From a Binomial Model to a Continuous-Time Price
In a binomial model, a stock moves up or down over each time step, and an option price can be obtained by discounting its expected payoff under risk-neutral probabilities. The Black–Scholes formula gives the corresponding price in a continuous-time model. Its compact expression can make the result seem like a formula to memorize; its derivation shows instead how the payoff, the stock-price distribution, and discounting fit together.
Assume a probability space \((\Omega,\mathcal{F},\mathbb{Q})\), where \(\mathbb{Q}\) is the risk-neutral pricing measure, and a filtration \((\mathcal{F}_t)_{t\ge0}\) describing the information available through time \(t\). Under \(\mathbb{Q}\), let \((B_t)_{t\ge0}\) be standard Brownian motion. The stock price \((S_t)_{t\ge0}\) is modeled by geometric Brownian motion with constant rate \(r\) and volatility \(\sigma\):
Here \(r\) is the continuously compounded risk-free rate, \(\sigma>0\) is the stock’s annualized volatility, and \(T>0\) is the time to maturity in years. The option is a European call: its holder may exercise only at time \(T\), receiving \(\max(S_T-K,0)\), where \(K>0\) is the strike price. The stock pays no dividends in this model.
The formula values the payoff under the pricing model, not by inserting an investor’s forecast probability of a rise into the calculation. The risk-neutral model assigns the stock the drift \(r\); its role in valuation is to make the discounted stock price a martingale under \(\mathbb{Q}\). In the earlier binomial setting, replication and no-arbitrage lead to the same distinction between pricing probabilities and forecast probabilities.
Deriving the Formula
The geometric Brownian motion solution has the distribution $$S_T=S_0\exp\big((r-\sigma^2/2)T+\sigma\sqrt{T}Z\big),$$ where \(Z\) has the standard normal distribution under \(\mathbb{Q}\). Risk-neutral valuation discounts the expected payoff: $$C=e^{-rT}\mathbb{E}_{\mathbb{Q}}\big[\max(S_T-K,0)\big].$$ We now evaluate this expectation, showing why the two normal probabilities in the formula have different arguments.
Proof. Since \(K>0\), the call payoff is positive exactly when \(S_T>K\). Substituting the expression for \(S_T\), this event is \(Z>-d_2\), because $$ \ln(S_T/K)=\ln(S_0/K)+(r-\sigma^2/2)T+\sigma\sqrt{T}Z $$ and solving \(\ln(S_T/K)>0\) for \(Z\) gives \(Z>-d_2\). Thus $$ C=e^{-rT}\mathbb{E}_{\mathbb{Q}}\big[S_T\mathbf{1}_{\{Z>-d_2\}}\big] -Ke^{-rT}\mathbb{Q}(Z>-d_2). $$ By symmetry of the standard normal density, \(\mathbb{Q}(Z>-d_2)=\Phi(d_2)\). For the stock term, put \(a=\sigma\sqrt T\). Completing the square gives $$ e^{az}\frac{e^{-z^2/2}}{\sqrt{2\pi}} =e^{a^2/2}\frac{e^{-(z-a)^2/2}}{\sqrt{2\pi}}. $$ Therefore $$ \mathbb{E}_{\mathbb{Q}}\big[S_T\mathbf{1}_{\{Z>-d_2\}}\big] =S_0e^{(r-\sigma^2/2)T}e^{a^2/2} \int_{-d_2}^{\infty}\frac{e^{-(z-a)^2/2}}{\sqrt{2\pi}}\,dz. $$ The factor before the integral is \(S_0e^{rT}\). Changing variables to \(y=z-a\), the lower limit becomes \(-d_2-a=-d_1\), so the integral is \(\Phi(d_1)\). Discounting gives \(S_0\Phi(d_1)\). Combining the stock and strike terms proves \(C=S_0\Phi(d_1)-Ke^{-rT}\Phi(d_2)\). \(\square\)
This derivation requires the stated lognormal model and risk-neutral valuation. It is not a claim that real stock prices always have constant volatility or normally distributed log returns. The assumptions that make the formula a benchmark are considered below.
Worked Example: Evaluating a Call with Simple \(d_1\) and \(d_2\)
Take \(S_0=\sqrt e\), \(K=1\), \(r=0\), \(\sigma=1\), and \(T=1\). These are mathematical inputs chosen to simplify the calculation. First, $$ d_1=\frac{\ln(\sqrt e/1)+(0+1/2)\cdot1}{1\cdot\sqrt1} =\frac{1/2+1/2}{1}=1,\qquad d_2=1-1\cdot1=0. $$ The call price is consequently $$ C=\sqrt e\,\Phi(1)-1\cdot e^0\Phi(0) =\sqrt e\,\Phi(1)-\frac12, $$ using \(\Phi(0)=1/2\).
Check by the payoff expectation. Here \(S_T=\sqrt e\,e^{-1/2+Z}=e^Z\), so \(S_T>1\) exactly when \(Z>0\). The undiscounted expected call payoff is $$ \int_0^\infty(e^z-1)\frac{e^{-z^2/2}}{\sqrt{2\pi}}\,dz =e^{1/2}\Phi(1)-\frac12 =\sqrt e\,\Phi(1)-\frac12. $$ The rate is zero, so discounting does not change this value. This agrees with the formula.
Interpretation and a Useful Check
Volatility describes the scale of uncertainty in log stock returns in the model; it is not the probability that the stock will rise. The quantity \(\sigma\sqrt T\) is the standard deviation of the log-price change over the interval. Increasing \(\sigma\) generally raises a call’s value because the call benefits from large upward moves while its loss at maturity cannot be less than zero. This statement can be made precise for fixed \(S_0,K,r,T\): differentiating the formula gives $$ \frac{\partial C}{\partial \sigma}=S_0\phi(d_1)\sqrt T>0, $$ where \(\phi(x)=(2\pi)^{-1/2}e^{-x^2/2}\). The inequality follows from \(S_0>0\), \(T>0\), and \(\phi(x)>0\) for every real \(x\). The derivative formula follows by the chain rule and the identity \(S_0\phi(d_1)=Ke^{-rT}\phi(d_2)\), obtained by substituting the definitions of \(d_1,d_2\) into the normal density.
Time to maturity \(T\) changes both the uncertainty \(\sigma\sqrt{T}\) and the discount factor \(e^{-rT}\), as well as the distribution of \(S_T\). Its effect on a particular call is therefore not summarized by “more time always increases value” without specifying the other inputs and the setting. In this no-dividend model, \(T\) is a contractual input; for an option that can be exercised early or a stock with dividends, the valuation problem differs.
A second check is put-call parity. If \(P_{\text{put}}\) denotes the price of a European put with the same \(S_0,K,T,r,\sigma\), then
Proof. For every possible terminal stock price \(S_T\), the identity $$\max(S_T-K,0)-\max(K-S_T,0)=S_T-K$$ holds: if \(S_T\ge K\), the left side is \(S_T-K\); if \(S_T<K\), it is \(0-(K-S_T)=S_T-K\). Discounting the risk-neutral expectations of both payoffs therefore gives $$C-P_{\text{put}}=e^{-rT}\mathbb{E}_{\mathbb{Q}}[S_T-K].$$ The model gives \(\mathbb{E}_{\mathbb{Q}}[S_T]=S_0e^{rT}\), as follows from the normal exponential moment in the derivation. Also, \(K\) is constant. Hence the right side is \(e^{-rT}(S_0e^{rT}-K)=S_0-Ke^{-rT}\), as asserted. \(\square\)
Worked Example: Checking the Matching European Put
Use the previous inputs \(S_0=\sqrt e\), \(K=1\), \(r=0\), \(\sigma=1\), and \(T=1\). We found \(C=\sqrt e\,\Phi(1)-1/2\). Put-call parity gives $$ P_{\text{put}}=C-S_0+Ke^{-rT} =\sqrt e\,\Phi(1)-\frac12-\sqrt e+1 =\sqrt e\big(\Phi(1)-1\big)+\frac12. $$ Because \(\Phi(1)-1=-\Phi(-1)\), this is also \(1/2-\sqrt e\,\Phi(-1)\).
Check directly. With these inputs \(S_T=e^Z\). A put pays \(1-e^Z\) when \(Z<0\), so its expected payoff is $$ \int_{-\infty}^0(1-e^z)\frac{e^{-z^2/2}}{\sqrt{2\pi}}\,dz =\frac12-\sqrt e\,\Phi(-1). $$ The rate is zero, and this equals the parity result.
Worked Example: Volatility Changes the Call Value
Fix \(S_0=K=1\), \(r=0\), and \(T=1\). For any \(\sigma>0\), $$ d_1=\frac{\sigma^2/2}{\sigma}=\frac{\sigma}{2},\qquad d_2=-\frac{\sigma}{2}, $$ so \(C(\sigma)=\Phi(\sigma/2)-\Phi(-\sigma/2)=2\Phi(\sigma/2)-1\). At \(\sigma=1\), \(C(1)=2\Phi(1/2)-1\). At \(\sigma=2\), \(C(2)=2\Phi(1)-1\). Since the normal density is positive, \(\Phi\) is strictly increasing; as \(1>1/2\), \(C(2)>C(1)\).
Check. The difference is \(C(2)-C(1)=2\int_{1/2}^{1}\phi(x)\,dx>0\). This verifies the comparison without requiring decimal approximations.
How Fine-Step Binomial Pricing Connects
The connection uses the same risk-neutral logic as binomial valuation, with shorter and shorter time steps. In the Cox–Ross–Rubinstein binomial model, divide \(T\) into \(n\) intervals of length \(\Delta t=T/n\), and take the up and down factors to be \(u=e^{\sigma\sqrt{\Delta t}}\) and \(d=e^{-\sigma\sqrt{\Delta t}}\). The risk-neutral up probability is $$ p_n=\frac{e^{r\Delta t}-d}{u-d}. $$ It is a valid probability when \(d<e^{r\Delta t}<u\); for fixed \(r\) and \(\sigma>0\), this holds for all sufficiently small \(\Delta t\). The down probability is \(1-p_n\), so the two probabilities sum to one. This choice makes the expected one-step stock multiplier \(p_nu+(1-p_n)d=e^{r\Delta t}\).
As the steps become fine, the accumulated log returns converge in distribution to a normal random variable with mean \((r-\sigma^2/2)T\) and variance \(\sigma^2T\). This is the standard central-limit connection for the binomial approximation; its proof is not given here. Under this convergence, the discounted European call values converge to the Black–Scholes value. The payoff is unbounded, so convergence in distribution alone is not sufficient to pass expectations to the limit; the usual binomial approximation result also uses control of the tails (uniform integrability of the discounted payoffs). The formula is thus connected to fine-step binomial pricing under a specified model and limiting argument, not by treating every finite binomial tree as exactly equal to the formula.
Worked Example: A One-Step Risk-Neutral Binomial Price
Take one year, \(S_0=100\), \(K=100\), \(r=0\), and choose \(\sigma=\ln 2\) in a one-step Cox–Ross–Rubinstein tree. Then \(\Delta t=1\), \(u=e^{\ln2}=2\), and \(d=e^{-\ln2}=1/2\). The risk-neutral up probability is $$ p_1=\frac{e^0-1/2}{2-1/2}=\frac{1/2}{3/2}=\frac13, $$ and the down probability is \(1-p_1=2/3\). They sum to \(1/3+2/3=1\), and the expected stock multiplier is $$ \frac13\cdot2+\frac23\cdot\frac12=\frac23+\frac13=1=e^{r\Delta t}. $$ The up-state stock price is \(100\cdot2=200\), giving call payoff \(200-100=100\); the down-state price is \(100\cdot(1/2)=50\), giving payoff \(0\). With zero interest, the price is \((1/3)\cdot100+(2/3)\cdot0=100/3\).
Check. The one-step expected terminal stock price is \((1/3)\cdot200+(2/3)\cdot50=200/3+100/3=100\), consistent with the zero risk-free rate. The tree price is exact for this one-step model; the Black–Scholes formula describes the continuous-time model, approached through refinement rather than asserted to equal this coarse-tree price.
Assumptions and a Common Mistake
The Black–Scholes call formula is a benchmark under a collection of assumptions: trading is frictionless, there are no transaction costs or taxes, the stock can be shorted, and borrowing and lending occur at a constant risk-free rate. Trading and hedging can be adjusted continuously; the stock pays no dividends; volatility and the risk-free rate are constant; and the option is European. The model also assumes the geometric Brownian stock-price dynamics stated above. Changing these assumptions can change the appropriate pricing model or formula.
Common mistake: using a forecast probability as the pricing probability. In the one-step tree above, suppose someone substitutes a forecast up probability of \(3/5\) into the discounted payoff calculation. They obtain $$ \frac35\cdot100+\frac25\cdot0=60, $$ which is not the no-arbitrage price of this tree. Its risk-neutral probability is \(1/3\), determined by the stock returns and risk-free rate, and the price is \(100/3\). The forecast probability may describe someone’s belief about the next move, but it does not replace the probability required for risk-neutral valuation and replication.
When using the formula, first check the contract (European exercise and no dividends in this version), put all rates and times on consistent annual units, and distinguish the risk-free rate from an expected stock return. Then calculate \(d_1\) and \(d_2\), evaluate the two normal probabilities, and check the result against a relevant identity such as put-call parity. These checks do not validate the model assumptions, but they can reveal input or arithmetic errors.
Check Your Understanding
Use the definitions and assumptions in this tutorial to answer the following questions.
- In the Black–Scholes formula, what do \(S_0\), \(K\), \(T\), \(r\), and \(\sigma\) represent?
- Why does the event that a call pays a positive amount correspond to \(Z>-d_2\) in the derivation?
- For a European put and call with matching inputs, state put-call parity and identify the identity between their terminal payoffs that gives it.
- In the one-step tree with \(u=2\), \(d=1/2\), and \(r=0\), compute the risk-neutral up probability and explain why a forecast probability need not equal it.
- Name two assumptions that would need reconsideration before applying this particular formula to an option with early exercise or a dividend-paying stock.