The Black-Scholes formula for a European Call
The mean rate of growth of all assets under the risk-neutral measure \(\mathbb{Q}\) is risk-free rate \(r\).
The stock price process has the \(\mathbb{Q}\)-dynamics:
\[dS_t = r S_t dt + \sigma S_t dW^{\mathbb{Q}}(t) \tag{1}\]
The solution to this SDE is:
\[S(t) = S(0)\exp \left[ \left(r - \frac{\sigma^2}{2}\right)t + \sigma W^{\mathbb{Q}}(t)\right]\tag{2}\]
Consider a call option with maturity time \(T\). Then, the stock price at \(T\) is:
\[S(T) = S(0)\exp \left[ \left(r - \frac{\sigma^2}{2}\right)T + \sigma W^{\mathbb{Q}}(T)\right]\tag{3}\]
Denoting \(\tau = T - t\), we have:
\[S(T) = S(t)\exp \left[ \left(r - \frac{\sigma^2}{2}\right)\tau + \sigma (W^{\mathbb{Q}}(T)-W^{\mathbb{Q}}(t))\right]\tag{4}\]
Since, \(W^{\mathbb{Q}}(T)-W^{\mathbb{Q}}(t)\) is a gaussian random variable with mean \(0\) and variance \(\tau = T-t\), we can write \(-(W^{\mathbb{Q}}(T)-W^{\mathbb{Q}}(t)) = \sqrt{\tau}Z\), where \(Z\) is a standard normal random variable. Thus,
\[S(T) = S(t)\exp \left[ \left(r - \frac{\sigma^2}{2}\right)\tau - \sigma \sqrt{\tau}Z\right]\tag{5}\]
By the risk-neutral pricing formula, the time-\(t\) price of the European call option is:
\[ \begin{align*} V(t) &= \mathbb{E}^{\mathbb{Q}}\left[e^{-r(T-t)}\max(S(T) - K,0)|\mathcal{F}_t\right] \\ &= e^{-r(T-t)}\mathbb{E}^{\mathbb{Q}}\left[\left(S(t)\exp \left\{ \left(r - \frac{\sigma^2}{2}\right)\tau - \sigma \sqrt{\tau}Z\right\} - K\right)\cdot 1_{S(T)>K}|\mathcal{F}_t\right]\\ &= e^{-r(T-t)}\mathbb{E}^{\mathbb{Q}}\left[\left(S(t)\exp \left\{ \left(r - \frac{\sigma^2}{2}\right)\tau - \sigma \sqrt{\tau}Z\right\} - K\right)\cdot 1_{S_t e^{(r-\sigma^2/2) - \sigma\tau Z}>K}\right] \end{align*} \]
In the last-but-one step, everything is \(\mathcal{F}_t\)-measurable.
The domain of integration is all \(z\) satisfying:
\[ \begin{align*} S(t)\exp \left[ \left(r - \frac{\sigma^2}{2}\right)\tau - \sigma \sqrt{\tau}Z\right] &> K\\ \log \frac{S(t)}{K} + \left(r - \frac{\sigma^2}{2}\right)\tau &> \sigma \sqrt{\tau}Z \end{align*} \]
Define \(d_{-} = \frac{\log \frac{S(t)}{K} +(r-\sigma^2/2)\tau}{\sigma\sqrt{\tau}}\).
Then, the region \(D\) is:
\[Z < d_{-}\]
So, we can expand the expectation in (6) as:
\[ \begin{align*} V(t) &= \int_{-\infty}^{d_{-}} e^{-r\tau}\left(S(t)\exp \left\{\left(r - \frac{\sigma^2}{2}\right)\tau - \sigma \sqrt{\tau}z \right\} - K\right)d\mathbb{Q} \\ &=\int_{-\infty}^{d_{-}} e^{-r\tau}\left(S(t)\exp \left\{ \left(r - \frac{\sigma^2}{2}\right)\tau - \sigma \sqrt{\tau}z\right\} - K\right) f_Z^{\mathbb{Q}}(z) dz \\ &=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{d_{-}}e^{-r\tau} \left(S(t)\exp \left\{ \left(r - \frac{\sigma^2}{2}\right)\tau - \sigma \sqrt{\tau}z\right\} - K\right) e^{-\frac{z^2}{2}} dz \\ &=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{d_{-}} e^{-r\tau}S(t)\exp \left\{ \left(r - \frac{\sigma^2}{2}\right)\tau - \sigma \sqrt{\tau}z\right\}e^{-\frac{z^2}{2}} dz \\ &- Ke^{-r\tau}\cdot \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{d_{-}} e^{-\frac{z^2}{2}} dz \\ &=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{d_{-}} e^{-r\tau}S(t)\exp \left\{ \left(r - \frac{\sigma^2}{2}\right)\tau - \sigma \sqrt{\tau}z\right\}e^{-\frac{z^2}{2}} dz - Ke^{-r\tau}\Phi(d_{-})\tag{7} \end{align*} \]
We have:
\[ \begin{align*} &\exp \left[-\frac{\sigma^2}{2}\tau - \sigma\sqrt{\tau} z - \frac{z^2}{2}\right]\\ =&\exp\left[-\frac{\sigma^2 \tau + 2\sigma \sqrt{\tau}z + z^2}{2}\right]\\ =&\exp\left[-\frac{(z+\sigma\sqrt{\tau})^2}{2}\right] \tag{8} \end{align*} \]
Substituting (8) into (7), we get:
\[ \begin{align*} V(t) &=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{d_{-}} S(t)\exp\left[-\frac{(z+\sigma\sqrt{\tau})^2}{2}\right] dz - Ke^{-r\tau}\Phi(d_{-}) \tag{9} \end{align*} \]
Put \(u = z + \sigma \sqrt{\tau}\). Then, \(dz = du\). The upper limit of integration is \(d_{+} = d_{-} + \sigma \sqrt{\tau}\), which is:
\[ \begin{align*} d_{+} &=\frac{\log \frac{S(t)}{K} + (r-\sigma^2/2)\tau}{\sigma \sqrt{\tau}} + \sigma \sqrt{\tau}\\ &= \frac{\log \frac{S(t)}{K} + (r+\sigma^2/2)\tau}{\sigma \sqrt{\tau}} \end{align*} \]
So, the equation (9) can be written as:
\[ \begin{align*} V(t) &=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{d_{+}} S(t)e^{-\frac{u^2}{2}} du - Ke^{-r\tau}\Phi(d_{-}) \\ &= S(t)\Phi(d_{+}) - Ke^{-r\tau} \Phi(d_{-}) \end{align*} \]
Appendix
Lemma. The discounted stock-price process \((D(t)S(t),t\geq 0)\) is a \(\mathbb{Q}\)-martingale.
Suppose we have a risk-free money-market account with the dynamics:
\[dM(t) = rM(t)dt\]
and the dynamics of the stock-price process is:
\[dS(t) = \mu S(t) dt + \sigma S(t) dW^\mathbb{P}(t)\]
Thus, the discounting process is:
\[dD(t) = -rD(t)dt\]
where the instantaneous interest rate \(r\) is a constant.
By Ito’s product rule:
\[ \begin{align*} d(D(t)S(t)) &= dD(t) S(t) + D(t)dS(t)\\ &= -rD(t)S(t)dt + D(t)(\mu S(t) dt + \sigma S(t)dW^\mathbb{P}(t))\\ &= D(t)S(t)((\mu - r)dt + \sigma dW^\mathbb{P}(t))\\ \end{align*} \]
We are interested to write:
\[ \begin{align*} d(D(t)S(t)) &= D(t)S(t)\sigma dW^\mathbb{Q}(t) \end{align*} \]
Comparing the right hand sides, we have: \[ \begin{align*} \sigma dW^\mathbb{Q}(t) &= (\mu - r)dt + \sigma dW^\mathbb{P}(t) \end{align*} \]
Let’s define:
\[dW^\mathbb{Q}(t) = \theta dt + dW^\mathbb{P}(t)\]
where \(\theta = (\mu - r)/\sigma\) and the Radon-Nikodym derivative \(Z\) as:
\[Z = \exp\left[-\int_0^T \theta dW^\mathbb{P}(u) - \frac{1}{2}\int_0^T \theta^2 du \right]\]
By the Girsanov theorem, \(W^\mathbb{Q}(t)\) is a \(\mathbb{Q}\)-standard brownian motion. Hence, we can write:
\[ \begin{align*} d(D(t)S(t)) &= D(t)S(t)\sigma dW^\mathbb{Q}(t) \end{align*} \]
Since the Ito integral is a martingale, \(D(t)S(t)\) is a \(\mathbb{Q}\)-martingale. This closes the proof.
Claim. The \(\mathbb{Q}\)-dynamics of \(S_t\) satisfy :
\[dS(t) = rS(t) dt + \sigma S(t) dW^{\mathbb{Q}}(t)\]
Proof.
We have:
\[ \begin{align*}dS(t) &= d(S(t)D(t)M(t))\\ &= d(S(t)D(t))M(t) + S(t)D(t)dM(t)\\ &= D(t)M(t) S(t)\sigma dW^\mathbb{Q}(t) + S(t)D(t)r M(t)dt\\ &= S(t)(rdt + \sigma dW^\mathbb{Q}(t)) \end{align*} \]
We can easily solve this linear SDE; its solution is:
\[S(t) = S(0)\exp\left[\left(\mu - \frac{\sigma^2}{2}\right)dt + \sigma W^\mathbb{Q}(t)\right]\]