Similarity Solutions and Scaling Structure¶

Similarity solutions are fundamentally geometric: they say that a whole one-parameter family of solutions to a PDE is generated by a single universal profile, glued together by a rescaling. The Black–Scholes formula contains exactly such a hidden profile — once we know what to look for, the appearance of $\ln(S/K) / (\sigma\sqrt{\tau})$ and $r\tau$ becomes geometric necessity rather than algebraic coincidence.

We build this in the order a reader naturally builds intuition: a concrete visual toy first, then the dimensional bookkeeping that explains why the toy works, and only then the Black–Scholes specifics.

1. A Toy Picture: Heat Kernel Snapshots Collapse¶

Take the simplest possible diffusion problem: the heat equation $u_t = u_{xx}$ with point-source initial condition $u(x, 0) = \delta(x)$. The solution at any positive time is

\[ G(x, t) = \frac{1}{\sqrt{4\pi t}}\, \exp\!\left(-\frac{x^2}{4 t}\right) \]

— a Gaussian with width $\sqrt{2t}$ and peak height $1 / \sqrt{4\pi t}$. Plot snapshots at $t = 0.05, 0.2, 1.0$ side by side in the original $(x, G)$ plane and they look like three completely different curves: a tall narrow spike, a moderate bump, a low flat hill.

1.1 The Collapse¶

Now rescale. Introduce the similarity variable and rescaled height

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

The right-hand side does not depend on $t$ at all. Every snapshot — at every time — lies on the same universal Gaussian curve when plotted in $(\eta, \sqrt{t}\, G)$. The time-dependence has been completely absorbed into the rescaling.

Imagine: left panel, three Gaussians at $t = 0.05, 0.2, 1.0$ in raw coordinates — tall narrow spike, moderate bump, low hill. Right panel, the same three curves in $(\eta, \sqrt{t}\, G)$ — they lie exactly on top of one another, indistinguishable from a single Gaussian $\tfrac{1}{\sqrt{4\pi}} e^{-\eta^2 / 4}$.

This collapse is the geometric content of self-similarity: a single profile $\tilde G(\eta)$ generates the entire one-parameter family of solutions $G(x, t)$, by scaling alone.

1.2 Why $x / \sqrt{t}$ and Not Some Other Combination?¶

Two clean reasons converge on the same answer.

Dimensional. The heat equation $u_t = u_{xx}$ has no parameter with dimensions of time. The only dimensional inputs are $x$ (length) and $t$ (time), and the unique dimensionless combination is $x / \sqrt{t}$. Anything we compute can depend on $(x, t)$ only through this ratio.

Symmetry. The heat equation is invariant under the parabolic rescaling $(x, t) \mapsto (\lambda x, \lambda^2 t)$ — stretching space by $\lambda$ and time by $\lambda^2$ leaves the PDE unchanged. The combination $\eta = x / \sqrt{t}$ is the unique invariant of this rescaling, so any solution that also respects the rescaling can depend on $(x, t)$ only through $\eta$.

Why Gaussian? Substituting $G(x, t) = t^{-1/2} \tilde G(\eta)$ into the heat equation reduces it to the ODE $\tilde G''(\eta) + \tfrac{1}{2}\eta\, \tilde G'(\eta) + \tfrac{1}{2}\tilde G(\eta) = 0$ (Exercise 2 below). The unique decaying solution of integral $1$ is $\tilde G(\eta) = \tfrac{1}{\sqrt{4\pi}} e^{-\eta^2/4}$. The Gaussian kernel is therefore forced by similarity alone — independent of the kernel formula derived in § Heat Equation.

1.3 How Much of This Survives for Black–Scholes?¶

This subsubsection answers that question. Briefly:

The scaling logic survives in full. Option prices are dimensionless functions of $S/K$, $\sigma\sqrt{\tau}$, and $r\tau$ — no other combinations appear — and the variables $d_1, d_2$ are drift-adjusted versions of $\eta$.
The strict ODE reduction does not survive. The interest rate $r$ has dimensions of $[T^{-1}]$ and supplies an intrinsic time scale, breaking the parabolic-rescaling symmetry that made the heat-equation collapse possible. A naive single-variable ansatz $V = K\, g(\xi)$ fails, exactly because of this $r$-clock.

The plan of the subsubsection is therefore: dimensional analysis (§2), scale invariance of the BS PDE (§3), the natural similarity variable (§4), why the naive ansatz fails (§5), modified similarity variables that recover the BS structure (§6), and a compact recap of the heat-equation case (§7).

The Black-Scholes PDE does not admit classical similarity reduction

In the strict Lie-symmetry sense, the Black-Scholes PDE is not self-similar: the interest rate $r$ has dimensions of $[T^{-1}]$ and therefore introduces an intrinsic time scale that breaks the scaling symmetry. No single-variable ansatz $V = K\,g(\xi)$ reduces the PDE to a genuine ODE. What this subsubsection calls "similarity analysis" is therefore a structural / dimensional-analysis perspective, not a canonical Lie-group reduction. The canonical derivations of $(d_1, d_2)$ live in § Heat Equation and § Feynman--Kac; here we only reinterpret them through the lens of scaling.

2. 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:

\[\begin{array}{lllll} \pi_1 &=&\displaystyle \frac{S}{K} && \text{(moneyness)}\\ \pi_2 &=&\displaystyle \sigma\sqrt{T-t} = \sigma\sqrt{\tau} && \text{(scaled time)}\\ \pi_3 &=&\displaystyle r\tau && \text{(discount factor exponent)}\\ \pi_4 &=&\displaystyle \frac{V}{K} && \text{(normalized value)} \end{array}\]

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.

3. 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.

4. The Black-Scholes Similarity Variable¶

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.

5. Why the Naive Ansatz Fails¶

Substituting the similarity ansatz

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

into the Black-Scholes PDE produces a relation that mixes $\xi$ and $\tau$ explicitly --- the $r$-terms refuse to be absorbed into a function of $\xi$ alone. The presence of $\tau$ confirms what the dimensional argument already suggested: $r$ has dimensions of $[T^{-1}]$ and therefore breaks scaling, so no reduction of the form $V = K\,g(\xi)$ eliminates time entirely.

Detailed substitution

From $\xi = x/(\sigma\sqrt{\tau})$ with $x = \ln(S/K)$:

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

Chain rule gives

\[ \frac{\partial V}{\partial t} = K g'(\xi)\frac{\xi}{2\tau}, \quad \frac{\partial V}{\partial S} = \frac{K g'(\xi)}{S\sigma\sqrt{\tau}}, \quad \frac{\partial^2 V}{\partial S^2} = \frac{K g''(\xi)}{S^2\sigma^2\tau} - \frac{K g'(\xi)}{S^2\sigma\sqrt{\tau}} \]

Substituting into the PDE and multiplying by $\tau/K$:

\[ \frac{g''(\xi)}{2} + \frac{\xi\,g'(\xi)}{2} + \frac{(r - \tfrac{1}{2}\sigma^2)\sqrt{\tau}}{\sigma}\,g'(\xi) - r\tau\,g(\xi) = 0 \]

The $\sqrt{\tau}$ and $\tau$ coefficients show explicitly that the equation is not an ODE in $\xi$.

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

Heuristic vs. exact

The reduction in this subsubsection is not a classical similarity reduction. It is a structural observation: once one absorbs $r$ into a drift-adjusted coordinate and writes the solution as a sum of two scaling pieces, each piece satisfies an ODE in the appropriate variable. This explains why the canonical derivations (see § Heat Equation and § Feynman--Kac) produce $N(d_1)$ and $N(d_2)$ --- but it is not itself a rigorous symmetry-group reduction of the Black-Scholes PDE.

Drift-adjusted coordinates. Absorbing the discount-rate scale into the similarity variable produces

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

so that the solution decomposes as

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

with the same function $h$ in each term. From § Heat Equation, recall that the canonical derivation identifies $\xi_1 = d_1$ and $\xi_2 = d_2$, with $h = N$ the standard normal CDF; we do not re-derive these formulas here.

The ODE for $h$. A structural calculation (compressed below) shows that $h$ satisfies

\[ h''(\xi) + \xi\,h'(\xi) = 0 \]

whose general solution $h'(\xi) = C e^{-\xi^2/2}$ integrates to $h(\xi) = C_1 \sqrt{2\pi}\,N(\xi) + C_2$. Imposing the call payoff fixes $h = N$, recovering the Black-Scholes formula.

Detailed reduction

Plugging $V = S\,h(\xi_1) - Ke^{-r\tau}\,h(\xi_2)$ into the BS PDE, the cross-terms cancel by virtue of the algebraic identity $S\,N'(d_1) = Ke^{-r\tau}\,N'(d_2)$ (equivalently, $d_1 - d_2 = \sigma\sqrt{\tau}$), and what remains, after multiplying through by $\tau$, is

\[ h''(\xi_i) + \xi_i\,h'(\xi_i) = 0 \qquad (i = 1, 2). \]

Integrating once gives $h'(\xi) = C e^{-\xi^2/2}$; integrating again recovers $N(\xi)$ up to affine constants. The constants are pinned down by the terminal payoff $V(S,T) = (S-K)^+$.

7. Heat Equation Similarity (Compact Recap)¶

The BS → heat-equation transformation $w_\tau = w_{xx}$ (with $x = \ln(S/K)$, $\tau = T - t$, and drift/decay removed) is carried out in detail in § Heat Equation. The takeaway for the present perspective is exactly the collapse of §1: applied to the transformed problem, where there is no $[T^{-1}]$ parameter to break the symmetry, the similarity ansatz $w(x, \tau) = \tau^{-1/2} \tilde G(x / \sqrt{\tau})$ reduces the PDE to a genuine ODE whose unique decaying solution is the Gaussian kernel. The §1 picture is, in this sense, literally the spectral signature of the heat-equation half of the BS calculation. The $r$-clock that breaks the BS reduction lives only in the inverse transformation back to the original $(S, t)$ coordinates.

8. Explicit Example: European Call¶

The Terminal Condition Is Not Self-Similar¶

In the similarity variable $\xi = x/(\sigma\sqrt{\tau})$ with $x = \ln(S/K)$, the call payoff reads $V(x,0) = K(e^{\sigma\sqrt{\tau}\,\xi} - 1)^+$ at $\tau = 0$. As $\tau \to 0$ this does not collapse to a function of $\xi$ alone — the payoff has its own scale, $K$ — which is a second, independent reason (beyond the $r$-obstruction of §5) why a naive single-variable ansatz cannot capture the full solution. The resolution is to superpose the self-similar kernel against the payoff (convolution with $G$, as in § Heat Equation); the kernel is the building block, but the superposition is not.

Black-Scholes Formula¶

Recall: canonical $(d_1, d_2)$

From § Heat Equation (and equivalently § Feynman--Kac), the call price is

\[ C(S,t) = S\,N(d_1) - Ke^{-r\tau}\,N(d_2), \qquad d_{1,2} = \frac{\ln(S/K) + (r \pm \tfrac{1}{2}\sigma^2)\tau}{\sigma\sqrt{\tau}}. \]

The similarity perspective adds interpretation, not derivation: $d_1$ and $d_2$ are drift-adjusted similarity coordinates, and the structure $N(d_1), N(d_2)$ inherits the self-similar reduction of the underlying heat equation, shifted to account for the drift $r$.

9. 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.

10. Connection to Probability Theory¶

The heat kernel $G(x,\tau)$ (see § Heat Equation) is a Gaussian density of variance $2\tau$, and the similarity variable $\xi = x/\sqrt{2\tau}$ is exactly the standardized CLT variable. So the similarity reduction of the heat equation is the PDE counterpart of the CLT scaling under which fluctuations of size $\sqrt{\tau}$ are universal.

Two further regimes follow immediately:

Large deviations ($\tau \to \infty$, $x/\tau = v$ fixed): $-\tau^{-1}\ln G(x,\tau) \to v^2/4$, so the rate function $I(v) = v^2/4$ governs exponential decay of tail events.
Small-noise limit ($\tau \to 0$, $\xi$ fixed): the kernel concentrates on $\{\xi = 0\}$, i.e., on the strike $S = K$ — the zero-diffusion limit.

Both regimes are statements about the same Gaussian kernel under different scalings of $(x, \tau)$.

11. 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 canonical formula $C = SN(d_1) - Ke^{-r\tau}N(d_2)$, derived rigorously in § Heat Equation and § Feynman--Kac, can be reinterpreted as a decomposition into two drift-adjusted similarity coordinates. The structural ODE $h'' + \xi h' = 0$ explains why $N$ (rather than some other function) appears, even though this is not a strict similarity reduction.

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 $1/\sqrt{2\pi}$. $\square$