Similarity Solutions and Scaling Structure¶

The heat equation and separation of variables sections both transformed the Black-Scholes PDE into standard forms and solved them. But neither explained why the final formula depends on the particular combinations $\ln(S/K)/(\sigma\sqrt{\tau})$ and $r\tau$ rather than on $S$, $K$, $\sigma$, $\tau$, and $r$ independently. The answer lies in scale invariance and dimensional analysis.

Similarity solutions identify combinations of variables that respect the symmetry of a PDE, potentially reducing it to an ODE. Unlike the previous methods, this section does not provide a standalone derivation of the Black-Scholes formula. Instead, it reveals the structural reason why the formula takes the form it does: $d_1$ and $d_2$ are modified similarity coordinates, and the dependence on $S/K$, $\sigma\sqrt{\tau}$, and $r\tau$ is forced by dimensional analysis. The naive similarity ansatz does not fully reduce the Black-Scholes PDE to an ODE --- the interest rate $r$ breaks exact self-similarity --- but the scaling structure it uncovers is essential for understanding asymptotic behavior and building intuition.

1. Dimensional Analysis Foundation¶

Physical Dimensions¶

In the Black-Scholes problem, the dimensional quantities are:

$S$: stock price $[\$]$
$K$: strike price $[\$]$
$t, T$: time $[T]$
$\sigma$: volatility $[T^{-1/2}]$ (since $dW_t$ scales as $\sqrt{dt}$)
$r$: interest rate $[T^{-1}]$
$V$: option value $[\$]$

Buckingham Pi Theorem¶

With $n = 6$ variables and $m = 2$ fundamental dimensions ($[\$], [T]$), we get $n - m = 4$ dimensionless groups.

Dimensionless Variables¶

Define:

\[ \pi_1 = \frac{S}{K} \quad \text{(moneyness)} \]

\[ \pi_2 = \sigma\sqrt{T-t} = \sigma\sqrt{\tau} \quad \text{(scaled time)} \]

\[ \pi_3 = r\tau \quad \text{(discount factor exponent)} \]

\[ \pi_4 = \frac{V}{K} \quad \text{(normalized value)} \]

Dimensional Reduction¶

The option value must have the form:

\[ V(S,K,t,\sigma,r,T) = K \cdot f\!\left(\frac{S}{K},\, \sigma\sqrt{\tau},\, r\tau\right) \]

This is the most general form consistent with dimensional analysis.

2. Scale Invariance of the Black-Scholes PDE¶

The PDE¶

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

Scaling Transformation¶

Consider the one-parameter family of transformations:

\[ S \to \lambda S, \quad K \to \lambda K, \quad V \to \lambda V \]

with $t, \sigma, r$ unchanged.

Invariance¶

Substituting into the PDE:

\[ \frac{\partial(\lambda V)}{\partial t} + r(\lambda S)\frac{\partial(\lambda V)}{\partial(\lambda S)} + \frac{\sigma^2(\lambda S)^2}{2}\frac{\partial^2(\lambda V)}{\partial(\lambda S)^2} - r(\lambda V) \]

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

The PDE is homogeneous of degree 1 in $(S, K, V)$.

Economic Interpretation¶

If all dollar amounts scale by $\lambda$ (change of currency), the form of the PDE does not change. This is the scale invariance or homogeneity of the market.

3. Similarity Variable for the Black-Scholes PDE¶

Log-Moneyness Variable¶

The most natural similarity variable combines space and time:

\[ \xi = \frac{\ln(S/K)}{\sigma\sqrt{\tau}} = \frac{x}{\sigma\sqrt{\tau}} \]

where $x = \ln(S/K)$ and $\tau = T - t$.

Physical Meaning¶

Numerator: log-moneyness (how far from strike in log terms)
Denominator: volatility times the time scale (uncertainty measure)
$\xi$: standardized distance from strike

For $|\xi| \approx 1$: option is at-the-money over the relevant time scale. For $|\xi| \gg 1$: option is deep in- or out-of-the-money.

Alternative Variables¶

Other common choices:

\[ \eta = \frac{\ln(S/K) + (r \pm \frac{\sigma^2}{2})\tau}{\sigma\sqrt{\tau}} \quad \text{(drift-adjusted)} \]

\[ \zeta = \frac{S}{Ke^{-r\tau}} \quad \text{(forward moneyness)} \]

Each has advantages for different problems.

4. Attempt to Reduce the Black-Scholes PDE to an ODE¶

Similarity Ansatz¶

Seek a solution of the form:

\[ V(S,t) = K \cdot g(\xi) = K \cdot g\!\left(\frac{\ln(S/K)}{\sigma\sqrt{\tau}}\right) \]

Computing Derivatives¶

Partial derivatives of $\xi$:

\[ \frac{\partial\xi}{\partial\tau} = -\frac{\ln(S/K)}{2\sigma\tau^{3/2}} = -\frac{\xi}{2\tau} \]

\[ \frac{\partial\xi}{\partial S} = \frac{1}{S\sigma\sqrt{\tau}} \]

\[ \frac{\partial^2\xi}{\partial S^2} = -\frac{1}{S^2\sigma\sqrt{\tau}} \]

Chain rule:

\[ \frac{\partial V}{\partial t} = -\frac{\partial V}{\partial\tau} = K g'(\xi)\frac{\xi}{2\tau} \]

\[ \frac{\partial V}{\partial S} = K g'(\xi) \cdot \frac{1}{S\sigma\sqrt{\tau}} \]

\[ \frac{\partial^2 V}{\partial S^2} = K g''(\xi) \cdot \frac{1}{S^2\sigma^2\tau} - K g'(\xi) \cdot \frac{1}{S^2\sigma\sqrt{\tau}} \]

Substituting into the PDE¶

\[ Kg'(\xi)\frac{\xi}{2\tau} + rS \cdot Kg'(\xi)\frac{1}{S\sigma\sqrt{\tau}} + \frac{\sigma^2 S^2}{2}\!\left[Kg''(\xi)\frac{1}{S^2\sigma^2\tau} - Kg'(\xi)\frac{1}{S^2\sigma\sqrt{\tau}}\right] - rKg(\xi) = 0 \]

Simplifying:

\[ \frac{Kg'(\xi)\xi}{2\tau} + \frac{rKg'(\xi)}{\sigma\sqrt{\tau}} + \frac{Kg''(\xi)}{2\tau} - \frac{Kg'(\xi)}{2\sigma\sqrt{\tau}} - rKg(\xi) = 0 \]

Multiply by $\tau / K$:

\[ \frac{g'(\xi)\xi}{2} + \frac{rg'(\xi)\sqrt{\tau}}{\sigma} + \frac{g''(\xi)}{2} - \frac{g'(\xi)\sqrt{\tau}}{2\sigma} - rg(\xi)\tau = 0 \]

The Ansatz Fails: Non-Similarity¶

The presence of $\tau$ prevents complete reduction to an ODE. The interest rate $r$ introduces a scale that breaks perfect similarity. This is a key structural observation: the Black-Scholes PDE is not purely self-similar, and no single-variable reduction of the form $V = Kg(\xi)$ eliminates time entirely.

5. Modified Similarity Variables and the Black-Scholes Structure¶

Dimensionless Time¶

Including the interest rate scaling yields the drift-adjusted similarity variable:

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

This is related to $d_2$ in the Black-Scholes formula.

Modified Ansatz¶

Instead of a single similarity function, try:

\[ V(S,t) = Se^{-q\tau}h(\xi_1) - Ke^{-r\tau}h(\xi_2) \]

where:

\[ \xi_1 = \frac{\ln(S/K) + (r + \frac{\sigma^2}{2})\tau}{\sigma\sqrt{\tau}} = d_1 \]

\[ \xi_2 = \frac{\ln(S/K) + (r - \frac{\sigma^2}{2})\tau}{\sigma\sqrt{\tau}} = d_2 \]

This is the Black-Scholes structure: the formula decomposes into two terms, each involving a similarity coordinate evaluated through the same function $h$.

The ODE for h¶

Both $h(\xi_1)$ and $h(\xi_2)$ satisfy the same second-order ODE:

\[ \frac{d^2h}{d\xi^2} + \xi\frac{dh}{d\xi} = 0 \]

Solution¶

Integrate once:

\[ \frac{dh}{d\xi} = Ce^{-\xi^2/2} \]

Integrate again:

\[ h(\xi) = C_1\int_{-\infty}^{\xi}e^{-s^2/2}\,ds + C_2 = C_1\sqrt{2\pi}\,N(\xi) + C_2 \]

where $N(\xi) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\xi}e^{-s^2/2}\,ds$ is the standard normal CDF. The Black-Scholes formula emerges with $h = N$ (after applying boundary conditions to fix the constants).

6. Heat Equation Similarity¶

Transformed PDE¶

In variables $x = \ln(S/K)$ and $\tau = T - t$, after removing drift and decay:

\[ \frac{\partial w}{\partial \tau} = \frac{\partial^2 w}{\partial x^2} \]

This is the heat equation.

Self-Similar Solution¶

The fundamental solution (heat kernel) has the form:

\[ w(x,\tau) = \frac{1}{\sqrt{\tau}}\,G\!\left(\frac{x}{\sqrt{\tau}}\right) \]

where $\eta = x / \sqrt{\tau}$ is the similarity variable.

Reduction to ODE¶

Substitute:

\[ \frac{\partial w}{\partial\tau} = -\frac{1}{2\tau^{3/2}}G(\eta) - \frac{\eta}{2\tau^{3/2}}G'(\eta) \]

\[ \frac{\partial^2 w}{\partial x^2} = \frac{1}{\tau^{3/2}}G''(\eta) \]

The heat equation becomes:

\[ -\frac{1}{2\tau^{3/2}}G - \frac{\eta}{2\tau^{3/2}}G' = \frac{1}{\tau^{3/2}}G'' \]

Multiply by $\tau^{3/2}$:

\[ G''(\eta) + \frac{\eta}{2}G'(\eta) + \frac{1}{2}G(\eta) = 0 \]

Solution: Gaussian¶

The solution is:

\[ G(\eta) = Ce^{-\eta^2/4} \]

Therefore:

\[ w(x,\tau) = \frac{C}{\sqrt{4\pi\tau}}\,e^{-x^2/(4\tau)} \]

This is the heat kernel (fundamental solution). Unlike the Black-Scholes PDE, the heat equation admits an exact self-similar reduction because there is no interest-rate parameter to break the scaling.

7. Explicit Example: European Call¶

Terminal Condition¶

\[ V(S,T) = (S - K)^+ = K(e^x - 1)^+ \]

In the similarity variable $\xi = x / (\sigma\sqrt{\tau})$:

\[ V(x,0) = K(e^{\sigma\sqrt{\tau}\,\xi} - 1)^+ \quad \text{at } \tau = 0 \]

The Terminal Condition Is Not Self-Similar¶

As $\tau \to 0$, the payoff $(S - K)^+$ does not collapse to a function of $\xi$ alone. The terminal condition is not self-similar, which is another reason the naive similarity ansatz cannot capture the full solution.

Resolution via Superposition¶

The full solution is a superposition over the self-similar fundamental solution:

\[ V(x,\tau) = \int_{-\infty}^{\infty}G(x-y,\tau)\,\Phi(y)\,dy \]

where $G$ is the heat kernel. The self-similar kernel acts as the building block, even though the overall solution is not self-similar.

Black-Scholes Formula¶

After transformation and integration:

\[ C(S,t) = SN(d_1) - Ke^{-r\tau}N(d_2) \]

where $d_1, d_2$ are the modified similarity variables:

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

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

The structure $N(d_1), N(d_2)$ reflects the underlying similarity symmetry of the heat equation, adapted to account for the drift introduced by the interest rate.

8. Asymptotics and Similarity¶

Short-Time Asymptotics¶

As $\tau \to 0$, the similarity variable $\xi = \ln(S/K) / (\sigma\sqrt{\tau})$ becomes:

$\xi \to +\infty$ if $S > K$ (ITM)
$\xi \to -\infty$ if $S < K$ (OTM)
$\xi = O(1)$ if $S \approx K$ (ATM)

ATM Expansion¶

For $S \approx K$ (so $\xi = O(1)$):

\[ V \approx K\left[N(d_1) - N(d_2)\right] \approx \frac{K\sigma\sqrt{\tau}}{\sqrt{2\pi}}\left[1 + O(\tau)\right] \]

This is the ATM approximation: $V \sim \sigma\sqrt{\tau}$ (time-value decay).

Deep OTM and ITM¶

For $|\xi| \gg 1$:

\[ N(d_i) \approx \begin{cases} 1 & d_i \to +\infty \\ 0 & d_i \to -\infty \end{cases} \]

Using Mill's ratio:

\[ 1 - N(x) \approx \frac{e^{-x^2/2}}{x\sqrt{2\pi}} \quad \text{for } x \gg 1 \]

This gives exponential decay in $\xi$: deep out-of-the-money options have prices that vanish exponentially in the squared similarity variable.

9. Connection to Probability Theory¶

Central Limit Theorem¶

The heat kernel:

\[ G(x,\tau) = \frac{1}{\sqrt{4\pi\tau}}\,e^{-x^2/(4\tau)} \]

is the Gaussian density with variance $2\tau$.

The similarity variable:

\[ \xi = \frac{x}{\sqrt{2\tau}} \]

is the standardized variable for the CLT.

Large Deviations¶

For $\tau \to \infty$ with $x/\tau = v$ fixed:

\[ -\frac{1}{\tau}\ln G(x,\tau) \approx \frac{v^2}{4} + \frac{1}{2}\ln(4\pi\tau) \]

The rate function $I(v) = v^2/4$ governs large deviations.

Scaling Limits¶

As $\tau \to 0$ with $\xi = x/\sqrt{\tau}$ fixed, the distribution concentrates on $\{\xi = 0\}$, i.e., $S = K$.

This is the zero-diffusion limit (small-noise asymptotics).

10. Summary¶

Key Insights¶

Black-Scholes is scale-invariant: Multiply all dollar amounts by $\lambda$, and the solution scales proportionally.
Natural similarity variable: $\xi = \ln(S/K) / (\sigma\sqrt{\tau})$ is the standardized log-moneyness.
The naive ansatz fails: The interest rate $r$ introduces a scale that prevents a pure similarity reduction of the Black-Scholes PDE to a single ODE.
Heat equation succeeds: After removing drift and discounting, the transformed PDE admits an exact self-similar reduction whose fundamental solution is the Gaussian kernel.
Black-Scholes formula as modified similarity: The formula $C = SN(d_1) - Ke^{-r\tau}N(d_2)$ decomposes into two terms, each evaluated at a drift-adjusted similarity coordinate. The function $N$ arises from the ODE $h'' + \xi h' = 0$.

The Similarity Perspective¶

Similarity solutions reveal that option prices depend on ratios, not absolutes:

Not $S$ and $K$ separately, but $S/K$ (moneyness)
Not $\tau$ and $\sigma$ separately, but $\sigma\sqrt{\tau}$ (total volatility)
Not absolute prices, but dimensionless combinations

This is the geometric essence of the Black-Scholes pricing structure. In the operator framework of the introduction, similarity analysis reveals the invariant structure of the pricing semigroup $\mathcal{P}_\tau = e^{\tau\mathcal{L}}$: its action depends only on dimensionless combinations of the problem's parameters.

Exercises¶

Exercise 1. Using the Buckingham Pi theorem, show that the Black-Scholes call price must have the form $C = K \cdot f(S/K, \sigma^2\tau, r\tau)$ for some dimensionless function $f$. Verify this by examining the Black-Scholes formula and identifying $f$ explicitly.

Solution to Exercise 1

The Black-Scholes call price depends on the dimensional quantities $S$, $K$, $\tau = T - t$, $\sigma$, and $r$. These involve two fundamental dimensions: currency $[\$]$ and time $[T]$.

By the Buckingham Pi theorem, any physical law relating $n$ dimensional quantities with $m$ independent dimensions can be expressed in terms of $n - m$ dimensionless groups. Here $n = 6$ (including $C$) and $m = 2$, giving 4 dimensionless groups:

\[ \pi_1 = \frac{S}{K}, \quad \pi_2 = \sigma^2 \tau, \quad \pi_3 = r\tau, \quad \pi_4 = \frac{C}{K} \]

The Buckingham Pi theorem requires $\pi_4 = f(\pi_1, \pi_2, \pi_3)$, i.e.,

\[ C = K \cdot f\!\left(\frac{S}{K},\, \sigma^2\tau,\, r\tau\right) \]

Now verify with the Black-Scholes formula $C = S\mathcal{N}(d_1) - Ke^{-r\tau}\mathcal{N}(d_2)$. Dividing by $K$:

\[ \frac{C}{K} = \frac{S}{K}\mathcal{N}(d_1) - e^{-r\tau}\mathcal{N}(d_2) \]

The arguments $d_1$ and $d_2$ are

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

Both depend only on $S/K$, $\sigma^2\tau$, and $r\tau$. Therefore the dimensionless function is

\[ f(x, v, \rho) = x\,\mathcal{N}\!\left(\frac{\ln x + \rho + v/2}{\sqrt{v}}\right) - e^{-\rho}\,\mathcal{N}\!\left(\frac{\ln x + \rho - v/2}{\sqrt{v}}\right) \]

where $x = S/K$, $v = \sigma^2\tau$, $\rho = r\tau$, confirming the Buckingham Pi prediction exactly.

Exercise 2. The similarity variable for the heat equation is $\xi = x / \sqrt{\tau}$. Show that if $F(x, \tau) = g(\xi)$, then substituting into $\frac{\partial F}{\partial \tau} = \frac{1}{2}\sigma^2 \frac{\partial^2 F}{\partial x^2}$ yields the ODE $-\frac{\xi}{2}g'(\xi) = \frac{\sigma^2}{2}g''(\xi)$. Solve this ODE and relate the solution to the error function.

Solution to Exercise 2

Let $\xi = x / \sqrt{\tau}$ and assume $F(x, \tau) = g(\xi)$. We compute the partial derivatives using the chain rule. Since $\xi = x\tau^{-1/2}$:

\[ \frac{\partial \xi}{\partial \tau} = -\frac{x}{2\tau^{3/2}} = -\frac{\xi}{2\tau}, \quad \frac{\partial \xi}{\partial x} = \frac{1}{\sqrt{\tau}} \]

Therefore:

\[ \frac{\partial F}{\partial \tau} = g'(\xi)\cdot\left(-\frac{\xi}{2\tau}\right) \]

\[ \frac{\partial F}{\partial x} = g'(\xi)\cdot\frac{1}{\sqrt{\tau}}, \quad \frac{\partial^2 F}{\partial x^2} = g''(\xi)\cdot\frac{1}{\tau} \]

Substituting into $\frac{\partial F}{\partial \tau} = \frac{1}{2}\sigma^2\frac{\partial^2 F}{\partial x^2}$:

\[ -\frac{\xi}{2\tau}g'(\xi) = \frac{\sigma^2}{2\tau}g''(\xi) \]

Multiplying both sides by $\tau$ yields:

\[ -\frac{\xi}{2}g'(\xi) = \frac{\sigma^2}{2}g''(\xi) \]

For the standard heat equation with $\sigma = 1$, this becomes $g'' + \xi g' = 0$. To solve, let $h = g'$:

\[ h' + \xi h = 0 \implies h(\xi) = A e^{-\xi^2/2} \]

Integrating:

\[ g(\xi) = A\int_0^{\xi} e^{-s^2/2}\,ds + B = A\sqrt{\frac{\pi}{2}}\,\mathrm{erf}\!\left(\frac{\xi}{\sqrt{2}}\right) + B \]

For general $\sigma$, the substitution $\eta = \xi/\sigma$ gives $g(\xi) = A\,\mathrm{erf}\!\left(\frac{\xi}{\sigma\sqrt{2}}\right) + B$. This is precisely the error function solution. The fundamental solution of the heat equation, $\frac{1}{\sigma\sqrt{2\pi\tau}}e^{-x^2/(2\sigma^2\tau)}$, is recovered by differentiating $g$ with respect to $x$, confirming that the Gaussian kernel arises naturally from the similarity reduction.

Exercise 3. The Black-Scholes formula depends on $S$ and $K$ only through the ratio $S/K$ (moneyness). Explain this using the scale invariance of GBM: if $S_t$ satisfies the GBM SDE, show that $\lambda S_t$ also satisfies the same SDE with the same $\mu$ and $\sigma$, and conclude that the call price must be homogeneous of degree 1 in $(S, K)$.

Solution to Exercise 3

Let $S_t$ satisfy the GBM SDE under the risk-neutral measure:

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

For any constant $\lambda > 0$, define $\tilde{S}_t = \lambda S_t$. Then:

\[ d\tilde{S}_t = \lambda\,dS_t = r(\lambda S_t)\,dt + \sigma(\lambda S_t)\,dW_t = r\tilde{S}_t\,dt + \sigma\tilde{S}_t\,dW_t \]

So $\tilde{S}_t$ satisfies the same GBM SDE with the same drift $r$ and volatility $\sigma$, confirming scale invariance.

Now consider the European call price $C(S, K, \tau) = e^{-r\tau}\mathbb{E}[(S_T - K)^+]$. For any $\lambda > 0$:

\[ C(\lambda S, \lambda K, \tau) = e^{-r\tau}\mathbb{E}[(\lambda S_T - \lambda K)^+] = \lambda\, e^{-r\tau}\mathbb{E}[(S_T - K)^+] = \lambda\, C(S, K, \tau) \]

where we used the fact that $\lambda S_T$ has the same distribution as $S_T$ starting from $\lambda S$ (by scale invariance of GBM), and homogeneity of the payoff.

This shows $C$ is homogeneous of degree 1 in $(S, K)$: $C(\lambda S, \lambda K, \tau) = \lambda C(S, K, \tau)$. Setting $\lambda = 1/K$:

\[ C(S, K, \tau) = K \cdot C\!\left(\frac{S}{K}, 1, \tau\right) \]

Therefore $C$ depends on $S$ and $K$ only through the moneyness ratio $S/K$. This is a direct consequence of the scale invariance of the underlying GBM dynamics.

Exercise 4. Show that the Black-Scholes call price satisfies Euler's identity for homogeneous functions: $C = S\frac{\partial C}{\partial S} + K\frac{\partial C}{\partial K}$. Verify this directly using the Black-Scholes formula.

Solution to Exercise 4

Euler's theorem states that if $f$ is homogeneous of degree $k$, then $\sum_i x_i \frac{\partial f}{\partial x_i} = k f$. From Exercise 3, $C(S, K, \tau)$ is homogeneous of degree 1 in $(S, K)$, so:

\[ S\frac{\partial C}{\partial S} + K\frac{\partial C}{\partial K} = C \]

We verify directly using the Black-Scholes formula $C = S\mathcal{N}(d_1) - Ke^{-r\tau}\mathcal{N}(d_2)$.

Step 1: Compute $\frac{\partial C}{\partial S}$. Note that $\frac{\partial d_1}{\partial S} = \frac{\partial d_2}{\partial S} = \frac{1}{S\sigma\sqrt{\tau}}$. Then:

\[ \frac{\partial C}{\partial S} = \mathcal{N}(d_1) + S\mathcal{N}'(d_1)\frac{1}{S\sigma\sqrt{\tau}} - Ke^{-r\tau}\mathcal{N}'(d_2)\frac{1}{S\sigma\sqrt{\tau}} \]

Using the identity $S\mathcal{N}'(d_1) = Ke^{-r\tau}\mathcal{N}'(d_2)$ (which follows from $d_1 - d_2 = \sigma\sqrt{\tau}$ and the log-normal relationship), the last two terms cancel:

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

Step 2: Compute $\frac{\partial C}{\partial K}$. Since $\frac{\partial d_1}{\partial K} = \frac{\partial d_2}{\partial K} = -\frac{1}{K\sigma\sqrt{\tau}}$:

\[ \frac{\partial C}{\partial K} = S\mathcal{N}'(d_1)\!\left(-\frac{1}{K\sigma\sqrt{\tau}}\right) - e^{-r\tau}\mathcal{N}(d_2) - Ke^{-r\tau}\mathcal{N}'(d_2)\!\left(-\frac{1}{K\sigma\sqrt{\tau}}\right) \]

Again using $S\mathcal{N}'(d_1) = Ke^{-r\tau}\mathcal{N}'(d_2)$, the first and third terms cancel:

\[ \frac{\partial C}{\partial K} = -e^{-r\tau}\mathcal{N}(d_2) \]

Step 3: Verify Euler's identity:

\[ S\frac{\partial C}{\partial S} + K\frac{\partial C}{\partial K} = S\mathcal{N}(d_1) - Ke^{-r\tau}\mathcal{N}(d_2) = C \]

This confirms Euler's identity for the degree-1 homogeneous Black-Scholes call price. $\square$

Exercise 5. The ATM approximation $C \approx 0.4\, S\sigma\sqrt{T}$ can be understood as a similarity scaling. Show that for $S = K$ and $r = 0$, the call price depends on $S$, $\sigma$, and $T$ only through the combination $S\sigma\sqrt{T}$, and determine the proportionality constant from the Black-Scholes formula.

Solution to Exercise 5

Set $S = K$ (ATM) and $r = 0$. The Black-Scholes formula becomes:

\[ C = S\mathcal{N}(d_1) - S\mathcal{N}(d_2) \]

where $d_1 = \frac{\sigma\sqrt{T}}{2}$ and $d_2 = -\frac{\sigma\sqrt{T}}{2}$.

By symmetry of the normal distribution, $\mathcal{N}(-x) = 1 - \mathcal{N}(x)$, so:

\[ C = S\mathcal{N}(d_1) - S(1 - \mathcal{N}(d_1)) = S(2\mathcal{N}(d_1) - 1) \]

Dimensional analysis: With $r = 0$ and $S = K$, the only remaining dimensional quantities are $S$ $[\$]$, $\sigma$ $[T^{-1/2}]$, and $T$ $[T]$. The unique dimensionless combination from $\sigma$ and $T$ is $\sigma\sqrt{T}$, so $C$ must have the form:

\[ C = S \cdot \psi(\sigma\sqrt{T}) \]

where $\psi(z) = 2\mathcal{N}(z/2) - 1$.

Proportionality constant: For small $z = \sigma\sqrt{T}$, expand $\mathcal{N}(z/2)$ using $\mathcal{N}(x) \approx \frac{1}{2} + \frac{x}{\sqrt{2\pi}}$ for small $x$:

\[ C \approx S\left(2\left(\frac{1}{2} + \frac{\sigma\sqrt{T}}{2\sqrt{2\pi}}\right) - 1\right) = S\cdot\frac{\sigma\sqrt{T}}{\sqrt{2\pi}} \]

Since $\frac{1}{\sqrt{2\pi}} \approx 0.3989 \approx 0.4$:

\[ C \approx 0.4\, S\sigma\sqrt{T} \]

This confirms the ATM approximation as a similarity scaling result, with the proportionality constant being exactly $1/\sqrt{2\pi}$. $\square$