]

Write z=WTQ/Tz = W_T^{\mathbb{Q}} / \sqrt{T}, so z∼N(0,1)z \sim \mathcal{N}(0, 1) and WTQ=TzW_T^{\mathbb{Q}} = \sqrt{T}z. Then

ST=S0e(r−σ2/2)T+σTzS_T = S_0 e^{(r - \sigma^2/2)T + \sigma\sqrt{T}z}

and the event {ST>K}\{S_T > K\} is equivalent to {z>−d2}\{z > -d_2\} (same manipulation as Part 1).

EQ[ST1{ST>K}]=∫−d2∞S0e(r−σ2/2)T+σTz⋅12πe−z2/2 dz\mathbb{E}^{\mathbb{Q}}[S_T \mathbf{1}_{\{S_T > K\}}] = \int_{-d_2}^{\infty} S_0 e^{(r - \sigma^2/2)T + \sigma\sqrt{T}z} \cdot \frac{1}{\sqrt{2\pi}} e^{-z^2/2}\,dz

Pull the S0e(r−σ2/2)TS_0 e^{(r - \sigma^2/2)T} out and combine exponents inside the integral:

=S0e(r−σ2/2)T∫−d2∞12πeσTz−z2/2 dz= S_0 e^{(r - \sigma^2/2)T} \int_{-d_2}^{\infty} \frac{1}{\sqrt{2\pi}} e^{\sigma\sqrt{T}z - z^2/2}\,dz

Complete the square in the exponent:

σT z−z22=−12(z−σT)2+σ2T2\sigma\sqrt{T}\,z - \frac{z^2}{2} = -\frac{1}{2}\bigl(z - \sigma\sqrt{T}\bigr)^2 + \frac{\sigma^2 T}{2}

Substitute back:

=S0e(r−σ2/2)Teσ2T/2∫−d2∞12πe−(z−σT)2/2 dz= S_0 e^{(r - \sigma^2/2)T} e^{\sigma^2 T / 2} \int_{-d_2}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-(z - \sigma\sqrt{T})^2/2}\,dz =S0erT∫−d2∞12πe−(z−σT)2/2 dz= S_0 e^{rT} \int_{-d_2}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-(z - \sigma\sqrt{T})^2/2}\,dz

Substitute u=z−σTu = z - \sigma\sqrt{T}, du=dzdu = dz. The lower limit becomes u=−d2−σTu = -d_2 - \sigma\sqrt{T}. Note that d1=d2+σTd_1 = d_2 + \sigma\sqrt{T}, so −d2−σT=−d1-d_2 - \sigma\sqrt{T} = -d_1. Hence

=S0erT∫−d1∞12πe−u2/2 du=S0erT⋅P(Z>−d1)=S0erTΦ(d1)= S_0 e^{rT} \int_{-d_1}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-u^2/2}\,du = S_0 e^{rT} \cdot \mathbb{P}(Z > -d_1) = S_0 e^{rT} \Phi(d_1)

So EQ[ST1{ST>K}]=S0erTΦ(d1)\mathbb{E}^{\mathbb{Q}}[S_T \mathbf{1}_{\{S_T > K\}}] = S_0 e^{rT}\Phi(d_1).

Part 3: combine

C0=e−rT(EQ[ST1{ST>K}]−K Q(ST>K))C_0 = e^{-rT}\bigl(\mathbb{E}^{\mathbb{Q}}[S_T \mathbf{1}_{\{S_T > K\}}] - K\,\mathbb{Q}(S_T > K)\bigr) =e−rT(S0erTΦ(d1)−KΦ(d2))= e^{-rT}\bigl(S_0 e^{rT}\Phi(d_1) - K\Phi(d_2)\bigr) =S0Φ(d1)−Ke−rTΦ(d2)= S_0\Phi(d_1) - K e^{-rT}\Phi(d_2)
This is the Black-Scholes call price. □\square

Part 4: absence of μ\mu

The entire calculation was performed under Q\mathbb{Q}, where the stock drift is rr. The real-world drift μ\mu never appeared. This is expected: risk-neutral pricing replaces the real-world drift with rr, so no formula derived as a Q\mathbb{Q}-expectation can depend on μ\mu. The market's view of the equity risk premium is irrelevant to the option's no-arbitrage price — only the replicability (here, through σ\sigma) matters.

Takeaways

  • The call price splits into two pieces. Φ(d1)\Phi(d_1) is the risk-neutral expected in-the-money payoff per unit of stock, and Φ(d2)\Phi(d_2) is the risk-neutral exercise probability. Both are outputs of the same log-normal integral, separated by the completing-the-square step.
  • d1d_1 and d2d_2 differ by σT\sigma\sqrt{T}. Not by accident: the shift comes from the Gaussian change of variable in the STS_T-weighted integral, which is exactly the "change of numeraire" from the bank account to the stock — a measure change in its own right.
  • Risk-neutral calculations are drift-free. The computation went through without any assumption about μ\mu. This is the computational payoff of risk-neutral pricing: a single measure change makes the derivative price a deterministic function of S0,K,r,T,σS_0, K, r, T, \sigma.
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