Probabilistic Interpretation of the Black-Scholes Formula¶

The Black-Scholes formula is not merely a mathematical expression—it has deep probabilistic meaning. The terms $\mathcal{N}(d_1)$ and $\mathcal{N}(d_2)$ represent probabilities under different measures, and the formula can be understood as a weighted average of terminal payoffs.

Toy mechanism: two probabilities, one event

Imagine a single stock that ends up at $S_T = 120$ with probability $p$ and at $S_T = 80$ with probability $1-p$. The call payoff at $K = 100$ is $20\,\mathbf{1}_{\{S_T = 120\}}$, so under the risk-neutral measure $\mathbb{Q}$ the discounted price is $C_0 = e^{-rT}\cdot 20\,q$, where $q = \mathbb{Q}(S_T = 120)$. Now rewrite the same expression by splitting the expectation into a stock-receipt term and a strike-payment term: $C_0 = e^{-rT}\mathbb{E}^{\mathbb{Q}}[S_T\,\mathbf{1}_{\{S_T>K\}}] - Ke^{-rT}\mathbb{Q}(S_T>K)$. The strike term carries the bare probability $q$; the stock term carries the stock-weighted probability $\bar q = 120 q / \mathbb{E}^{\mathbb{Q}}[S_T]$. Two probabilities, one event — the discrete picture is exactly what $\mathcal{N}(d_2)$ and $\mathcal{N}(d_1)$ generalise in the lognormal model below.

This section reveals the probabilistic structure underlying the option pricing formula.

Where this fits

Roadmap row(s): Probability, Measure change.
Builds on: The Black-Scholes formula (statement and notation).
Feeds into: Properties and bounds (delta as $\mathcal{N}(d_1)$) and Digital option pricing (strike differentiation extracts the implied density).

Risk-Neutral Expectation¶

Recall (see § Risk-Neutral Valuation Principle and § Feynman-Kac): under $\mathbb{Q}$, $C_0 = e^{-rT}\mathbb{E}^{\mathbb{Q}}[(S_T-K)^+]$ with $S_T = S_0 e^{(r-\frac{1}{2}\sigma^2)T + \sigma W_T}$ and $\log S_T$ Gaussian with mean $\log S_0 + (r-\tfrac{1}{2}\sigma^2)T$ and variance $\sigma^2 T$. (Throughout this section, $W_t$ is Brownian under $\mathbb{Q}$; the measure change to $\mathbb{Q}^S$ below uses $\widetilde{W}_t = W_t - \sigma t$.)

1. Decomposition of Call Expectation¶

\[ \begin{aligned} \mathbb{E}^{\mathbb{Q}}[(S_T - K)^+] &= \mathbb{E}^{\mathbb{Q}}[S_T \cdot \mathbf{1}_{\{S_T > K\}}] - K\mathbb{E}^{\mathbb{Q}}[\mathbf{1}_{\{S_T > K\}}] \\ &= \mathbb{E}^{\mathbb{Q}}[S_T | S_T > K] \cdot \mathbb{Q}(S_T > K) - K \cdot \mathbb{Q}(S_T > K) \end{aligned} \]

This reveals two key probabilities we need to evaluate.

The Meaning of N(d_2)¶

Section goal: $\mathcal{N}(d_2)$ as the risk-neutral exercise probability $\mathbb{Q}(S_T > K)$.

1. Exercise Probability Under Q¶

\[ \boxed{\mathcal{N}(d_2) = \mathbb{Q}(S_T > K)} \]

Interpretation: $\mathcal{N}(d_2)$ is the risk-neutral probability that the option expires in-the-money.

2. Derivation¶

Under $\mathbb{Q}$:

\[\begin{array}{lll} S_T > K &\iff&\displaystyle \log S_T > \log K\\ &\iff&\displaystyle S_0 e^{(r - \frac{1}{2}\sigma^2)T + \sigma W_T} > K\\ &\iff&\displaystyle \sigma W_T > \log(K/S_0) - \left(r - \frac{1}{2}\sigma^2\right)T\\ &\iff&\displaystyle \frac{W_T}{\sqrt{T}} > \frac{\log(K/S_0) - (r - \frac{1}{2}\sigma^2)T}{\sigma\sqrt{T}} \end{array}\]

Since $\frac{W_T}{\sqrt{T}} \sim \mathcal{N}(0,1)$ under $\mathbb{Q}$:

\[\begin{array}{lll} \mathbb{Q}(S_T > K) &=&\displaystyle \mathbb{Q}\left(Z > \frac{\log(K/S_0) - (r - \frac{1}{2}\sigma^2)T}{\sigma\sqrt{T}}\right) \\ &=&\displaystyle \mathbb{Q}\left(Z < - \frac{\log(K/S_0) - (r - \frac{1}{2}\sigma^2)T}{\sigma\sqrt{T}}\right) \\ &=&\displaystyle \mathcal{N}(d_2) \end{array}\]

where $Z \sim \mathcal{N}(0,1)$.

3. Intuition¶

If $d_2 > 0$: More than 50% probability of finishing ITM
If $d_2 = 0$: Exactly 50% probability (ATM forward)
If $d_2 < 0$: Less than 50% probability of finishing ITM

Factors increasing $\mathcal{N}(d_2)$:

Higher current stock price $S_0$ (already ITM)
Lower strike $K$ (easier to exceed)
Higher risk-free rate $r$ (stock grows faster)
Longer time $T$ (more drift accumulation)
Higher volatility $\sigma$ (but weaker effect due to $-\frac{1}{2}\sigma^2$ term)

The Meaning of N(d_1)¶

Section goal: $\mathcal{N}(d_1)$ as the same exercise event measured under the stock-numéraire $\mathbb{Q}^S$.

1. Stock Measure Probability¶

$\mathcal{N}(d_1)$ is the probability of exercise under the stock measure $\mathbb{Q}^S$ (where the stock is used as numeraire):

\[ \boxed{\mathcal{N}(d_1) = \mathbb{Q}^S(S_T > K)} \]

2. Derivation via Measure Change¶

Three pieces are now in play: the risk-neutral measure $\mathbb{Q}$, the stock-numéraire measure $\mathbb{Q}^S$, and the Radon–Nikodym derivative $d\mathbb{Q}^S/d\mathbb{Q}$ that links them. Writing $B_t = e^{rt}$ for the deterministic bond-account numéraire, the change-of-numéraire weight takes the canonical ratio-of-numéraires form

\[ \frac{d\mathbb{Q}^S}{d\mathbb{Q}} = \frac{S_T/S_0}{B_T/B_0} = \frac{S_T\, e^{-rT}}{S_0} \]

The numerator $S_T/S_0$ is stock-denominated growth; the denominator $B_T/B_0$ is bond-denominated growth; their ratio is exactly the relative-performance weight that reweights the measure. The same template generalises to any numéraire pair $(N, M)$ via $\dfrac{d\mathbb{Q}^N}{d\mathbb{Q}^M} = \dfrac{N_T/M_T}{N_0/M_0}$. Substituting the explicit form of $S_T$ under $\mathbb{Q}$:

\[ \frac{d\mathbb{Q}^S}{d\mathbb{Q}} = \frac{S_0 e^{(r - \frac{1}{2}\sigma^2)T + \sigma W_T} \cdot e^{-rT}}{S_0} = \exp\!\left(\sigma W_T - \frac{1}{2}\sigma^2 T\right) \]

This is the exponential martingale $\mathcal{E}(\sigma W)_T$. By Girsanov's theorem, the process $\tilde{W}_t = W_t - \sigma t$ is a Brownian motion under $\mathbb{Q}^S$, which shifts the stock price drift from $r$ to $r + \sigma^2$:

\[ S_T = S_0 e^{(r + \frac{1}{2}\sigma^2)T + \sigma \tilde{W}_T} \]

where $\tilde{W}_T \sim \mathcal{N}(0, T)$ under $\mathbb{Q}^S$. Following the same standardization as for $d_2$:

\[ \begin{aligned} S_T > K &\iff S_0 e^{(r + \frac{1}{2}\sigma^2)T + \sigma\tilde{W}_T} > K \\[2mm] &\iff \sigma\tilde{W}_T > \log(K/S_0) - \left(r + \frac{1}{2}\sigma^2\right)T \\[2mm] &\iff \frac{\tilde{W}_T}{\sqrt{T}} > \frac{\log(K/S_0) - \left(r + \frac{1}{2}\sigma^2\right)T}{\sigma\sqrt{T}} \end{aligned} \]

Since $Z = \tilde{W}_T/\sqrt{T} \sim \mathcal{N}(0,1)$ under $\mathbb{Q}^S$:

\[ \begin{aligned} \mathbb{Q}^S(S_T > K) &= \mathbb{Q}^S\!\left(Z > \frac{\log(K/S_0) - \left(r + \frac{1}{2}\sigma^2\right)T}{\sigma\sqrt{T}}\right) \\[2mm] &= \mathbb{Q}^S\!\left(Z < - \frac{\log(K/S_0) - \left(r + \frac{1}{2}\sigma^2\right)T}{\sigma\sqrt{T}}\right) \\[2mm] &= \mathcal{N}(d_1) \end{aligned} \]

3. Alternative Interpretation: Delta¶

$\mathcal{N}(d_1)$ also equals the option's delta (hedge ratio):

\[ \Delta = \frac{\partial C}{\partial S} = \mathcal{N}(d_1) \]

Two distinct interpretations, one number. Delta is a replication coefficient: the share count that locally replicates the option, defined by the self-financing condition. $\mathbb{Q}^S(S_T > K)$ is a probability under the stock-numéraire measure, defined by Girsanov's theorem. They happen to be equal in the Black–Scholes model — this is an algebraic fact, not a definition — but they live in different conceptual categories. Treating delta as "literally a probability" is a useful mnemonic but conflates a hedging quantity with a measure-theoretic one. The agreement is what makes the numéraire-change derivation of the formula clean; it is not the reason delta works as a hedge.

4. Relationship Between d_1 and d_2¶

\[ d_1 = d_2 + \sigma\sqrt{T} \]

The difference $\sigma\sqrt{T}$ represents the volatility-time effect that shifts the probability when changing from risk-neutral to stock measure.

Effect: Since $d_1 > d_2$, we always have $\mathcal{N}(d_1) > \mathcal{N}(d_2)$ (higher probability under stock measure).

The Two-Term Structure¶

Section goal: how $S\mathcal{N}(d_1)$ and $Ke^{-rT}\mathcal{N}(d_2)$ split a conditional expectation across two measures.

1. Call Option¶

\[ C_0 = \underbrace{S_0\mathcal{N}(d_1)}_{\text{Expected stock receipt}} - \underbrace{Ke^{-rT}\mathcal{N}(d_2)}_{\text{Expected strike payment}} \]

Interpretation:

Strike term $Ke^{-rT}\mathcal{N}(d_2)$: discounted strike weighted by the risk-neutral exercise probability — the expected present value of payment if exercised under $\mathbb{Q}$.
Stock term $S_0\mathcal{N}(d_1)$: equals $e^{-rT}\mathbb{E}^{\mathbb{Q}}[S_T \cdot \mathbf{1}_{\{S_T > K\}}]$ (not the simpler $\mathbb{E}^{\mathbb{Q}}[S_T | S_T > K]$).

2. Conditional Expectation Decomposition¶

Define:

$p = \mathbb{Q}(S_T > K) = \mathcal{N}(d_2)$
$\bar{S} = \mathbb{E}^{\mathbb{Q}}[S_T | S_T > K]$ = conditional expected stock price

Then:

\[ \mathbb{E}^{\mathbb{Q}}[S_T \cdot \mathbf{1}_{\{S_T > K\}}] = \bar{S} \cdot p \]

After calculation (completing the square in the Gaussian integral, exactly as in Exercise 4), it turns out:

\[ \bar{S} \cdot p = \mathbb{E}^{\mathbb{Q}}[S_T \cdot \mathbf{1}_{\{S_T > K\}}] = S_0\, e^{rT}\, \mathcal{N}(d_1) \]

Note that this $\mathcal{N}(d_1)$ is not the same probability $p = \mathcal{N}(d_2)$ — the conditional-mean factor $\bar S$ shifts the relevant standardization by exactly $\sigma\sqrt{T}$, which is the Girsanov drift to the stock measure. Substituting both pieces back:

\[ C_0 = e^{-rT}\!\left[\bar{S} \cdot p - K \cdot p\right] = \underbrace{e^{-rT}\cdot S_0 e^{rT}\mathcal{N}(d_1)}_{S_0\,\mathcal{N}(d_1)} - Ke^{-rT}\mathcal{N}(d_2) = S_0\,\mathcal{N}(d_1) - Ke^{-rT}\,\mathcal{N}(d_2) \]

This is the Black–Scholes formula. The two-term structure thus has a clean reading: the strike term carries $\mathcal{N}(d_2)$ (the bare exercise probability), and the stock term carries $\mathcal{N}(d_1)$ (the same exercise probability re-weighted by the conditional expected stock value).

3. Put Option¶

\[ P_0 = \underbrace{Ke^{-rT}\mathcal{N}(-d_2)}_{\text{Expected strike receipt}} - \underbrace{S_0\mathcal{N}(-d_1)}_{\text{Expected stock surrender}} \]

Interpretation:

Strike term $Ke^{-rT}\mathcal{N}(-d_2)$:
Discounted strike received if $S_T < K$
$\mathcal{N}(-d_2) = \mathbb{Q}(S_T < K)$ = probability of exercise
Stock term $S_0\mathcal{N}(-d_1)$:
Value of stock surrendered if exercised
$\mathcal{N}(-d_1) = \mathbb{Q}^S(S_T < K)$ under stock measure

Summary of Probabilities¶

Term	Formula	Meaning	Measure
$\mathcal{N}(d_2)$	$\mathbb{Q}(S_T > K)$	Prob. call finishes ITM	Risk-neutral $\mathbb{Q}$
$\mathcal{N}(-d_2)$	$\mathbb{Q}(S_T < K)$	Prob. put finishes ITM	Risk-neutral $\mathbb{Q}$
$\mathcal{N}(d_1)$	$\mathbb{Q}^S(S_T > K)$	Prob. call finishes ITM	Stock measure $\mathbb{Q}^S$
$\mathcal{N}(-d_1)$	$\mathbb{Q}^S(S_T < K)$	Prob. put finishes ITM	Stock measure $\mathbb{Q}^S$

Key insight: Different measures give different probabilities for the same event. The stock measure "biases" probabilities toward higher stock prices because it uses the stock as numeraire.

Numerical Example¶

Section goal: a worked illustration that $\mathcal{N}(d_1) \neq \mathcal{N}(d_2)$ and quantification of the gap.

Consider:

$S_0 = 100$, $K = 100$ (ATM)
$r = 5\%$, $\sigma = 20\%$, $T = 1$ year

Step 1: Compute $d_1$ and $d_2$

\[ d_1 = \frac{\ln(100/100) + (0.05 + 0.5 \times 0.04) \times 1}{0.2 \times 1} = \frac{0 + 0.07}{0.2} = 0.35 \]

\[ d_2 = 0.35 - 0.2 = 0.15 \]

Step 2: Evaluate probabilities

\[ \mathcal{N}(d_1) = \mathcal{N}(0.35) \approx 0.6368 \]

\[ \mathcal{N}(d_2) = \mathcal{N}(0.15) \approx 0.5596 \]

Interpretation:

Risk-neutral probability of call finishing ITM: 55.96%
Stock measure probability of call finishing ITM: 63.68%
The 7.72 percentage point difference reflects the measure change

Step 3: Call price

\[ C_0 = 100 \times 0.6368 - 100 \times e^{-0.05} \times 0.5596 \approx 10.46 \]

Why Two Different Probabilities?¶

Section goal: the Girsanov drift shift and why $d_1 - d_2 = \sigma\sqrt{T}$ encodes it.

1. The Measure Change Effect¶

The shift from $\mathbb{Q}$ to $\mathbb{Q}^S$ raises the expected return of the stock from $r$ to $r + \sigma^2$:

\[ dS_t = rS_t dt + \sigma S_t dW_t \quad\longrightarrow\quad dS_t = (r + \sigma^2)S_t dt + \sigma S_t d\tilde{W}_t \]

The likelihood ratio $\frac{d\mathbb{Q}^S}{d\mathbb{Q}} = \frac{S_T e^{-rT}}{S_0}$ is exactly proportional to the terminal stock value, so $\mathbb{Q}^S$ tilts $\mathbb{Q}$ by a factor linear in $S_T$ — and that linear weighting is precisely what Girsanov converts into the drift shift from $r$ to $r + \sigma^2$.

2. Girsanov's Theorem¶

The Brownian motions under the two measures are related by $d\tilde{W}_t = dW_t + \sigma dt$, and this $\sigma dt$ adjustment is what produces $d_1 = d_2 + \sigma\sqrt{T}$.

3. Economic Intuition¶

Risk-neutral measure: Used for pricing—treats all investors as risk-neutral
Stock measure: Used for hedging—gives the delta (hedge ratio)

The difference captures the risk premium embedded in the stock.

Connection to the SDF / state-price density

The Radon–Nikodym derivative $\frac{d\mathbb{Q}}{d\mathbb{P}}$ is closely related to the stochastic discount factor $M_T$ (also called the pricing kernel or state-price density): up to discounting, $M_T = e^{-rT}\,\frac{d\mathbb{Q}}{d\mathbb{P}}$. So pricing under $\mathbb{Q}$ and pricing by SDF, $V_0 = \mathbb{E}^{\mathbb{P}}[M_T \cdot \text{Payoff}]$, are the same operation written in two languages — the first absorbs discounting and risk-adjustment into a measure change, the second into a per-state weight. The numéraire change to $\mathbb{Q}^S$ then corresponds to multiplying the SDF by $S_T$ before pricing — equivalently, expressing prices in units of stock.

Connection to Delta Hedging¶

Recall (see § Properties and Bounds and § Greeks in Black-Scholes): $\Delta_{\text{call}} = \partial C/\partial S = \mathcal{N}(d_1)$, so the same number plays two roles — replication coefficient (shares to hedge) and stock-measure exercise probability — the algebraic coincidence that makes the numéraire-change derivation clean.

Limiting Cases¶

Recall (see § Asymptotic Behavior): in the deep-ITM, deep-OTM, and ATM-forward regimes, both $\mathcal{N}(d_1)$ and $\mathcal{N}(d_2)$ collapse to $1$, $0$, or symmetrically about $\tfrac{1}{2}$ — see the summary table there for the precise limits.

Summary¶

The Black-Scholes formula has a deep probabilistic structure:

$\mathcal{N}(d_2) = \mathbb{Q}(S_T > K)$ is the risk-neutral exercise probability; $\mathcal{N}(d_1) = \mathbb{Q}^S(S_T > K)$ is the stock-measure exercise probability and equals delta.
The two-term structure pairs each measure with its natural quantity: the strike term carries $\mathcal{N}(d_2)$ (bare exercise probability), the stock term carries $\mathcal{N}(d_1)$ (exercise probability re-weighted by conditional expected stock value).
The gap $d_1 - d_2 = \sigma\sqrt{T}$ encodes the Girsanov drift between $\mathbb{Q}$ (used for pricing) and $\mathbb{Q}^S$ (used for hedging).

Option pricing is fundamentally about weighted expectations under carefully chosen probability measures.

Exercises¶

Exercise 1. Let $S_0 = 120$, $K = 110$, $r = 4\%$, $\sigma = 30\%$, $T = 0.75$ years. Compute $d_1$ and $d_2$, then evaluate $\mathcal{N}(d_1)$ and $\mathcal{N}(d_2)$. Interpret the numerical difference $\mathcal{N}(d_1) - \mathcal{N}(d_2)$ in terms of the measure change between $\mathbb{Q}$ and $\mathbb{Q}^S$.

Solution to Exercise 1

Parameters: $S_0 = 120$, $K = 110$, $r = 0.04$, $\sigma = 0.30$, $T = 0.75$.

Compute $d_1$:

\[ d_1 = \frac{\ln(120/110) + (0.04 + 0.5 \times 0.09) \times 0.75}{0.30\sqrt{0.75}} \]

\[ = \frac{0.08701 + (0.04 + 0.045) \times 0.75}{0.25981} = \frac{0.08701 + 0.06375}{0.25981} = \frac{0.15076}{0.25981} = 0.5803 \]

Compute $d_2$:

\[ d_2 = 0.5803 - 0.25981 = 0.3205 \]

Evaluate probabilities:

\[ \mathcal{N}(d_1) = \mathcal{N}(0.5803) \approx 0.7191 \]

\[ \mathcal{N}(d_2) = \mathcal{N}(0.3205) \approx 0.6257 \]

Difference: $\mathcal{N}(d_1) - \mathcal{N}(d_2) = 0.7191 - 0.6257 = 0.0934$.

Interpretation: The 9.34 percentage point gap represents the effect of changing from the risk-neutral measure $\mathbb{Q}$ to the stock measure $\mathbb{Q}^S$. Under the stock measure, the stock's drift is $r + \sigma^2 = 0.04 + 0.09 = 0.13$ instead of $r = 0.04$, which tilts the distribution toward higher stock prices and increases the probability of finishing ITM. This gap is $\mathcal{N}(d_1) - \mathcal{N}(d_2) = \mathcal{N}(d_2 + \sigma\sqrt{T}) - \mathcal{N}(d_2)$, which is approximately $\phi(d_2) \cdot \sigma\sqrt{T} \approx 0.3790 \times 0.2598 \approx 0.0985$ (close to our exact value). The gap is larger when volatility and time to maturity are large, since these amplify the drift difference between the two measures.

Exercise 2. Prove that $\mathcal{N}(d_2) = \mathbb{Q}(S_T > K)$ by starting from the log-normal distribution of $S_T$ under $\mathbb{Q}$ and standardizing the inequality $S_T > K$ to obtain a standard normal probability. Show every algebraic step.

Solution to Exercise 2

Under $\mathbb{Q}$, $S_T = S_0 \exp\left((r - \frac{1}{2}\sigma^2)T + \sigma W_T\right)$ where $W_T \sim \mathcal{N}(0, T)$.

The event $S_T > K$ is equivalent to:

\[ S_0 \exp\left(\left(r - \frac{1}{2}\sigma^2\right)T + \sigma W_T\right) > K \]

Taking logarithms:

\[ \ln S_0 + \left(r - \frac{1}{2}\sigma^2\right)T + \sigma W_T > \ln K \]

Isolating $W_T$:

\[ \sigma W_T > \ln K - \ln S_0 - \left(r - \frac{1}{2}\sigma^2\right)T \]

\[ W_T > \frac{\ln(K/S_0) - (r - \frac{1}{2}\sigma^2)T}{\sigma} \]

Dividing both sides by $\sqrt{T}$ to standardize (since $Z = W_T/\sqrt{T} \sim \mathcal{N}(0,1)$):

\[ Z > \frac{\ln(K/S_0) - (r - \frac{1}{2}\sigma^2)T}{\sigma\sqrt{T}} \]

Now define:

\[ d_2 = \frac{\ln(S_0/K) + (r - \frac{1}{2}\sigma^2)T}{\sigma\sqrt{T}} \]

The right-hand side of the inequality is:

\[ \frac{\ln(K/S_0) - (r - \frac{1}{2}\sigma^2)T}{\sigma\sqrt{T}} = \frac{-[\ln(S_0/K) + (r - \frac{1}{2}\sigma^2)T]}{\sigma\sqrt{T}} = -d_2 \]

Therefore:

\[ \mathbb{Q}(S_T > K) = \mathbb{Q}(Z > -d_2) = 1 - \mathcal{N}(-d_2) = \mathcal{N}(d_2) \]

where the last step uses the symmetry $1 - \mathcal{N}(-x) = \mathcal{N}(x)$.

Exercise 3. The stock measure $\mathbb{Q}^S$ is defined by the Radon–Nikodym derivative $\frac{d\mathbb{Q}^S}{d\mathbb{Q}} = \frac{S_T e^{-rT}}{S_0}$. Show that under $\mathbb{Q}^S$, the drift of the stock price becomes $r + \sigma^2$ instead of $r$. Use Girsanov's theorem to identify the new Brownian motion $\tilde{W}_t = W_t - \sigma t$.

Solution to Exercise 3

The Radon–Nikodym derivative is $\frac{d\mathbb{Q}^S}{d\mathbb{Q}} = \frac{S_T e^{-rT}}{S_0}$. Under $\mathbb{Q}$:

\[ S_T = S_0 e^{(r - \frac{1}{2}\sigma^2)T + \sigma W_T} \]

So:

\[ \frac{d\mathbb{Q}^S}{d\mathbb{Q}} = \frac{S_0 e^{(r - \frac{1}{2}\sigma^2)T + \sigma W_T} \cdot e^{-rT}}{S_0} = e^{-\frac{1}{2}\sigma^2 T + \sigma W_T} \]

This has the form of an exponential martingale $\mathcal{E}(\sigma W)_T = \exp(\sigma W_T - \frac{1}{2}\sigma^2 T)$.

By Girsanov's theorem, under $\mathbb{Q}^S$, the process:

\[ \tilde{W}_t = W_t - \sigma t \]

is a Brownian motion. Substituting $W_t = \tilde{W}_t + \sigma t$ into the risk-neutral SDE:

\[ dS_t = rS_t\,dt + \sigma S_t\,dW_t = rS_t\,dt + \sigma S_t(d\tilde{W}_t + \sigma\,dt) \]

\[ = (r + \sigma^2)S_t\,dt + \sigma S_t\,d\tilde{W}_t \]

Therefore under $\mathbb{Q}^S$, the drift of $S_t$ is $r + \sigma^2$ instead of $r$.

Exercise 4. Verify the conditional expectation identity

\[ e^{-rT}\mathbb{E}^{\mathbb{Q}}[S_T \cdot \mathbf{1}_{\{S_T > K\}}] = S_0 \mathcal{N}(d_1) \]

by writing $S_T = S_0 e^{(r - \frac{1}{2}\sigma^2)T + \sigma\sqrt{T} Z}$ with $Z \sim \mathcal{N}(0,1)$, substituting into the expectation, and completing the square in the exponent.

Solution to Exercise 4

We compute $e^{-rT}\mathbb{E}^{\mathbb{Q}}[S_T \cdot \mathbf{1}_{\{S_T > K\}}]$.

Write $S_T = S_0 e^{(r - \frac{1}{2}\sigma^2)T + \sigma\sqrt{T}Z}$ with $Z \sim \mathcal{N}(0,1)$.

The condition $S_T > K$ becomes $Z > -d_2$ (from Exercise 2).

\[ e^{-rT}\mathbb{E}^{\mathbb{Q}}[S_T \cdot \mathbf{1}_{\{Z > -d_2\}}] = e^{-rT} S_0 \int_{-d_2}^{\infty} e^{(r-\frac{1}{2}\sigma^2)T + \sigma\sqrt{T}z} \frac{1}{\sqrt{2\pi}}e^{-z^2/2}\,dz \]

\[ = S_0 \int_{-d_2}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}\sigma^2 T + \sigma\sqrt{T}z - z^2/2}\,dz \]

Complete the square in the exponent:

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

Substituting $u = z - \sigma\sqrt{T}$, so $dz = du$ and when $z = -d_2$, $u = -d_2 - \sigma\sqrt{T} = -d_1$:

\[ = S_0 \int_{-d_1}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-u^2/2}\,du = S_0 \mathcal{N}(d_1) \]

This confirms:

\[ e^{-rT}\mathbb{E}^{\mathbb{Q}}[S_T \cdot \mathbf{1}_{\{S_T > K\}}] = S_0\mathcal{N}(d_1) \]

Exercise 5. For a European put option, show that $\mathcal{N}(-d_2) = \mathbb{Q}(S_T < K)$ is the risk-neutral probability that the put finishes in-the-money. Using the parameters $S_0 = 90$, $K = 100$, $r = 2\%$, $\sigma = 25\%$, $T = 1$, compute the risk-neutral exercise probability for the put and compare it to the stock-measure probability $\mathcal{N}(-d_1)$.

Solution to Exercise 5

By the complementary probability, $\mathcal{N}(-d_2) = 1 - \mathcal{N}(d_2) = 1 - \mathbb{Q}(S_T > K) = \mathbb{Q}(S_T \leq K)$.

Since $S_T$ has a continuous distribution (log-normal), $\mathbb{Q}(S_T = K) = 0$, so:

\[ \mathcal{N}(-d_2) = \mathbb{Q}(S_T < K) \]

Numerical computation with $S_0 = 90$, $K = 100$, $r = 0.02$, $\sigma = 0.25$, $T = 1$:

\[ d_1 = \frac{\ln(90/100) + (0.02 + 0.03125) \times 1}{0.25} = \frac{-0.10536 + 0.05125}{0.25} = \frac{-0.05411}{0.25} = -0.2164 \]

\[ d_2 = -0.2164 - 0.25 = -0.4664 \]

Risk-neutral exercise probability for the put: $\mathcal{N}(-d_2) = \mathcal{N}(0.4664) = 0.6795$.

Stock-measure probability: $\mathcal{N}(-d_1) = \mathcal{N}(0.2164) = 0.5857$.

The risk-neutral probability ($67.95\%$) exceeds the stock-measure probability ($58.57\%$). This is because the stock measure tilts the distribution toward higher stock prices (drift $r + \sigma^2$ vs. $r$), making it less likely that $S_T < K$. The difference of $8.38$ percentage points reflects the measure change effect, consistent with $d_1 - d_2 = \sigma\sqrt{T} = 0.25$.

Exercise 6. Show that the difference $d_1 - d_2 = \sigma\sqrt{T}$ implies that the gap between stock-measure and risk-neutral exercise probabilities is always positive and increasing in both $\sigma$ and $T$. Under what market conditions does this gap become negligible? When does it become large?

Solution to Exercise 6

Since $d_1 = d_2 + \sigma\sqrt{T}$, we have:

\[ \mathcal{N}(d_1) - \mathcal{N}(d_2) = \mathcal{N}(d_2 + \sigma\sqrt{T}) - \mathcal{N}(d_2) \]

By the mean value theorem, this equals $\phi(c) \cdot \sigma\sqrt{T}$ for some $c \in (d_2, d_1)$, where $\phi > 0$. Since $\sigma > 0$ and $T > 0$, the gap is always strictly positive.

Increasing in $\sigma$: The gap $\sigma\sqrt{T}$ between $d_1$ and $d_2$ increases linearly in $\sigma$. Since $\phi$ is bounded and positive, the gap $\mathcal{N}(d_1) - \mathcal{N}(d_2)$ generally increases with $\sigma$ (especially for moderate values of $d_2$).

Increasing in $T$: Similarly, $\sigma\sqrt{T}$ increases with $T$, widening the gap.

When the gap is negligible: When $\sigma\sqrt{T} \ll 1$ (either very low volatility or very short time to maturity), $d_1 \approx d_2$ and the two probabilities are nearly equal. The two measures are "close" because the Girsanov drift adjustment $\sigma\,dt$ has little cumulative effect over a short time or with small volatility.

When the gap is large: When $\sigma\sqrt{T} \gg 1$ (high volatility, long maturity), the gap between $d_1$ and $d_2$ is large, and the stock measure assigns significantly more probability to high stock prices than the risk-neutral measure. This occurs in practice for long-dated options on volatile stocks, where the hedging ratio $\mathcal{N}(d_1)$ can substantially exceed the risk-neutral exercise probability $\mathcal{N}(d_2)$.

Exercise 7. Consider a deep out-of-the-money call with $S_0 = 50$, $K = 100$, $r = 5\%$, $\sigma = 40\%$, $T = 2$. Compute $\mathcal{N}(d_1)$, $\mathcal{N}(d_2)$, and the call price. Despite the option being far OTM, explain why the call still has significant value by referring to the probabilistic interpretation and the log-normal distribution of $S_T$.

Solution to Exercise 7

Parameters: $S_0 = 50$, $K = 100$, $r = 0.05$, $\sigma = 0.40$, $T = 2$.

\[ d_1 = \frac{\ln(50/100) + (0.05 + 0.08) \times 2}{0.40\sqrt{2}} = \frac{-0.6931 + 0.26}{0.5657} = \frac{-0.4331}{0.5657} = -0.7655 \]

\[ d_2 = -0.7655 - 0.5657 = -1.3312 \]

\[ \mathcal{N}(d_1) = \mathcal{N}(-0.7655) \approx 0.2220 \]

\[ \mathcal{N}(d_2) = \mathcal{N}(-1.3312) \approx 0.0916 \]

Call price:

\[ C_0 = 50 \times 0.2220 - 100 \times e^{-0.10} \times 0.0916 = 11.10 - 90.48 \times 0.0916 = 11.10 - 8.29 = 2.81 \]

Despite the stock being at half the strike price, the call is worth $\$2.81$ (about $5.6\%$ of the stock price).

Why the call has significant value: The risk-neutral probability of finishing ITM is $\mathcal{N}(d_2) = 9.16\%$, which is far from negligible. This is because $S_T$ follows a log-normal distribution, which is right-skewed. With $\sigma = 40\%$ and $T = 2$ years, the total uncertainty is $\sigma\sqrt{T} = 56.57\%$, meaning the stock price can easily double or more. Specifically, a $2$-standard-deviation upward move gives $S_T = 50 \times e^{0.26 + 2 \times 0.5657} \approx 50 \times e^{1.39} \approx 201$, far above the strike. The log-normal distribution places substantial probability on extreme upward moves, and these contribute disproportionately to the option's expected payoff since the payoff is linear in $S_T - K$ for $S_T > K$. The stock-measure probability $\mathcal{N}(d_1) = 22.2\%$ is even higher, reflecting the additional upward bias of the stock numéraire measure.

Term	Formula	Meaning	Measure
\(\mathcal{N}(d_2)\)	\(\mathbb{Q}(S_T > K)\)	Prob. call finishes ITM	Risk-neutral \(\mathbb{Q}\)
\(\mathcal{N}(-d_2)\)	\(\mathbb{Q}(S_T < K)\)	Prob. put finishes ITM	Risk-neutral \(\mathbb{Q}\)
\(\mathcal{N}(d_1)\)	\(\mathbb{Q}^S(S_T > K)\)	Prob. call finishes ITM	Stock measure \(\mathbb{Q}^S\)
\(\mathcal{N}(-d_1)\)	\(\mathbb{Q}^S(S_T < K)\)	Prob. put finishes ITM	Stock measure \(\mathbb{Q}^S\)