Itô Dynamics Of The Reciprocal Fundamental Solution

Why the Reciprocal Has a Corrected Drift

A scalar linear stochastic differential equation has a fundamental solution that multiplies the initial value. Its reciprocal is useful because it can undo that multiplicative factor. But taking the reciprocal is not an ordinary algebraic operation on differentials: Brownian fluctuations contribute a second-order term. In particular, the drift of the reciprocal is not just the negative of the original drift.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$. Fix a finite horizon $T$, and let $a,b:[0,T]\to\mathbb{R}$ be bounded deterministic measurable functions. The fundamental solution considered here starts from $\Phi_0=1$ and satisfies

d\Phi_t=a(t)\Phi_t\,dt+b(t)\Phi_t\,dB_t.

Every stochastic integral in this tutorial is an Itô integral. Its integrand is adapted and progressively measurable, and belongs to $\mathcal{L}^2$ on $[0,T]$, meaning $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. In particular, the deterministic integrand $b$ is in $\mathcal{L}^2$ because it is bounded on a finite interval. The explicit fundamental solution is

\Phi_t=\exp\!\left(\int_0^t\left(a(s)-\tfrac12 b(s)^2\right)ds+\int_0^t b(s)\,dB_s\right).

The first integral in the exponent is an ordinary Lebesgue integral; the second is an Itô integral with deterministic $\mathcal{L}^2$ integrand $b$. In particular, $\Phi_t>0$ for every $t$, almost surely, so $1/\Phi_t$ is defined. We will derive its dynamics and then check them by multiplying the two differentials.

Definition: The reciprocal process is $R_t=1/\Phi_t$. The drift coefficient of an SDE is the coefficient of $dt$, while its diffusion coefficient is the coefficient of $dB_t$. These are distinct terms; the reciprocal’s drift includes a contribution from quadratic variation.

Itô’s Formula for the Reciprocal

Theorem: Under the assumptions above, $R_t=1/\Phi_t$ satisfies $$ dR_t=\bigl(b(t)^2-a(t)\bigr)R_t\,dt-b(t)R_t\,dB_t,\qquad R_0=1. $$ On each finite horizon, the stochastic integral with integrand $b(t)R_t$ is an Itô integral with an adapted, progressively measurable $\mathcal{L}^2$ integrand.

Proof. The explicit formula for $\Phi_t$ shows that it is strictly positive. Apply Itô’s formula to $f(x)=x^{-1}$ on $(0,\infty)$. The derivatives are $f'(x)=-x^{-2}$ and $f''(x)=2x^{-3}$. Since the diffusion coefficient of $\Phi_t$ is $b(t)\Phi_t$, its squared diffusion contributes $\tfrac12 f''(\Phi_t)b(t)^2\Phi_t^2\,dt$. Thus

\begin{aligned} dR_t &=f'(\Phi_t)\,d\Phi_t+\tfrac12 f''(\Phi_t)b(t)^2\Phi_t^2\,dt\\ &=-\Phi_t^{-2}\left(a(t)\Phi_t\,dt+b(t)\Phi_t\,dB_t\right) +\tfrac12\left(2\Phi_t^{-3}\right)b(t)^2\Phi_t^2\,dt\\ &=\bigl(-a(t)+b(t)^2\bigr)\Phi_t^{-1}\,dt -b(t)\Phi_t^{-1}\,dB_t\\ &=\bigl(b(t)^2-a(t)\bigr)R_t\,dt-b(t)R_t\,dB_t. \end{aligned}

For completeness, the integrability required to write the last stochastic integral as an $\mathcal{L}^2$ Itô integral follows from the explicit exponential. Put $A_t=\int_0^t a(s)\,ds$ and $C_t=\int_0^t b(s)^2\,ds$. Since $\int_0^t b(s)\,dB_s$ is centered normal with variance $C_t$, the normal exponential moment formula gives $\mathbb{E}[R_t^2]=\exp(-2A_t+3C_t)$. Boundedness of $a,b$ makes this expectation bounded over $0\le t\le T$. Therefore $\mathbb{E}\int_0^T b(t)^2R_t^2\,dt<\infty$, and $bR$ is progressively measurable because $R$ is adapted and continuous. The initial value is $R_0=1/\Phi_0=1$. $\square$

The term $b(t)^2R_t\,dt$ is the drift correction. It comes from the second derivative in Itô’s formula and the quadratic-variation rule $dB_t\,dB_t=dt$. In the multiplication table, $dB_t\,dt=0$ and $dt\,dt=0$; they do not remove the $dt$ contribution from the squared Brownian term.

Worked Calculations

Worked Example: Constant Drift and Unit Diffusion

Take $a(t)=\tfrac12$ and $b(t)=1$. Compute $\Phi_t$, derive $R_t$, and check the reciprocal drift using its expectation.

The exponent of the fundamental solution is obtained term by term: $a-\tfrac12b^2=\tfrac12-\tfrac12=0$, and $\int_0^t b(s)\,dB_s=B_t$. Hence $\Phi_t=e^{B_t}$, so $R_t=e^{-B_t}$. The theorem gives drift $b^2-a=1-\tfrac12=\tfrac12$ and diffusion coefficient $-b=-1$, multiplied by $R_t$:

dR_t=\tfrac12R_t\,dt-R_t\,dB_t.

As an independent check, $B_t\sim N(0,t)$, so the normal moment-generating formula gives $\mathbb{E}[R_t]=\mathbb{E}[e^{-B_t}]=e^{t/2}$. The drift equation predicts the same mean: $\mathbb{E}[\tfrac12R_t]=\tfrac12e^{t/2}$, which is the derivative of $e^{t/2}$. This expectation check is not a claim that $R_t$ is a martingale; in fact its mean varies with time.

Worked Example: Time-Varying Coefficients

Let $a(t)=t$ and $b(t)=\tfrac12$. Find the reciprocal explicitly and compute the coefficients of its SDE, including their values at $t=1$.

First, $\int_0^t a(s)\,ds=\int_0^t s\,ds=t^2/2$, and $\tfrac12b(s)^2=\tfrac12\cdot\tfrac14=\tfrac18$, so $\int_0^t\tfrac12b(s)^2\,ds=t/8$. The Brownian integral is $\int_0^t\tfrac12\,dB_s=\tfrac12B_t$. Therefore

\Phi_t=\exp\!\left(\frac{t^2}{2}-\frac{t}{8}+\frac{B_t}{2}\right), \qquad R_t=\exp\!\left(-\frac{t^2}{2}+\frac{t}{8}-\frac{B_t}{2}\right).

The reciprocal drift coefficient multiplying $R_t$ is $b(t)^2-a(t)=\tfrac14-t$, and the diffusion coefficient multiplying $R_t$ is $-b(t)=-\tfrac12$. Thus $dR_t=(\tfrac14-t)R_t\,dt-\tfrac12R_t\,dB_t$. At $t=1$, the drift coefficient is $\tfrac14-1=-\tfrac34$. This also follows by differentiating the deterministic part of the exponent of $R_t$: its derivative is $-t+\tfrac18$, while the Itô correction for diffusion $-\tfrac12R_t$ is $\tfrac12(\tfrac14)R_t=\tfrac18R_t$; their sum is $-t+\tfrac14$, equal to the stated drift coefficient.

Check: at $t=0$, the drift coefficient is $\tfrac14$, matching $b(0)^2-a(0)=\tfrac14-0$. At $t=1$, the explicit exponent has deterministic derivative $-1+\tfrac18=-\tfrac78$; adding the Itô correction $\tfrac18$ gives $-\tfrac78+\tfrac18=-\tfrac34$, as required.

Worked Example: Product Check with Exact Coefficients

Take constant coefficients $a=\tfrac13$ and $b=\tfrac23$. Compute both processes and verify directly that their product differential vanishes.

Here $b^2=\tfrac49$, and $\tfrac12b^2=\tfrac29$. The exponent of $\Phi_t$ has drift $\tfrac13-\tfrac29=\tfrac19$, so $\Phi_t=\exp(\tfrac19t+\tfrac23B_t)$. Consequently $R_t=\exp(-\tfrac19t-\tfrac23B_t)$. The reciprocal equation has drift coefficient $b^2-a=\tfrac49-\tfrac13=\tfrac19$ and diffusion coefficient $-\tfrac23$:

d\Phi_t=\tfrac13\Phi_t\,dt+\tfrac23\Phi_t\,dB_t, \qquad dR_t=\tfrac19R_t\,dt-\tfrac23R_t\,dB_t.

Use the Itô product rule $d(R_t\Phi_t)=R_t\,d\Phi_t+\Phi_t\,dR_t+d[R,\Phi]_t$. The covariation term is the product of the diffusion coefficients times $dt$, so $d[R,\Phi]_t=(-\tfrac23R_t)(\tfrac23\Phi_t)\,dt=-\tfrac49R_t\Phi_t\,dt$. Substitution gives

\begin{aligned} d(R_t\Phi_t) &=\left(\tfrac13+\tfrac19-\tfrac49\right)R_t\Phi_t\,dt +\left(\tfrac23-\tfrac23\right)R_t\Phi_t\,dB_t\\ &=0\,dt+0\,dB_t=0. \end{aligned}

Since $R_0\Phi_0=1$, the product remains equal to $1$. This independently confirms the reciprocal relation and, in particular, shows why the positive $\tfrac49$ contribution in the reciprocal drift is needed: it cancels the negative covariation in the product rule.

The Product Identity and Its Use

Proposition: If $\Phi$ and $R$ satisfy the two equations in the theorem with $R_0=\Phi_0=1$, then $R_t\Phi_t=1$ for all $t\in[0,T]$, almost surely.

Proof. The Itô product rule applies to the continuous semimartingales $\Phi$ and $R$. Their diffusion coefficients are $b(t)\Phi_t$ and $-b(t)R_t$, so their covariation differential is $-b(t)^2R_t\Phi_t\,dt$. The $dt$ coefficient of $d(R_t\Phi_t)$ is therefore $(b(t)^2-a(t))R_t\Phi_t+a(t)R_t\Phi_t-b(t)^2R_t\Phi_t=0$. The $dB_t$ coefficient is $-b(t)R_t\Phi_t+b(t)R_t\Phi_t=0$. Thus the product has zero differential and is constant. Its initial value is $1$, proving the claim. $\square$

The proposition is a verification, not a replacement for Itô’s formula. Starting with $R=1/\Phi$, Itô’s formula derives the correction; the product rule then checks the resulting equation. Conversely, when a candidate reciprocal is constructed, the product identity is a useful test: both the drift and diffusion terms must cancel. The statement is pathwise up to indistinguishability under the stated continuous-process assumptions; it is not merely an equality of expectations.

Common Mistake: Negating the Drift

A tempting but incorrect calculation is to set $d(1/\Phi_t)=-\Phi_t^{-2}d\Phi_t$ and stop there. For the exact coefficients $a=\tfrac13$, $b=\tfrac23$ above, that calculation would give the drift $-\tfrac13R_t$ and diffusion $-\tfrac23R_t$. It omits the second derivative contribution. The corrected drift is $-\tfrac13+\tfrac49=\tfrac19$, so the wrong value differs by $\tfrac49R_t$.

The product check exposes the error immediately. With the incorrect drift, the $dt$ coefficient in $d(R_t\Phi_t)$ would be $-\tfrac13+\tfrac13-\tfrac49=-\tfrac49$, rather than zero. The error is exactly the unaccounted-for covariation. Ordinary differentiation would be appropriate for a differentiable path with no quadratic variation; for Brownian-driven paths, $dB_t\,dB_t=dt$ must be included.

In applications, the reciprocal fundamental solution can remove a multiplicative homogeneous factor from a linear equation. The calculation here supplies that inverse factor and its correct dynamics; it does not license treating stochastic differentials as ordinary fractions. Keep the drift correction, the negative diffusion coefficient, and the product covariation together.

Check Your Understanding

Use the formula and product rule above. When checking a proposed answer, compare both its $dt$ and $dB_t$ coefficients.

For $a=1$, $b=1$, find the drift and diffusion coefficients of $R_t$.
For $a(t)=0$, $b(t)=t$, compute the reciprocal drift coefficient at $t=2$.
For constant $a,b$, identify the term in the Itô formula that would be lost by ordinary differentiation of $x^{-1}$.
Suppose a proposed reciprocal equation has diffusion coefficient $+b(t)R_t$. What happens to the $dB_t$ coefficient in the product with $\Phi_t$?
Explain why bounded deterministic $b$ and a finite time horizon ensure that $bR$ is an admissible Itô-integral integrand here.