The Black-Scholes Formula¶

The Black-Scholes formula, derived independently by Fischer Black, Myron Scholes (1973), and Robert Merton (1973), provides a closed-form solution for pricing European options. It remains one of the most influential results in financial economics, earning Scholes and Merton the 1997 Nobel Prize in Economics.

This section presents the formula, its underlying assumptions, and the meaning of its components.

Model Setup and Assumptions¶

1. Asset Price Dynamics¶

The underlying asset price $S_t$ follows geometric Brownian motion under the risk-neutral measure $\mathbb{Q}$:

\[ dS_t = rS_t dt + \sigma S_t dW_t \]

where: - $r$ = risk-free interest rate (continuously compounded, constant) - $\sigma$ = volatility (constant) - $W_t$ = standard Brownian motion under $\mathbb{Q}$

Equivalent representation:

\[ S_t = S_0 \exp\left(\left(r - \frac{1}{2}\sigma^2\right)t + \sigma W_t\right) \]

2. Fundamental Assumptions¶

Frictionless markets: No transaction costs, taxes, or restrictions on short selling
Continuous trading: Assets can be traded continuously in time
Constant parameters: Risk-free rate $r$ and volatility $\sigma$ are known constants
No dividends: The underlying asset pays no dividends during the option's life
European exercise: Options can only be exercised at maturity $T$
Complete markets: No arbitrage opportunities exist
Log-normal distribution: Asset returns are normally distributed

3. Contract Specifications¶

$S_0$ = current asset price (at time $t=0$)
$K$ = strike price (exercise price)
$T$ = time to maturity (in years)
$\tau = T - t$ = time remaining (for pricing at time $t$)

The Black-Scholes Formulas¶

1. European Call Option¶

The price of a European call option at time $t$ is:

\[ \boxed{C(S,t) = S\mathcal{N}(d_1) - Ke^{-r(T-t)}\mathcal{N}(d_2)} \]

where:

\[ \boxed{d_1 = \frac{\ln(S/K) + (r + \frac{1}{2}\sigma^2)(T-t)}{\sigma\sqrt{T-t}}} \]

\[ \boxed{d_2 = d_1 - \sigma\sqrt{T-t} = \frac{\ln(S/K) + (r - \frac{1}{2}\sigma^2)(T-t)}{\sigma\sqrt{T-t}}} \]

Notation: $\mathcal{N}(x)$ denotes the cumulative distribution function of the standard normal distribution:

\[ \mathcal{N}(x) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^x e^{-\frac{z^2}{2}} dz = \mathbb{P}(Z \leq x) \quad \text{where } Z \sim \mathcal{N}(0,1) \]

2. European Put Option¶

The price of a European put option at time $t$ is:

\[ \boxed{P(S,t) = Ke^{-r(T-t)}\mathcal{N}(-d_2) - S\mathcal{N}(-d_1)} \]

where $d_1$ and $d_2$ are defined identically as for the call.

Alternative form using $\mathcal{N}(-x) = 1 - \mathcal{N}(x)$:

\[ P(S,t) = Ke^{-r(T-t)}[1 - \mathcal{N}(d_2)] - S[1 - \mathcal{N}(d_1)] \]

Component Analysis¶

1. The d_1 Parameter¶

\[ d_1 = \frac{\ln(S/K) + (r + \frac{1}{2}\sigma^2)(T-t)}{\sigma\sqrt{T-t}} \]

Structure: - Numerator: Log-moneyness $\ln(S/K)$ plus drift-adjusted growth $(r + \frac{1}{2}\sigma^2)(T-t)$ - Denominator: Volatility-adjusted time $\sigma\sqrt{T-t}$

Interpretation: Measures how many standard deviations the expected log-price is above the strike (in standardized units).

Components: 1. $\ln(S/K)$: Current log-moneyness - $> 0$ if $S > K$ (in-the-money) - $= 0$ if $S = K$ (at-the-money) - $< 0$ if $S < K$ (out-of-the-money)

$(r + \frac{1}{2}\sigma^2)(T-t)$: Expected growth under risk-neutral measure
$r(T-t)$: Risk-free growth
$\frac{1}{2}\sigma^2(T-t)$: Volatility adjustment (Itô correction)
$\sigma\sqrt{T-t}$: Total uncertainty (volatility × $\sqrt{\text{time}}$)

2. The d_2 Parameter¶

\[ d_2 = d_1 - \sigma\sqrt{T-t} \]

Relationship to $d_1$: - $d_2$ is always less than $d_1$ by exactly the "volatility-time product" - As volatility or time increases, the gap $d_1 - d_2$ widens

Interpretation: Related to the risk-neutral probability of exercise (explained in next section).

3. The Normal CDF N(·)¶

Properties: - $\mathcal{N}(0) = 0.5$ (median) - $\mathcal{N}(x) \to 1$ as $x \to \infty$ - $\mathcal{N}(x) \to 0$ as $x \to -\infty$ - $\mathcal{N}(-x) = 1 - \mathcal{N}(x)$ (symmetry)

Computational note: $\mathcal{N}(x)$ has no closed-form expression but is efficiently computed using polynomial approximations or built-in functions in numerical software.

Standard values: | $x$ | $\mathcal{N}(x)$ | |-----|------------------| | -3 | 0.0013 | | -2 | 0.0228 | | -1 | 0.1587 | | 0 | 0.5000 | | 1 | 0.8413 | | 2 | 0.9772 | | 3 | 0.9987 |

Formula Structure¶

1. Call Option Decomposition¶

\[ C = \underbrace{S\mathcal{N}(d_1)}_{\text{Stock term}} - \underbrace{Ke^{-r(T-t)}\mathcal{N}(d_2)}_{\text{Strike term}} \]

Interpretation: - Stock term $S\mathcal{N}(d_1)$: Expected present value of receiving the stock if exercised - Strike term $Ke^{-r(T-t)}\mathcal{N}(d_2)$: Expected present value of paying the strike if exercised

The call value is the difference: value received minus value paid.

2. Put Option Decomposition¶

\[ P = \underbrace{Ke^{-r(T-t)}\mathcal{N}(-d_2)}_{\text{Strike term}} - \underbrace{S\mathcal{N}(-d_1)}_{\text{Stock term}} \]

Interpretation: - Strike term $Ke^{-r(T-t)}\mathcal{N}(-d_2)$: Expected present value of receiving the strike if exercised - Stock term $S\mathcal{N}(-d_1)$: Expected present value of giving up the stock if exercised

The put value is the difference: value received minus value given up.

3. Symmetry¶

Notice the structural symmetry: - Call uses $\mathcal{N}(d_1)$ and $\mathcal{N}(d_2)$ - Put uses $\mathcal{N}(-d_1)$ and $\mathcal{N}(-d_2)$ - Both involve the same $d_1$ and $d_2$ parameters

This symmetry reflects the underlying put-call parity relationship.

Moneyness Classification¶

1. Definitions¶

For a call option with strike $K$:

In-the-money (ITM): $S > K$
Intrinsic value = $S - K > 0$
Would have positive payoff if exercised immediately
At-the-money (ATM): $S \approx K$
Intrinsic value $\approx 0$
Most sensitive to volatility changes
Out-of-the-money (OTM): $S < K$
Intrinsic value = $0$
Pure time value

For a put: ITM when $S < K$, OTM when $S > K$.

2. Relationship to d_1 and d_2¶

Moneyness	$\ln(S/K)$	$d_1, d_2$	$\mathcal{N}(d_1), \mathcal{N}(d_2)$
Deep OTM	$\ll 0$	Large negative	Close to 0
OTM	$< 0$	Negative	$< 0.5$
ATM	$\approx 0$	Small	$\approx 0.5$
ITM	$> 0$	Positive	$> 0.5$
Deep ITM	$\gg 0$	Large positive	Close to 1

Special Cases¶

1. At-the-Money Forward (ATMF)¶

When $S = Ke^{-r(T-t)}$ (current price equals discounted strike):

\[ d_1 = \frac{\frac{1}{2}\sigma^2(T-t)}{\sigma\sqrt{T-t}} = \frac{\sigma\sqrt{T-t}}{2} \]

\[ d_2 = -\frac{\sigma\sqrt{T-t}}{2} \]

The call and put have symmetric probabilities around $\mathcal{N}(0) = 0.5$.

2. At Maturity (T - t = 0)¶

As $t \to T$: - If $S > K$: $d_1, d_2 \to +\infty$, so $C \to S - K$, $P \to 0$ - If $S < K$: $d_1, d_2 \to -\infty$, so $C \to 0$, $P \to K - S$ - If $S = K$: $d_1, d_2$ undefined, but limits give $C \to 0$, $P \to 0$

This recovers the terminal payoff: $C(S,T) = (S-K)^+$, $P(S,T) = (K-S)^+$.

3. Zero Volatility (σ → 0)¶

The formulas reduce to the intrinsic value discounted at the risk-free rate:

\[ C = \max(S - Ke^{-r(T-t)}, 0) \]

\[ P = \max(Ke^{-r(T-t)} - S, 0) \]

This is the forward value with no uncertainty premium.

Comparison with Binomial Model¶

The Black-Scholes formula is the continuous-time limit of the binomial model:

Binomial Model	Black-Scholes Model
Discrete time steps	Continuous time
Two possible outcomes per step	Infinitesimal changes
Risk-neutral probability $q$	Risk-neutral measure $\mathbb{Q}$
Backward induction	PDE or expectation
$\mathcal{N}(d_2) \approx$ binomial probability	$\mathcal{N}(d_2) = \lim_{n\to\infty}$ binomial

As the number of time steps $n \to \infty$ in the binomial model with appropriate parameter scaling:

\[ \text{Binomial price} \to \text{Black-Scholes price} \]

The Black-Scholes PDE¶

1. PDE Formulation¶

The Black-Scholes formula is the solution to a partial differential equation:

\[ \boxed{\frac{\partial V}{\partial t} + \frac{1}{2}\sigma^2 S^2 \frac{\partial^2 V}{\partial S^2} + rS\frac{\partial V}{\partial S} - rV = 0} \]

with terminal condition:

\[ V(S,T) = \text{Payoff}(S) = \begin{cases} (S-K)^+ & \text{for call} \\ (K-S)^+ & \text{for put} \end{cases} \]

Interpretation: Any replicable derivative whose value is $V(S,t)$ must satisfy this PDE. The equation describes how the option value evolves over time given the stock price dynamics.

2. Trivial Solutions¶

Before solving for option prices, observe that simple portfolios satisfy the PDE:

1. The stock itself: $V = S$

Verify:

\[ \frac{\partial S}{\partial t} = 0, \quad \frac{\partial S}{\partial S} = 1, \quad \frac{\partial^2 S}{\partial S^2} = 0 \]

\[ 0 + \frac{1}{2}\sigma^2 S^2 \cdot 0 + rS \cdot 1 - rS = 0 \quad \checkmark \]

2. The risk-free bond: $V = e^{rt}$

Verify:

\[ \frac{\partial e^{rt}}{\partial t} = re^{rt}, \quad \frac{\partial e^{rt}}{\partial S} = 0, \quad \frac{\partial^2 e^{rt}}{\partial S^2} = 0 \]

\[ re^{rt} + 0 + 0 - re^{rt} = 0 \quad \checkmark \]

These trivial solutions confirm that the PDE correctly describes basic traded assets. Any linear combination of stock and bond also satisfies the PDE, which forms the basis of the replication argument.

3. Connection to Option Pricing¶

The Black-Scholes call and put formulas are non-trivial solutions to this PDE with the appropriate terminal conditions. Section 2.4 will derive the formula by solving this PDE using various analytical techniques.

Why This Formula?¶

The Black-Scholes formula can be derived via multiple methods (covered in subsequent sections):

PDE approach (Section 2.4): Solve the Black-Scholes partial differential equation (introduced in Section 8 above)
Risk-neutral expectation (Section 2.5): Compute $e^{-rT}\mathbb{E}^{\mathbb{Q}}[(S_T - K)^+]$ using various analytical methods
Replication (Section 2.4): Construct a self-financing portfolio that replicates the option payoff
Martingale theory (Section 2.2): Apply Girsanov's theorem and martingale pricing

Each method provides different insights into why this particular formula emerges from the no-arbitrage principle.

Summary¶

The Black-Scholes formula for European options:

Call: $C = S\mathcal{N}(d_1) - Ke^{-r(T-t)}\mathcal{N}(d_2)$

Put: $P = Ke^{-r(T-t)}\mathcal{N}(-d_2) - S\mathcal{N}(-d_1)$

where

\[ d_1 = \frac{\ln(S/K) + (r + \frac{1}{2}\sigma^2)(T-t)}{\sigma\sqrt{T-t}}, \quad d_2 = d_1 - \sigma\sqrt{T-t} \]

Key features: - Closed-form solution (no numerical iteration needed) - Depends on five observable inputs: $S$, $K$, $T-t$, $r$, $\sigma$ - $\mathcal{N}(d_1)$ and $\mathcal{N}(d_2)$ have probabilistic interpretations - Decomposes into "stock term" and "strike term" - Satisfies put-call parity, boundary conditions, and no-arbitrage constraints

The formula's elegance and practical utility have made it ubiquitous in financial markets, despite its simplifying assumptions.

Exercises¶

Exercise 1. A European call option has the following parameters: $S_0 = 80$, $K = 85$, $r = 3\%$, $\sigma = 25\%$, and $T = 0.5$ years. Compute $d_1$, $d_2$, and the Black-Scholes call price $C_0$. Classify the option as ITM, ATM, or OTM.

Solution to Exercise 1

Parameters: $S_0 = 80$, $K = 85$, $r = 0.03$, $\sigma = 0.25$, $T = 0.5$.

Step 1: Compute $d_1$

\[ d_1 = \frac{\ln(80/85) + (0.03 + 0.5 \times 0.0625) \times 0.5}{0.25\sqrt{0.5}} \]

Numerator: $\ln(0.9412) + (0.03 + 0.03125) \times 0.5 = -0.06062 + 0.030625 = -0.02999$.

Denominator: $0.25 \times 0.7071 = 0.17678$.

\[ d_1 = \frac{-0.02999}{0.17678} = -0.1697 \]

Step 2: Compute $d_2$

\[ d_2 = -0.1697 - 0.17678 = -0.3465 \]

Step 3: Evaluate cumulative normal values

\[ \mathcal{N}(d_1) = \mathcal{N}(-0.1697) \approx 0.4326 \]

\[ \mathcal{N}(d_2) = \mathcal{N}(-0.3465) \approx 0.3645 \]

Step 4: Compute call price

\[ C_0 = 80 \times 0.4326 - 85 \times e^{-0.03 \times 0.5} \times 0.3645 \]

\[ = 34.61 - 85 \times 0.9851 \times 0.3645 = 34.61 - 30.52 = 4.09 \]

Classification: Since $S_0 = 80 < K = 85$, the call is out-of-the-money (OTM). The entire value of $4.09 is time value.

Exercise 2. Verify that the Black-Scholes call formula satisfies the lower bound $C \geq \max(S - Ke^{-rT}, 0)$ for the parameters $S_0 = 100$, $K = 90$, $r = 5\%$, $\sigma = 30\%$, $T = 1$. Compute both sides explicitly.

Solution to Exercise 2

Parameters: $S_0 = 100$, $K = 90$, $r = 0.05$, $\sigma = 0.30$, $T = 1$.

Lower bound: $\max(S_0 - Ke^{-rT}, 0) = \max(100 - 90e^{-0.05}, 0) = \max(100 - 85.61, 0) = 14.39$.

Compute Black-Scholes price:

\[ d_1 = \frac{\ln(100/90) + (0.05 + 0.045) \times 1}{0.30} = \frac{0.10536 + 0.095}{0.30} = \frac{0.20036}{0.30} = 0.6679 \]

\[ d_2 = 0.6679 - 0.30 = 0.3679 \]

\[ \mathcal{N}(0.6679) \approx 0.7479, \quad \mathcal{N}(0.3679) \approx 0.6436 \]

\[ C_0 = 100 \times 0.7479 - 90 \times e^{-0.05} \times 0.6436 = 74.79 - 85.61 \times 0.6436 = 74.79 - 55.10 = 19.69 \]

Verification:

Upper bound: $C_0 = 19.69 \leq 100 = S_0$ ✓
Lower bound: $C_0 = 19.69 \geq 14.39 = S_0 - Ke^{-rT}$ ✓

Both bounds are satisfied.

Exercise 3. Starting from the Black-Scholes call formula $C = S\mathcal{N}(d_1) - Ke^{-r(T-t)}\mathcal{N}(d_2)$, derive the put formula

\[ P = Ke^{-r(T-t)}\mathcal{N}(-d_2) - S\mathcal{N}(-d_1) \]

using put-call parity $C - P = S - Ke^{-r(T-t)}$ and the identity $\mathcal{N}(-x) = 1 - \mathcal{N}(x)$.

Solution to Exercise 3

Starting from put-call parity:

\[ C - P = S - Ke^{-r(T-t)} \]

Solving for $P$:

\[ P = C - S + Ke^{-r(T-t)} \]

Substituting the Black-Scholes call formula:

\[ P = S\mathcal{N}(d_1) - Ke^{-r(T-t)}\mathcal{N}(d_2) - S + Ke^{-r(T-t)} \]

Rearranging:

\[ P = -S[1 - \mathcal{N}(d_1)] + Ke^{-r(T-t)}[1 - \mathcal{N}(d_2)] \]

Using the identity $\mathcal{N}(-x) = 1 - \mathcal{N}(x)$:

\[ 1 - \mathcal{N}(d_1) = \mathcal{N}(-d_1), \quad 1 - \mathcal{N}(d_2) = \mathcal{N}(-d_2) \]

Therefore:

\[ P = Ke^{-r(T-t)}\mathcal{N}(-d_2) - S\mathcal{N}(-d_1) \]

This is exactly the Black-Scholes put formula.

Exercise 4. Consider the at-the-money forward case where $S = Ke^{-r(T-t)}$. Show that in this case $d_1 = \frac{\sigma\sqrt{T-t}}{2}$ and $d_2 = -\frac{\sigma\sqrt{T-t}}{2}$. Derive an approximate formula for the ATM forward call price when $\sigma\sqrt{T-t}$ is small, using the Taylor expansion $\mathcal{N}(x) \approx \frac{1}{2} + \frac{x}{\sqrt{2\pi}}$ for small $x$.

Solution to Exercise 4

When $S = Ke^{-r(T-t)}$, we have $\ln(S/K) = -r(T-t)$.

\[ d_1 = \frac{-r(T-t) + (r + \frac{1}{2}\sigma^2)(T-t)}{\sigma\sqrt{T-t}} = \frac{\frac{1}{2}\sigma^2(T-t)}{\sigma\sqrt{T-t}} = \frac{\sigma\sqrt{T-t}}{2} \]

\[ d_2 = d_1 - \sigma\sqrt{T-t} = \frac{\sigma\sqrt{T-t}}{2} - \sigma\sqrt{T-t} = -\frac{\sigma\sqrt{T-t}}{2} \]

Let $\epsilon = \frac{\sigma\sqrt{T-t}}{2}$, which is small when $\sigma\sqrt{T-t}$ is small. Using $\mathcal{N}(x) \approx \frac{1}{2} + \frac{x}{\sqrt{2\pi}}$:

\[ C = S\mathcal{N}(\epsilon) - Ke^{-r(T-t)}\mathcal{N}(-\epsilon) \]

Since $S = Ke^{-r(T-t)}$, denote this common value by $F$:

\[ C = F\left(\frac{1}{2} + \frac{\epsilon}{\sqrt{2\pi}}\right) - F\left(\frac{1}{2} - \frac{\epsilon}{\sqrt{2\pi}}\right) = \frac{2F\epsilon}{\sqrt{2\pi}} \]

Substituting back $\epsilon = \frac{\sigma\sqrt{T-t}}{2}$ and $F = Ke^{-r(T-t)}$:

\[ C_{\text{ATMF}} \approx \frac{Ke^{-r(T-t)} \sigma\sqrt{T-t}}{\sqrt{2\pi}} \approx 0.3989 \cdot Ke^{-r(T-t)} \cdot \sigma\sqrt{T-t} \]

Exercise 5. Show that the Black-Scholes formula recovers the correct terminal payoff. That is, prove that as $t \to T$:

\[ C(S, t) \to (S - K)^+ \quad \text{and} \quad P(S, t) \to (K - S)^+ \]

by analyzing the limits of $d_1$ and $d_2$ separately for the cases $S > K$, $S < K$, and $S = K$.

Solution to Exercise 5

Let $\tau = T - t \to 0$.

Case 1: $S > K$

\[ d_1 = \frac{\ln(S/K) + (r + \frac{1}{2}\sigma^2)\tau}{\sigma\sqrt{\tau}} \]

Since $\ln(S/K) > 0$ is fixed and $\sigma\sqrt{\tau} \to 0$, we get $d_1 \to +\infty$. Similarly $d_2 \to +\infty$.

\[ C \to S \cdot 1 - K \cdot e^0 \cdot 1 = S - K = (S-K)^+ \]

\[ P \to K \cdot 0 - S \cdot 0 = 0 = (K-S)^+ \]

Case 2: $S < K$

Now $\ln(S/K) < 0$, so $d_1 \to -\infty$ and $d_2 \to -\infty$.

\[ C \to S \cdot 0 - K \cdot 0 = 0 = (S-K)^+ \]

\[ P \to K \cdot 1 - S \cdot 1 = K - S = (K-S)^+ \]

Case 3: $S = K$

\[ d_1 = \frac{(r + \frac{1}{2}\sigma^2)\tau}{\sigma\sqrt{\tau}} = \frac{(r + \frac{1}{2}\sigma^2)\sqrt{\tau}}{\sigma} \to 0 \]

Similarly $d_2 \to 0$. So $\mathcal{N}(d_1), \mathcal{N}(d_2) \to \frac{1}{2}$.

\[ C \to K \cdot \frac{1}{2} - K \cdot 1 \cdot \frac{1}{2} = 0 = (S-K)^+ \]

\[ P \to K \cdot \frac{1}{2} - K \cdot \frac{1}{2} = 0 = (K-S)^+ \]

In all three cases, $C(S,t) \to (S-K)^+$ and $P(S,t) \to (K-S)^+$ as $t \to T$.

Exercise 6. Verify that $V(S, t) = S$ (holding the stock) and $V(S, t) = e^{rt}$ (the risk-free bond) both satisfy the Black-Scholes PDE

\[ \frac{\partial V}{\partial t} + \frac{1}{2}\sigma^2 S^2 \frac{\partial^2 V}{\partial S^2} + rS\frac{\partial V}{\partial S} - rV = 0 \]

by computing each partial derivative and substituting.

Solution to Exercise 6

For $V(S,t) = S$:

\[ \frac{\partial V}{\partial t} = 0, \quad \frac{\partial V}{\partial S} = 1, \quad \frac{\partial^2 V}{\partial S^2} = 0 \]

Substituting into the PDE:

\[ 0 + \frac{1}{2}\sigma^2 S^2 \cdot 0 + rS \cdot 1 - rS = rS - rS = 0 \quad \checkmark \]

For $V(S,t) = e^{rt}$:

\[ \frac{\partial V}{\partial t} = re^{rt}, \quad \frac{\partial V}{\partial S} = 0, \quad \frac{\partial^2 V}{\partial S^2} = 0 \]

Substituting into the PDE:

\[ re^{rt} + \frac{1}{2}\sigma^2 S^2 \cdot 0 + rS \cdot 0 - re^{rt} = re^{rt} - re^{rt} = 0 \quad \checkmark \]

Both trivial solutions satisfy the Black-Scholes PDE.

Exercise 7. In the zero-volatility limit $\sigma \to 0$, explain why the call price reduces to $C = \max(S - Ke^{-r(T-t)}, 0)$. What happens to $d_1$ and $d_2$ in this limit when $S > Ke^{-r(T-t)}$? When $S < Ke^{-r(T-t)}$? Relate your answer to the deterministic evolution of the stock price when $\sigma = 0$.

Solution to Exercise 7

When $\sigma = 0$, the stock evolves deterministically: $S_T = Se^{r(T-t)}$. There is no randomness, so the option payoff is known with certainty.

When $S > Ke^{-r(T-t)}$, equivalently $Se^{r(T-t)} > K$:

\[ d_1 = \frac{\ln(S/K) + r(T-t)}{\sigma\sqrt{T-t}} + \frac{\sigma\sqrt{T-t}}{2} \]

The numerator $\ln(S/K) + r(T-t) = \ln(Se^{r(T-t)}/K) > 0$ since $Se^{r(T-t)} > K$. Dividing by $\sigma\sqrt{T-t} \to 0^+$ gives $d_1 \to +\infty$. Similarly $d_2 \to +\infty$.

\[ C \to S \cdot 1 - Ke^{-r(T-t)} \cdot 1 = S - Ke^{-r(T-t)} \]

When $S < Ke^{-r(T-t)}$, equivalently $Se^{r(T-t)} < K$:

The numerator $\ln(Se^{r(T-t)}/K) < 0$, so dividing by $\sigma\sqrt{T-t} \to 0^+$ gives $d_1 \to -\infty$ and $d_2 \to -\infty$.

\[ C \to S \cdot 0 - Ke^{-r(T-t)} \cdot 0 = 0 \]

Combining: $C \to \max(S - Ke^{-r(T-t)}, 0)$.

This makes sense because with $\sigma = 0$, the stock grows deterministically at rate $r$, reaching $Se^{r(T-t)}$ at maturity. The call payoff is $(Se^{r(T-t)} - K)^+ = (S - Ke^{-r(T-t)})^+ \cdot e^{r(T-t)}$, and discounting back gives $\max(S - Ke^{-r(T-t)}, 0)$. Without randomness, there is no option premium beyond the forward intrinsic value.

Moneyness	\(\ln(S/K)\)	\(d_1, d_2\)	\(\mathcal{N}(d_1), \mathcal{N}(d_2)\)
Deep OTM	\(\ll 0\)	Large negative	Close to 0
OTM	\(< 0\)	Negative	\(< 0.5\)
ATM	\(\approx 0\)	Small	\(\approx 0.5\)
ITM	\(> 0\)	Positive	\(> 0.5\)
Deep ITM	\(\gg 0\)	Large positive	Close to 1