Black–Scholes Through Replication

1Model Stock Price as Geometric Brownian Motion

A standard baseline model for stock prices $(S_t{)}_{t\ge 0}$ is geometric Brownian motion (GBM), defined as the solution to the stochastic differential equation (SDE)

where $\mu$ is the drift (the instantaneous expected rate of return), $\sigma >0$ is the volatility, and $W_t$ is standard Brownian motion.

Why use Brownian motion to model the random fluctuations? Its increments have several convenient properties:

1. they have mean $0$;
2. increments over disjoint time intervals are independent;
3. the sample paths are continuous;
4. the variance of an increment is proportional to the length of the time interval.

The last two assumptions are particularly idealized from an empirical point of view: real prices can jump, and volatility is not constant over time.

As a simple extension, one can replace $\sigma$ by a deterministic function $\sigma (t)$ and consider the stochastic integral

Then $X_t$ still has mean $0$ and continuous paths, and it has independent increments because the integrand is deterministic and Brownian increments are independent. Its variance, however, is generally not linear in $t$:

Moreover, empirical return distributions often exhibit heavier tails than the Gaussian increments implied by Brownian motion. Despite these shortcomings, GBM remains a useful first model because it is analytically tractable and captures basic scaling behavior.

Equation (1) has an explicit solution. For $S_0>0$, applying Itô’s formula to $f(x)=\ln x$ gives

Integrating from $0$ to $t$ gives

and therefore

In particular, $S_t$ is lognormal and strictly positive almost surely whenever $S_0>0$.

2Risk-Neutral Measure under GBM-modeled Stock Price

The risk-neutral measure $\mathbb{Q}$ is a probability measure under which the discounted asset price is a martingale. Let $y_t=e^{-rt}{S}_t$ be the discounted stock price. Under the physical measure $\mathbb{P}$, if

then

Hence $y_t$ is a martingale under $\mathbb{P}$ if and only if $\mu =r$.

In the Black–Scholes model, we instead work under an equivalent risk-neutral measure $\mathbb{Q}$ under which

Thus the stock has drift $r$ under $\mathbb{Q}$, and the discounted stock price $e^{-rt}{S}_t$ is a martingale.

3Risk-neutral Pricing via Replication

Suppose the derivative pays $F(S_T)$ at maturity $T$, where $S_t$ is the stock price process. We consider a portfolio that at time $t$ holds:

• ${\mathrm{\Delta }}_t$ shares of stock (positive ${\mathrm{\Delta }}_t$ is a long position and negative ${\mathrm{\Delta }}_t$ is a short position);
• $A_t$ dollars in the bank account.

Its total value is $V_t=A_t+{\mathrm{\Delta }}_t{S}_t$.

We assume this portfolio is chosen so that it replicates the derivative payoff, meaning $V_T=F(S_T)$. By the no-arbitrage principle, if such a replicating portfolio exists, then its value must equal the derivative price at every time $t$. Otherwise, one could buy the cheaper one and sell the more expensive one, obtaining a riskless profit. Thus, once we identify the value process $V_t$, that is the option price.

The bank account grows at the risk-free rate $r$. Fix the portfolio holdings $(A_t,{\mathrm{\Delta }}_t)$ over the interval $(t,t+dt)$ and rebalance at the end of the interval. In the continuous-time limit, this yields the self-financing dynamics. During this interval, the bank position grows from $A_t$ to $A_t{e}^{rdt}\approx A_t(1+rdt)$, while the stock position changes from ${\mathrm{\Delta }}_t{S}_t$ to ${\mathrm{\Delta }}_t{S}_{t+dt}$. Therefore, the portfolio value at time $t+dt$ is

Subtracting $V_t=A_t+{\mathrm{\Delta }}_t{S}_t$, we obtain

In differential notation,

Since $V_t=A_t+{\mathrm{\Delta }}_t{S}_t$, we have

i.e.,

Define the discounted portfolio value $Y_t=e^{-rt}{V}_t$. Using the Itô product rule,

Since $e^{-rt}$ is deterministic, $d(e^{-rt})d{V}_t=0$. Thus

Substituting the expression for $d{V}_t$,

i.e.,

Under the risk-neutral measure $\mathbb{Q}$, $d{S}_t=r{S}_{t}dt+\sigma S_{t}d{W}_t^{\mathbb{Q}}$. Substituting,

The drift terms cancel:

Therefore, $Y_t$ is a martingale. Then by definition

That is,

Because the portfolio replicates the derivative payoff, $V_T=F(S_T)$. We have

This is the risk-neutral pricing formula in the Black–Scholes model.

4Closed-Form Black–Scholes Pricing Formula for a European Call

4.1Derivation of the Time-0 Formula

We derive a closed-form expression for the European call price by evaluating the risk-neutral expectation explicitly. For a European call,

Under the risk-neutral measure $\mathbb{Q}$, the stock price satisfies

Since $W_T^{\mathbb{Q}}\sim N(0,T)$, we may write $W_T^{\mathbb{Q}}=\sqrt{T}Z$ with $Z\sim N(0,1)$. Therefore,

and so

Using the standard normal density $\phi$, this becomes

The payoff is positive exactly when

Taking logarithms gives

so

Define

Then the condition for a positive payoff is $z>-d_2$. Hence

Split the integral into two terms:

The second integral is immediate:

where $\mathrm{\Phi }$ is the standard normal distribution function. It remains to compute

Using $\phi (z)=\frac{1}{\sqrt{2\pi }}{e}^{-z^2/2}$, we have

Complete the square in the exponent:

so

Therefore,

and thus

Now make the substitution $u=z-\sigma \sqrt{T}$. Then $du=dz$, and when $z=-d_2$, the lower limit becomes $u=-d_2-\sigma \sqrt{T}$. Define

So the lower limit is $-d_1$, and hence

Substituting back, we obtain

Since $e^{(r-\frac{1}{2}{\sigma }^2)T}{e}^{\frac{1}{2}{\sigma }^{2}T}=e^{rT}$, this simplifies to

Thus the Black–Scholes price of a European call option is

where

4.2Interpretation of $d_1$ and $d_2$

The two normal terms play somewhat different roles. The term $\mathrm{\Phi }(d_2)$ has the direct probability interpretation

so it is the risk-neutral probability that the call finishes in the money. The term $\mathrm{\Phi }(d_1)$ appears when we compute the stock-price contribution on that event.

To see this, recall the identity

For intuition, here is a brief derivation in the discrete case. For a discrete random variable,

Multiplying both sides by $\mathbb{P}(A)$ gives

But the right-hand side is exactly $\mathbb{E}[X{\mathbf{1}}_A]$, since $X{\mathbf{1}}_A$ equals $X$ on the event $A$ and $0$ outside $A$. Hence $\mathbb{E}[X{\mathbf{1}}_A]=\mathbb{E}[X\mid A]\mathbb{P}(A)$.

Applying this with $X=S_T-K$ and $A=\{S_T>K\}$, we obtain

So

Hence

Thus $\mathrm{\Phi }(d_1)$ is naturally associated with the stock-price-weighted part of the payoff, while $\mathrm{\Phi }(d_2)$ is the risk-neutral exercise probability.

4.3Black–Scholes Formula at an Intermediate Time

For any intermediate time $t$, recall

Because the future Brownian increment $W_T-W_t$ is independent of the information available at time $t$, the future evolution from $t$ to $T$ is a fresh Black–Scholes model starting at $S_t$ with time to maturity $T-t$. Thus

where

4.4Delta via Itô’s Formula

Let $v(t,S_t)$ denote the option price and define its discounted value by $Y_t=e^{-rt}v(t,S_t)$. Applying Itô’s formula under $\mathbb{Q}$ gives

Because $Y_t$ is a martingale under $\mathbb{Q}$, its drift is $0$. On the other hand, the discounted replicating portfolio satisfies

Comparing the diffusion coefficients gives

4.5Probabilistic Representation of Delta

Let $\tau :=T-t$. Under $\mathbb{Q}$, independent increments give

Let $Y:=S_T/S_t$. Since the distribution of $Y$ does not depend on the current stock price, we may write

Differentiating with respect to $x$ (for payoffs for which this interchange is valid), we get

Thus

For a call payoff $F(s)=(s-K{)}^{+}$, the derivative is $F^{\prime }(s)={\mathbf{1}}_{\{s>K\}}$ except at $s=K$. Substituting into the formula gives

In the Black–Scholes model this simplifies to

4.6Cash Position in the Replicating Portfolio

Thus the bond position is

Notice that

5American Options

Recall an American option may be exercised at any time in the interval $[0,T]$. This additional flexibility makes American options an optimal stopping problem rather than a fixed-horizon pricing problem.

Recall the definition of stopping time: a random time $\tau \in [0,T]$ is called a stopping time (with respect to the filtration $({\mathcal{F}}_t{)}_{t\ge 0}$) if for every $t\in [0,T]$, the event $\{\tau \le t\}$ belongs to ${\mathcal{F}}_t$.

If the payoff from exercising at time $t$ is $G(S_t)$, then the arbitrage-free price of the American claim at time $0$ is

where ${\mathcal{T}}_{[0,T]}$ denotes the set of stopping times taking values in $[0,T]$.

In general, this does not lead to a closed-form formula, and one typically needs free-boundary methods, variational inequalities, binomial trees, or numerical PDE methods.

5.1American Calls on Non-Dividend-Paying Stocks

In the Black–Scholes model with no dividends and $r\ge 0$, it is never optimal to exercise an American call early. Consequently, the American call price equals the European call.

A useful intuition is the following. Exercising early requires paying the strike $K$ immediately and destroys the remaining time value of the option. By waiting, the holder keeps the upside exposure to the stock while postponing payment of the strike. Since the stock pays no dividends in this model, there is no compensating benefit to early exercise.

5.2A Perpetual American Put

One important case that can be solved explicitly is the perpetual American put, where there is no maturity date. Assume $r>0$. In this case, one considers

where $S_0=s$. A natural class of stopping rules is given by threshold strategies:

If $s>L$, then exercising at ${\tau }_L$ yields payoff $K-L$, so

Using an exponential martingale and optional stopping, one finds

Hence the value corresponding to the threshold $L$ is

Optimizing over $L$ gives the optimal exercise boundary

Therefore, the perpetual American put value is

This example illustrates a general principle for American puts: unlike calls on non-dividend-paying stocks, early exercise can be optimal.