Itô Calculus at Work

Knowledge Map Atlas
First Computations Integration by Parts The Stratonovich Conversion

First Computations

The previous page proved the Itô formula and promised that it turns the evaluation of stochastic integrals into calculus. This page collects the payoff. We evaluate integrals that the definition alone would never surrender, prove the stochastic version of integration by parts, and settle the account the construction story left open, the correction separating the Itô and Stratonovich readings of a stochastic differential. Throughout, \(w_t\) is the standard one-dimensional Brownian motion of the last four pages, started at \(w_0 = 0\), with its Brownian filtration \(\{\mathcal{F}_t\}_{t \geq 0}\) containing all null sets, and Itô process notation \(dX_t = u\, dt + v\, dw_t\) carries its meaning from the previous page.

One discipline accompanies every application. The Itô formula carries a membership hypothesis, that the diffusion coefficient of the image, \(v\, \frac{\partial g}{\partial x}(s, X_s)\), lies in \(\mathcal{V}(0, T)\) for every \(T\). The check is one line in each case below, and we write the line each time, because a hypothesis that is never checked eventually gets forgotten.

The Integral That Started It

First, the formula must reproduce the one integral this track ever computed from the definition. Take \(X_t = w_t\), an Itô process with \(u = 0\) and \(v = 1\), and \(g(t, x) = \tfrac{1}{2} x^2\), which is \(C^2\) with \(\frac{\partial g}{\partial t} = 0\), \(\frac{\partial g}{\partial x} = x\), and \(\frac{\partial^2 g}{\partial x^2} = 1\). Membership asks that \(v\, \frac{\partial g}{\partial x}(s, X_s) = w_s\) belong to \(\mathcal{V}(0, T)\), and it does, being adapted with continuous paths, hence jointly measurable, and satisfying \(\mathbb{E}\bigl[ \int_0^T w_s^2\, ds \bigr] = \int_0^T s\, ds \lt \infty\), the exchange of expectation and integral being the Tonelli exchange for a nonnegative integrand. The formula returns, its initial term \(g(0, w_0) = 0\) vanishing,

\[ \begin{align*} \tfrac{1}{2} w_t^2 &= \int_0^t \tfrac{1}{2}\, ds + \int_0^t w_s\, dw_s \\\\ &= \tfrac{1}{2} t + \int_0^t w_s\, dw_s , \end{align*} \]

which is the integral of Brownian motion against itself read from the other side. What took a quadratic variation argument and a page of estimates is now the statement that the chain rule's correction term equals \(\tfrac{1}{2} \cdot 1^2 \cdot 1 = \tfrac{1}{2}\).

A First New Integral

Now an integral the definition never touched. What is \(\int_0^t s\, dw_s\)? Classical calculus suggests that a term \(t\, w_t\) should appear, so we aim the formula at it. Take \(X_t = w_t\) again and \(g(t, x) = t x\), which is \(C^2\) on \([0, \infty) \times \mathbb{R}\) with \(\frac{\partial g}{\partial t} = x\), \(\frac{\partial g}{\partial x} = t\), and all second partial derivatives either \(1\) or \(0\). Membership asks that \(v\, \frac{\partial g}{\partial x}(s, X_s) = s\) belong to \(\mathcal{V}(0, T)\), and the deterministic integrand \(s\) is adapted, jointly measurable, and square integrable on \([0, T]\), so it does. Since \(\frac{\partial^2 g}{\partial x^2} = 0\), the correction term vanishes, the initial term \(g(0, w_0) = 0\) vanishes as well, and the formula gives

\[ t\, w_t = \int_0^t w_s\, ds + \int_0^t s\, dw_s , \]

that is,

\[ \int_0^t s\, dw_s = t\, w_t - \int_0^t w_s\, ds . \]

The \(ds\)-term on the right is the pathwise time integral of a continuous adapted process, well defined path by path. The identity has exactly the shape classical integration by parts predicts, and no correction appears. That is not an accident. The function \(g(t, x) = t x\) is linear in \(x\), its second \(x\)-derivative vanishes, and the Itô correction is proportional to precisely that derivative. Whenever the noise enters linearly, Itô calculus and classical calculus agree. The next section turns this observation into a theorem.

Integration by Parts

The second computation of the last section is begging to be generalized. Nothing about the argument used the specific function \(t\), only that the noise entered linearly through \(g(t, x) = f(t)\, x\). Running the same argument with a general deterministic \(f\) yields the stochastic version of integration by parts.

Theorem: Integration by Parts

Let \(f \in C^2([0, \infty))\) be deterministic. Then, for every \(t \geq 0\), almost surely,

\[ \int_0^t f(s)\, dw_s = f(t)\, w_t - \int_0^t f'(s)\, w_s\, ds , \]

the \(ds\)-term being the pathwise time integral of the continuous adapted process \(f'(s)\, w_s\).

Proof.

Take \(X_t = w_t\), an Itô process with \(u = 0\) and \(v = 1\), and \(g(t, x) = f(t)\, x\). The partial derivatives are \(\frac{\partial g}{\partial t} = f'(t)\, x\), \(\frac{\partial g}{\partial x} = f(t)\), \(\frac{\partial^2 g}{\partial t^2} = f''(t)\, x\), \(\frac{\partial^2 g}{\partial t \partial x} = f'(t)\), and \(\frac{\partial^2 g}{\partial x^2} = 0\), all continuous on \([0, \infty) \times \mathbb{R}\) because \(f \in C^2\), so \(g \in C^2([0, \infty) \times \mathbb{R})\). Membership asks that \(v\, \frac{\partial g}{\partial x}(s, X_s) = f(s)\) belong to \(\mathcal{V}(0, T)\) for every \(T\), and the deterministic continuous integrand \(f\) is adapted, jointly measurable, and bounded on \([0, T]\), hence square integrable there. The Itô formula applies, the correction term vanishes with \(\frac{\partial^2 g}{\partial x^2}\), the initial term is \(g(0, w_0) = f(0) \cdot 0 = 0\), and

\[ f(t)\, w_t = \int_0^t f'(s)\, w_s\, ds + \int_0^t f(s)\, dw_s . \]

Rearranging gives the claim.

Two extensions are worth recording, both beyond what the formula's hypotheses deliver directly. First, the identity holds already for \(f \in C^1([0, \infty))\), by smoothing \(f\) with mollifiers and passing to the limit, the Itô isometry controlling the stochastic side and uniform convergence on \([0, t]\) the classical side. We record this without executing the limit passage. Second, in its strongest form the identity reads

\[ \int_0^t f(s, \omega)\, dw_s = f(t, \omega)\, w_t - \int_0^t w_s\, df_s \]

for adapted processes \(f\) whose paths are continuous and of bounded variation, with the integrability the left side requires, the last term a pathwise Riemann-Stieltjes integral against the path of \(f\). Determinism is not required. Our track has not built Riemann-Stieltjes integration, so this version stays a recorded signpost, to be collected if and when that machinery arrives. What is required, and cannot be weakened away, is bounded variation. It is exactly what makes a pathwise integral against \(df_s\) exist, and it is the hypothesis Brownian motion itself fails, on every interval.

Where the Classical Rule Breaks

The bounded variation hypothesis is doing real work, and the cleanest way to see it is to break the rule on purpose. Brownian motion has unbounded variation on every interval, so \(f = w\) sits outside every version of the identity above. Pretend otherwise. The symmetric product rule applied to \(f = w\) would predict \(w_t \cdot w_t = \int_0^t w_s\, dw_s + \int_0^t w_s\, dw_s\), that is,

\[ \int_0^t w_s\, dw_s = \tfrac{1}{2}\, w_t^2 , \]

while the true value carries the correction, \(\int_0^t w_s\, dw_s = \tfrac{1}{2} w_t^2 - \tfrac{1}{2} t\). The classical prediction is wrong by exactly \(\tfrac{1}{2} t\), half the quadratic variation of the path. This is the same half that the previous page located in the Taylor expansion, and the same half that the construction page located in the choice of the left endpoint. There is a third place to locate it, and it is the most illuminating one, because it explains what a calculus would have to give up to keep the classical rules. That is the business of the final section.

The Stratonovich Conversion

The half has now been found in two places, in the Taylor expansion behind the Itô formula and in the discrepancy of the broken product rule. The third place is the oldest. The construction of the integral committed to evaluating integrands at the left endpoint of each interval, the causal choice, and the previous page named the midpoint alternative the Stratonovich integral, computed its value for the test integrand at statement level, and displayed the drift correction \(\tfrac{1}{2} \sigma' \sigma\) that translates between the two calculi, with the derivation deferred. Everything needed to pay these debts is now on the table.

Definition: The Stratonovich Integral

Let \(\{Z_s\}_{s \geq 0}\) be an adapted process with continuous paths, and fix \(t \gt 0\). For a partition \(\pi : 0 = t_0 \lt \cdots \lt t_N = t\) write \(m_j = \tfrac{1}{2}(t_j + t_{j+1})\) for the midpoints and form the midpoint sums

\[ S_\pi(Z) = \sum_j Z_{m_j} \bigl( w_{t_{j+1}} - w_{t_j} \bigr) . \]

If there is a random variable \(I\) such that \(S_{\pi_n}(Z) \to I\) in probability along every sequence of partitions of \([0, t]\) with mesh tending to zero, we call \(I\) the Stratonovich integral of \(Z\) and write

\[ \int_0^t Z_s \circ dw_s = I . \]

The definition is an obligation, not a gift. Nothing guarantees that the midpoint sums converge, and existence must be proved integrand by integrand, or theorem by theorem. Both proofs below run on the factorization mechanism of the previous page's lemma, which we reassemble in one line for use here. If \(Y\) is integrable and measurable with respect to \(\mathcal{F}_a\), and \(h\) is a polynomial, then \(Y\) and \(h(w_b - w_a)\) are independent because the increment is independent of the past, and the product rule gives \(\mathbb{E}[Y\, h(w_b - w_a)] = \mathbb{E}[Y]\, \mathbb{E}[h(w_b - w_a)]\).

The Test Integrand, Settled

Proposition: The Stratonovich Integral of Brownian Motion

The Stratonovich integral of \(Z = w\) exists on every \([0, t]\), and

\[ \int_0^t w_s \circ dw_s = \tfrac{1}{2}\, w_t^2 . \]

Proof.

Fix a sequence of partitions with mesh tending to zero, and abbreviate \(\Delta w_j = w_{t_{j+1}} - w_{t_j}\), together with the half-increments \(\delta w_j = w_{m_j} - w_{t_j}\) and \(\delta w_j' = w_{t_{j+1}} - w_{m_j}\), so that \(\Delta w_j = \delta w_j + \delta w_j'\). Splitting the midpoint value against the left endpoint,

\[ \begin{align*} S_\pi(w) &= \sum_j w_{t_j}\, \Delta w_j + \sum_j \delta w_j \bigl( \delta w_j + \delta w_j' \bigr) \\\\ &= \sum_j w_{t_j}\, \Delta w_j + \sum_j (\delta w_j)^2 + \sum_j \delta w_j\, \delta w_j' . \end{align*} \]

The left-endpoint sums converge in \(L^2(\mathbb{P})\) to the Itô integral. They are the elementary integrals of \(\Phi_\pi = \sum_j w_{t_j} \mathbf{1}_{[t_j, t_{j+1})}\), and, exchanging expectation and integral by Tonelli for a nonnegative integrand and using the increment law \(\mathbb{E}[(w_{t_j(s)} - w_s)^2] = s - t_j(s)\), \[ \begin{align*} \mathbb{E}\bigl[ \int_0^t (\Phi_\pi - w)^2\, ds \bigr] &= \int_0^t (s - t_j(s))\, ds \\\\ &\leq t\, |\pi| , \end{align*} \] so L² continuity gives \(\sum_j w_{t_j} \Delta w_j \to \int_0^t w_s\, dw_s\).

The squared half-increments converge to \(\tfrac{1}{2} t\) in \(L^2(\mathbb{P})\). Consider the refined partition \(\pi'\) obtained by inserting every midpoint, and apply part (ii) of the weighted increment lemma over \(\pi'\) with weights equal to one on the first-half intervals and zero on the second-half intervals, which are measurable at the left endpoints and bounded by one. This gives \(\sum_j \bigl( (\delta w_j)^2 - \tfrac{1}{2} \Delta t_j \bigr) \to 0\) in \(L^2(\mathbb{P})\), and \(\sum_j \tfrac{1}{2} \Delta t_j = \tfrac{1}{2} t\).

The mixed sum dies. Expanding its square, the off-diagonal terms with \(i \lt j\) vanish by factorization, since \((\delta w_i\, \delta w_i')(\delta w_j)\) is \(\mathcal{F}_{m_j}\)-measurable and integrable while \(\mathbb{E}[\delta w_j'] = 0\), and the diagonal factorizes twice,

\[ \begin{align*} \mathbb{E}\Bigl[ \Bigl( \sum_j \delta w_j\, \delta w_j' \Bigr)^{2} \Bigr] &= \sum_j \mathbb{E}\bigl[ (\delta w_j)^2 \bigr]\, \mathbb{E}\bigl[ (\delta w_j')^2 \bigr] \\\\ &= \sum_j \tfrac{(\Delta t_j)^2}{4} \\\\ &\leq \tfrac{1}{4}\, t\, |\pi| \\\\ &\longrightarrow 0 . \end{align*} \]

Adding the three limits, \(S_\pi(w) \to \int_0^t w_s\, dw_s + \tfrac{1}{2} t\) in probability along every sequence with vanishing mesh, and the first computation identifies the limit as \(\tfrac{1}{2} w_t^2 - \tfrac{1}{2} t + \tfrac{1}{2} t = \tfrac{1}{2} w_t^2\).

The claim the previous page recorded at statement level is now a theorem. The midpoint calculus computes \(\int w \circ dw\) exactly as classical calculus would, and the entire Itô-Stratonovich discrepancy for this integrand is the half-quadratic-variation term made visible by the splitting above.

The Conversion Theorem

Theorem: The Itô-Stratonovich Conversion

Let \(X_t = X_0 + \int_0^t u\, ds + \int_0^t v\, dw_s\) be an Itô process whose coefficients \(u\) and \(v\) are, on \([0, t]\), elementary over a common partition \(R\) with all levels bounded by \(C\). Let \(\sigma : [0, \infty) \times \mathbb{R} \to \mathbb{R}\) have continuous partial derivatives \(\frac{\partial \sigma}{\partial t}\) and \(\frac{\partial \sigma}{\partial x}\), with \(\sigma\), \(\frac{\partial \sigma}{\partial t}\), and \(\frac{\partial \sigma}{\partial x}\) bounded by \(M\). Then, along every sequence of partitions of \([0, t]\) refining \(R\) with mesh tending to zero, the midpoint sums of \(Z_s = \sigma(s, X_s)\) converge in probability, and

\[ \int_0^t \sigma(s, X_s) \circ dw_s = \int_0^t \sigma(s, X_s)\, dw_s + \tfrac{1}{2} \int_0^t \frac{\partial \sigma}{\partial x}(s, X_s)\, v(s, \omega)\, ds , \]

the left side denoting the common limit along such sequences. That the restriction to refining sequences can be dropped, making the limit the Stratonovich integral in the unrestricted sense of the definition, travels with the approximation passage of the closing remark.

Proof.

Fix a sequence of partitions refining \(R\) with mesh tending to zero, keep the notation of the previous proof, and write \(X_j = X_{t_j}\), \(Z_j = \sigma(t_j, X_j)\), \(\sigma_x = \frac{\partial \sigma}{\partial x}\), and \(\sigma_t = \frac{\partial \sigma}{\partial t}\). Because the partition refines \(R\) and midpoints stay inside their intervals, \(u\) and \(v\) are constant on \([t_j, t_{j+1})\) with \(\mathcal{F}_{t_j}\)-measurable values \(u_j, v_j\) bounded by \(C\). The drift piece of the half-increment is the pathwise integral of a constant level, and the noise piece is the elementary integral of a constant level, identified with the increment of the integral process by additivity over intervals, so outside a single null set, for every partition of the sequence at once,

\[ X_{m_j} - X_j = u_j\, \tfrac{\Delta t_j}{2} + v_j\, \delta w_j . \]

Step 1 (Splitting at the left endpoint).
Decompose the midpoint value as \(Z_{m_j} = Z_j + \sigma_x(t_j, X_j)\,(X_{m_j} - X_j) + \rho_j\), where the remainder collects the time increment and the mean value error,

\[ \rho_j = \bigl[ \sigma(m_j, X_{m_j}) - \sigma(t_j, X_{m_j}) \bigr] + \bigl[ \sigma(t_j, X_{m_j}) - \sigma(t_j, X_j) - \sigma_x(t_j, X_j)\, (X_{m_j} - X_j) \bigr] . \]

Multiplying by \(\Delta w_j\) and summing,

\[ S_\pi(Z) = \sum_j Z_j\, \Delta w_j + \sum_j \sigma_x(t_j, X_j)\, u_j\, \tfrac{\Delta t_j}{2}\, \Delta w_j + \sum_j \sigma_x(t_j, X_j)\, v_j\, \delta w_j\, \Delta w_j + \sum_j \rho_j\, \Delta w_j . \]

Step 2 (The left sums and the drift-noise sum).
The left sums \(\sum_j Z_j \Delta w_j\) converge in \(L^2(\mathbb{P})\) to \(\int_0^t \sigma(s, X_s)\, dw_s\), by the argument of the previous page's proof run with \(\sigma\) in place of \(g_x v\). The approximants \(\Phi_\pi = \sum_j Z_j \mathbf{1}_{[t_j, t_{j+1})}\) are elementary with \(\mathcal{F}_{t_j}\)-measurable levels bounded by \(M\), they converge pointwise \(\lambda \otimes \mathbb{P}\)-almost everywhere to \(\sigma(s, X_s)\) by continuity of the paths, the constant \(2 M\) dominates, so the dominated convergence theorem on the finite product measure and L² continuity conclude. In passing, the same boundedness shows \(\sigma(s, X_s) \in \mathcal{V}(0, t)\), so the Itô integral on the right of the theorem is legitimate. The drift-noise sum is part (i) of the weighted increment lemma with weights \(\tfrac{1}{2} \sigma_x(t_j, X_j)\, u_j\) bounded by \(\tfrac{1}{2} M C\), and dies in \(L^2(\mathbb{P})\).

Step 3 (The main sum).
Write \(b_j = \sigma_x(t_j, X_j)\, v_j\), which is \(\mathcal{F}_{t_j}\)-measurable and bounded by \(M C\), and split \(\Delta w_j = \delta w_j + \delta w_j'\),

\[ \sum_j b_j\, \delta w_j\, \Delta w_j = \sum_j b_j\, (\delta w_j)^2 + \sum_j b_j\, \delta w_j\, \delta w_j' . \]

The first sum converges in probability to \(\tfrac{1}{2} \int_0^t \sigma_x(s, X_s)\, v(s, \omega)\, ds\). Centering it with part (ii) of the lemma over the midpoint-refined partition, with weights \(b_j\) on the first-half intervals and zero on the second-half intervals, leaves \(\sum_j b_j\, \tfrac{\Delta t_j}{2}\), a left-endpoint Riemann sum of the piecewise continuous path \(s \mapsto \tfrac{1}{2} \sigma_x(s, X_s)\, v(s)\), which converges pathwise interval by interval of \(R\) exactly as on the previous page. The second sum dies in \(L^2(\mathbb{P})\) by the computation of the proposition, decorated with the weights. Off-diagonal terms vanish by factorization, since \((b_i \delta w_i \delta w_i')(b_j \delta w_j)\) is \(\mathcal{F}_{m_j}\)-measurable and integrable while \(\mathbb{E}[\delta w_j'] = 0\), and the diagonal factorizes into \(\mathbb{E}[b_j^2 (\delta w_j)^2]\, \mathbb{E}[(\delta w_j')^2] \leq M^2 C^2 (\Delta t_j)^2 / 4\), summing to at most \(M^2 C^2\, t\, |\pi| / 4 \to 0\).

Step 4 (The remainder dies).
Write \(\rho_j^{\mathrm{time}}\) and \(\rho_j^{\mathrm{space}}\) for the two brackets of \(\rho_j\). Writing \(\rho_j^{\mathrm{time}}\) as the integral of \(\sigma_t\) over \([t_j, m_j]\) at frozen second argument, legitimate for a \(C^1\) time dependence with the one-sided derivative at the time origin, the bound on \(\sigma_t\) gives \(|\rho_j^{\mathrm{time}}| \leq M\, \tfrac{\Delta t_j}{2}\), and

\[ \begin{align*} \mathbb{E}\Bigl[ \sum_j \tfrac{M}{2}\, \Delta t_j\, |\Delta w_j| \Bigr] &\leq \tfrac{M}{2} \sum_j \Delta t_j\, (\Delta t_j)^{1/2} \\\\ &\leq \tfrac{M}{2}\, t\, |\pi|^{1/2} \\\\ &\longrightarrow 0 , \end{align*} \]

using \(\mathbb{E}|\Delta w_j| \leq \mathbb{E}[(\Delta w_j)^2]^{1/2} = (\Delta t_j)^{1/2}\) by the Cauchy-Schwarz inequality. For the space part, writing the increment of the \(C^1\) function \(\tau \mapsto \sigma(t_j, X_j + \tau (X_{m_j} - X_j))\) as the integral of its derivative over \([0, 1]\), the bracket equals \(\int_0^1 \bigl[ \sigma_x(t_j, X_j + \tau (X_{m_j} - X_j)) - \sigma_x(t_j, X_j) \bigr] d\tau \cdot (X_{m_j} - X_j)\). Fix \(\omega\) outside the null set of Step 1 and let \(\eta(\pi, \omega)\) be the modulus of continuity of \(\sigma_x\) on the compact \([0, t] \times [-R_\omega, R_\omega]\), with \(R_\omega = \max_{s \leq t} |X_s|\), evaluated at \(\max_j |X_{m_j} - X_j|\), which tends to zero by uniform continuity of the path on \([0, t]\). Then \(\eta \to 0\) pathwise, \(\eta \leq 2 M\), and with \(|X_{m_j} - X_j| \leq C \Delta t_j + C |\delta w_j|\) and \(2 |\delta w_j| |\Delta w_j| \leq 3 (\delta w_j)^2 + (\delta w_j')^2\),

\[ \sum_j \bigl| \rho_j^{\mathrm{space}} \bigr|\, |\Delta w_j| \leq \eta\, C \Bigl( \sum_j \Delta t_j |\Delta w_j| + \tfrac{3}{2}\, Q_{\pi'} \Bigr) , \]

where \(Q_{\pi'}\) is the quadratic sum over the midpoint-refined partition. The first term inside dies in \(L^1(\mathbb{P})\) by the display above, uniformly against \(\eta \leq 2 M\), and the second is handled exactly as on the previous page, by Cauchy-Schwarz, \(\mathbb{E}[\eta^2] \to 0\) from the dominated convergence theorem, and the \(L^2\)-boundedness of \(Q_{\pi'}\) from quadratic variation. So \(\sum_j \rho_j \Delta w_j \to 0\) in \(L^1(\mathbb{P})\).

Step 5 (Assembly).
Every piece of the splitting converges in probability, so the midpoint sums converge in probability to \(\int_0^t \sigma(s, X_s)\, dw_s + \tfrac{1}{2} \int_0^t \sigma_x(s, X_s)\, v\, ds\), and the limit does not depend on the chosen sequence, which is existence and the claimed identity at once.

Two remarks finish the account. First, the boundedness and elementary-coefficient hypotheses are the core-case regime of the Itô formula's proof, and the same trusted approximation passage recorded there carries the conversion to general coefficients. Second, granted that passage, take the equation-level case the previous page displayed, where the noise coefficient of \(X\) is itself \(v = \sigma(s, X_s)\). The correction becomes \(\tfrac{1}{2} \int_0^t \sigma_x\, \sigma\, ds\), which is exactly the drift correction \(\tfrac{1}{2} \sigma' \sigma\) of the promised display. Reading a stochastic differential in the Stratonovich sense and reading it in the Itô sense with that drift added are the same reading, and the derivation deferred there is the theorem above.

Two Calculi, One Division of Labor

The conversion identity means nothing is at stake mathematically in the choice of calculus, and everything is at stake ergonomically. The Itô integral evaluates at the left endpoint, never peeks ahead, and inherits the zero-mean and martingale structure that makes expectations computable, which is why probability, statistics, and finance live on the Itô side. The Stratonovich integral obeys the classical chain rule, which is why modeling limits of smooth physics and calculus on manifolds, where a chain rule that respects coordinate changes is not negotiable, live on the Stratonovich side. Machine learning uses both readings of the same equations. The forward noising process of a diffusion model is posed as an Itô equation because its analysis runs on expectations and martingales, while the conversion above is the dictionary that keeps the physics-flavored and probability-flavored formulations of the same model honest with each other.

The chapter of accounts is closed. The integral was built on the left endpoint, the formula priced the choice at half a quadratic variation, and the conversion theorem shows the price is a translation cost, not a loss. Either calculus reaches the other by a drift of size \(\tfrac{1}{2} \sigma' \sigma\), the discrepancy first seen in a single computation with \(\int w\, dw\) and now understood as the general exchange rate between causality and symmetry.