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.

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

Risk-Neutral Expectation¶

1. Fundamental Pricing Formula¶

Under the risk-neutral measure $\mathbb{Q}$, the option price is:

\[ V_0 = e^{-rT}\mathbb{E}^{\mathbb{Q}}[\text{Payoff at } T] \]

For a European call:

\[ C_0 = e^{-rT}\mathbb{E}^{\mathbb{Q}}[(S_T - K)^+] \]

2. Terminal Stock Price Distribution¶

Under $\mathbb{Q}$, the terminal stock price is:

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

where $W_T \sim \mathcal{N}(0, T)$ under $\mathbb{Q}$.

Log-normal distribution:

\[ \log S_T \sim \mathcal{N}\left(\log S_0 + \left(r - \frac{1}{2}\sigma^2\right)T, \sigma^2 T\right) \]

3. 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)¶

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 $\mathbb{Q}$:

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

where $Z \sim \mathcal{N}(0,1)$ and we used:

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

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)¶

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¶

Under the stock measure, the Radon-Nikodym derivative is:

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

Under $\mathbb{Q}^S$, the stock price dynamics shift:

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

where $\tilde{W}_T \sim \mathcal{N}(0, T)$ under $\mathbb{Q}^S$.

Following the same calculation as for $d_2$:

\[ \mathbb{Q}^S(S_T > K) = \mathcal{N}\left(\frac{\log(S_0/K) + (r + \frac{1}{2}\sigma^2)T}{\sigma\sqrt{T}}\right) = \mathcal{N}(d_1) \]

3. Alternative Interpretation: Delta¶

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

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

This is not a coincidence—the delta naturally emerges as the stock-measure probability through the replication argument.

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¶

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 $Ke^{-rT}$ multiplied by probability of exercise $\mathcal{N}(d_2)$
Expected present value of payment if exercised under $\mathbb{Q}$
Stock term $S_0\mathcal{N}(d_1)$:
Current stock price $S_0$ multiplied by $\mathcal{N}(d_1)$
This is not simply $\mathbb{E}^{\mathbb{Q}}[S_T | S_T > K]$
Rather, it equals $e^{-rT}\mathbb{E}^{\mathbb{Q}}[S_T \cdot \mathbf{1}_{\{S_T > K\}}]$ after evaluation

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), it turns out:

\[ \bar{S} \cdot p = S_0 e^{rT} \mathcal{N}(d_1) \]

Therefore:

\[ C_0 = e^{-rT}[\bar{S} \cdot p - K \cdot p] = e^{-rT}\bar{S} \cdot \mathcal{N}(d_2) - Ke^{-rT}\mathcal{N}(d_2) \]

which simplifies to the BS formula.

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¶

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

\[ \begin{aligned} C_0 &= 100 \times 0.6368 - 100 \times e^{-0.05} \times 0.5596 \\ &= 63.68 - 95.12 \times 0.5596 \\ &= 63.68 - 53.22 \\ &\approx 10.46 \end{aligned} \]

Why Two Different Probabilities?¶

1. The Measure Change Effect¶

The shift from $\mathbb{Q}$ to $\mathbb{Q}^S$ changes the drift of the stock price:

Under $\mathbb{Q}$ (risk-neutral):

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

Expected return = risk-free rate $r$

Under $\mathbb{Q}^S$ (stock measure):

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

Expected return = $r + \sigma^2$ (higher!)

The stock measure gives more weight to paths where $S_T$ is large, increasing the probability of $S_T > K$.

2. Girsanov's Theorem¶

The Brownian motions under the two measures are related by:

\[ d\tilde{W}_t = dW_t + \sigma dt \]

This drift adjustment of $\sigma dt$ shifts the distribution, causing:

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

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 Delta Hedging¶

The hedge ratio (delta) for a call is:

\[ \Delta_{\text{call}} = \frac{\partial C}{\partial S} = \mathcal{N}(d_1) \]

Interpretation: To hedge a short call position, hold $\mathcal{N}(d_1)$ shares of stock.

Why this equals the stock-measure probability: The delta emerges naturally from the replication argument, where we solve:

\[ V = \Delta S + \beta \]

The $\Delta$ that replicates the option is exactly the probability under the measure where $S$ is the numeraire.

Limiting Cases¶

1. Deep In-the-Money (S ≫ K)¶

$d_1, d_2 \to +\infty$
$\mathcal{N}(d_1), \mathcal{N}(d_2) \to 1$
Both probabilities approach 100% (certain exercise)

Call price:

\[ C \to S - Ke^{-rT} \]

(intrinsic value plus cost of carry)

2. Deep Out-of-the-Money (S ≪ K)¶

$d_1, d_2 \to -\infty$
$\mathcal{N}(d_1), \mathcal{N}(d_2) \to 0$
Both probabilities approach 0% (no exercise)

Call price:

\[ C \to 0 \]

3. At-the-Money Forward (S = Ke^-rT)¶

$d_1 = \frac{\sigma\sqrt{T}}{2}$, $d_2 = -\frac{\sigma\sqrt{T}}{2}$
$\mathcal{N}(d_1) = \mathcal{N}(-d_2) \approx 0.5 + \delta$
$\mathcal{N}(d_2) = \mathcal{N}(-d_1) \approx 0.5 - \delta$

Both probabilities are near 50%, symmetrically distributed around 0.5.

Summary¶

The Black-Scholes formula has a deep probabilistic structure:

$\mathcal{N}(d_2)$ = Risk-neutral probability of exercise = $\mathbb{Q}(S_T > K)$
$\mathcal{N}(d_1)$ = Stock-measure probability of exercise = $\mathbb{Q}^S(S_T > K)$ = Delta
Two-term structure:
Stock term: Expected stock value conditional on exercise (stock measure)
Strike term: Expected strike payment conditional on exercise (risk-neutral)
Difference $d_1 - d_2 = \sigma\sqrt{T}$: Reflects the measure change between $\mathbb{Q}$ and $\mathbb{Q}^S$
Economic meaning:
$\mathbb{Q}$ for pricing (neutral to risk)
$\mathbb{Q}^S$ for hedging (captures risk exposure)

This probabilistic interpretation reveals that 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 \(\mathbb{Q}^S\)