The Black-Scholes formula for a European Call

The mean rate of grownth 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:

$$S(t) = S(0)\exp\left[\left(\mu - \frac{\sigma^2}{2}\right)dt + \sigma W^\mathbb{Q}(t)\right]$$