Heat Equation Representations with Gaussian Terminal Data

From Brownian Increments to Heat Flow

For the heat equation, the diffusion is standard Brownian motion: over a time interval of length \(\tau\), its increment has mean zero and variance \(\tau\). Consequently, the expected value of a terminal payoff depends on the current spatial point and on the remaining time through a Gaussian distribution. For constant, linear, and quadratic payoffs, only the first two moments of that distribution are needed.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\), adapted to the filtration, with increments independent of the past. We consider the backward heat equation on \(0\le t\le T\), \(x\in\mathbb{R}\), with terminal data \(g\):

The drift coefficient is zero and the diffusion coefficient is one. Thus, starting at \(x\) at time \(t\), the terminal position has the distribution of \(x+B_T-B_t\), a normal random variable with mean \(x\) and variance \(T-t\). The relevant Feynman–Kac expectation is therefore \(\mathbb{E}[g(x+B_T-B_t)]\). No stochastic integral is needed in the computations below. When an Itô integral is used, its integrand is understood to be adapted and progressively measurable with \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\); no such integral is asserted here.

It is useful to write \(\tau=T-t\) for the remaining time. If \(Z\) is a standard normal random variable, the increment \(B_T-B_t\) has the same distribution as \(\sqrt{\tau}Z\). At \(\tau=0\), this increment is zero. The Gaussian moment facts needed below are \(\mathbb{E}[Z]=0\) and \(\mathbb{E}[Z^2]=1\).

The Gaussian Representation for Quadratic Data

Proof. Fix \(t\le T\), set \(\tau=T-t\), and use the equality in distribution \(B_T-B_t=\sqrt{\tau}Z\). The random terminal position \(Y=x+\sqrt{\tau}Z\) satisfies \(\mathbb{E}[Y]=x+\sqrt{\tau}\mathbb{E}[Z]=x\) and \(\mathbb{E}[Y^2]=x^2+2x\sqrt{\tau}\mathbb{E}[Z]+\tau\mathbb{E}[Z^2]=x^2+\tau\). Linearity of expectation now gives \(\mathbb{E}[aY^2+bY+c]=a(x^2+\tau)+bx+c\), including \(\tau=0\). For the resulting function, \(u_t=-a\), \(u_x=2ax+b\), and \(u_{xx}=2a\). Hence \(u_t+\tfrac12u_{xx}=-a+a=0\), and at \(t=T\) the expression is \(ax^2+bx+c=g(x)\).

For uniqueness in the stated class, let \(w(t,x)=A(t)x^2+D(t)x+E(t)\) be any other solution with the same terminal data. Its derivatives give \(w_t+\tfrac12w_{xx}=A'(t)x^2+D'(t)x+E'(t)+A(t)\). This polynomial in \(x\) vanishes for every \(x\), so its coefficients satisfy \(A'=0\), \(D'=0\), and \(E'=-A\). The zero terminal difference means \(A(T)=D(T)=E(T)=0\). The first two equations imply \(A=D=0\) throughout the interval, and the third then implies \(E'=0\), so \(E=0\) throughout as well. Thus the two solutions agree. \(\square\)

This direct calculation is the heat-equation specialization of the Feynman–Kac representation. It also makes the time direction transparent: increasing the remaining time adds \(\tau\) to the second moment, while increasing the calendar time \(t\) decreases it. For a quadratic payoff, that change in expected value is exactly balanced by the second-derivative term in the PDE.

Worked Gaussian Expectations

Why the Second-Derivative Term Matters

For a linear terminal payoff, Brownian noise does not change the expected payoff: the mean of the terminal position is the current spatial point. For a quadratic payoff, however, the expectation of the square changes because the variance is positive. The identity \(\mathbb{E}[(x+B_T-B_t)^2]=x^2+(T-t)\) is the source of the extra time term. In the PDE, its derivative is \(u_t=-a\); the curvature of the payoff contributes \(\tfrac12u_{xx}=a\), and the two terms cancel.

The same statement can be read in forward time. Define \(v(s,x)=u(T-s,x)\), where \(s\) is elapsed time from the terminal-data side of the time reversal. Then \(v_s=\tfrac12v_{xx}\), \(v(0,x)=g(x)\), and for quadratic data \(v(s,x)=a(x^2+s)+bx+c\). Thus the forward heat equation increases the quadratic component by \(as\). The factor \(\tfrac12\) in the PDE is essential: since the second derivative of \(ax^2\) is \(2a\), the diffusion term contributes \(\tfrac12(2a)=a\), exactly the rate of increase.

These formulas are useful as exact checks on more complicated calculations. A proposed expression for the heat evolution of polynomial data can be tested by its terminal value and by the PDE. The constant and linear cases test whether a spurious time correction has been introduced; the quadratic case tests the sign and coefficient of the variance correction.

Common Pitfalls

Wrong calculation: adding variance to a linear payoff. One might incorrectly write \(\mathbb{E}[x+B_T-B_t]=x+(T-t)\), treating the increment’s variance as its mean. The correction is that \(\mathbb{E}[B_T-B_t]=0\), so the linear expectation is \(x\), with no time term. Variance enters the expectation of a square, not the expectation of the increment itself.

Wrong time sign. For \(g(y)=y^2\), the correct backward solution is \(u(t,x)=x^2+T-t\). Its time derivative is \(-1\), while \(\tfrac12u_{xx}=1\). Writing \(x^2+t\) instead gives \(u_t=1\) and \(\tfrac12u_{xx}=1\), so the PDE left-hand side is \(2\), not zero. The terminal condition also fails unless the time interval has a specially chosen endpoint. Always check both the differential equation and the terminal value.

The formulas above apply to polynomial terminal data of degree at most two and have no growth or integrability difficulty: every Gaussian moment used is finite. They are not a claim about arbitrary terminal functions or about uniqueness among all possible growth classes. The uniqueness proved here is specifically within the class of quadratic polynomials in space with continuously differentiable time-dependent coefficients.