Exponential Martingales and Density in Brownian L2

Why Exponential Martingales Give a Dense Family

In a Brownian filtration, a terminal random variable can depend on the whole path up to time \(T\), not just on one Brownian increment. A useful way to study such variables is to begin with finite time grids. On any one grid, Brownian motion is described by finitely many independent Gaussian increments. Exponential functions of those increments form a dense family in the corresponding Gaussian \(L^2\) space. Refining the grids then captures the entire Brownian path.

Throughout, let \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carry a standard Brownian motion \(B_t\), and assume that \((\mathcal{F}_t)\) is the usual augmentation of the filtration generated by \(B\). Fix \(T>0\). For an integrand \(H\), the Ito integral \(\int_0^t H_s\,dB_s\) is taken when \(H\) is adapted and progressively measurable and \(\mathbb{E}\int_0^t H_s^2\,ds<\infty\); this class is denoted by \(\mathcal{L}^2\). A deterministic step function \(h\) on \([0,T]\) belongs to \(\mathcal{L}^2\).

These exponentials are defined under \(\mathbb{P}\). Although a positive mean-one terminal exponential can also be used as a change-of-measure density, the density result below is an \(L^2(\mathbb{P})\) statement and does not change the probability measure.

Why Each Exponential Is a True Martingale

Proof. Write \(I_{s,t}=\int_s^t h(u)\,dB_u\) and \(v_{s,t}=\int_s^t h(u)^2\,du\). Since \(h\) is deterministic, \(I_{s,t}\) is independent of \(\mathcal{F}_s\) and is normal with mean \(0\) and variance \(v_{s,t}\). Therefore \(\mathbb{E}[\exp(I_{s,t}-v_{s,t}/2)\mid\mathcal{F}_s]=1\), by the moment generating function of a centered normal variable. The exponential factors as \(Z_t(h)=Z_s(h)\exp(I_{s,t}-v_{s,t}/2)\), giving the asserted conditional expectation. Also \(I_{0,t}\) is centered normal with variance \(v_{0,t}\), so \(\mathbb{E}[Z_t(h)^2]=e^{-v_{0,t}}\mathbb{E}[e^{2I_{0,t}}]=e^{-v_{0,t}}e^{2v_{0,t}}=e^{v_{0,t}}\), which is finite. Thus the martingale is integrable and square-integrable; it is not merely a local martingale. \(\square\)

On an interval where \(h\) is constant, the Ito multiplication rules are \(dB_t\,dB_t=dt\), \(dB_t\,dt=0\), and \(dt\,dt=0\). Ito's formula also gives the familiar differential for the exponential. Specifically, for \(f(x)=e^x\), \(f'(x)=e^x\) and \(f''(x)=e^x\). With \(A_t=\int_0^t h(s)\,dB_s-\tfrac12\int_0^t h(s)^2\,ds\), the correction term in Ito's formula is \(\tfrac12 f''(A_t)h(t)^2\,dt\), which cancels the drift term \(f'(A_t)(-\tfrac12h(t)^2\,dt)\). Hence \(dZ_t(h)=Z_t(h)h(t)\,dB_t\) on each such interval. The conditional-expectation calculation above, rather than this differential alone, verifies the true martingale property.

Finite Grids: The Gaussian Density Result

Consider a partition \(0=t_0<t_1<\cdots<t_d=T\), and write \(\Delta t_j=t_j-t_{j-1}>0\) and \(G_j=(B_{t_j}-B_{t_{j-1}})/\sqrt{\Delta t_j}\). The vector \((G_1,\ldots,G_d)\) has the standard \(d\)-dimensional Gaussian distribution: its coordinates are independent \(N(0,1)\) variables. If \(a=(a_1,\ldots,a_d)\in\mathbb{R}^d\), choose \(h(t)=a_j/\sqrt{\Delta t_j}\) on \((t_{j-1},t_j]\). Then \(\int_0^T h\,dB=\sum_{j=1}^d a_jG_j\) and \(\int_0^T h^2\,dt=\sum_{j=1}^d a_j^2\). Thus grid-constant integrands produce precisely the normalized Gaussian exponentials \(\exp(a\cdot G-|a|^2/2)\).

Proof. Let \(\gamma_d\) denote the standard Gaussian probability measure on \(\mathbb{R}^d\). Suppose \(g\in L^2(\gamma_d)\) is orthogonal to every stated exponential. Define \(F(z)=\int_{\mathbb{R}^d}g(x)e^{z\cdot x}\,\gamma_d(dx)\) for \(z\in\mathbb{C}^d\). This integral exists and defines an entire function in each coordinate: on any bounded set of \(z\)'s, Cauchy–Schwarz bounds the integral by \(\|g\|_{L^2(\gamma_d)}\left(\int e^{2R\sum_j|x_j|}\,\gamma_d(dx)\right)^{1/2}<\infty\) for a sufficiently large finite \(R\). Orthogonality implies \(F(a)=0\) for every real \(a\), since the factor \(e^{-|a|^2/2}\) is nonzero. Applying the one-variable identity theorem successively in each coordinate shows \(F(z)=0\) for every \(z\in\mathbb{C}^d\).

In particular, \(F(i\xi)=0\) for every \(\xi\in\mathbb{R}^d\). This is the Fourier transform of the finite signed (or complex) measure \(g(x)\gamma_d(dx)\); the measure is finite because \(\int|g|\,d\gamma_d\le\|g\|_{L^2(\gamma_d)}\). By uniqueness of Fourier transforms of finite measures, that measure is zero, so \(g=0\) \(\gamma_d\)-almost everywhere. The identity theorem and Fourier uniqueness used here are standard results from analysis. We have proved that the orthogonal complement of the span contains only zero. In a Hilbert space, a linear subspace has dense closure exactly when its orthogonal complement is zero. This proves the density claim. \(\square\)

From Finite Grids to All Brownian Terminal Variables

Proof. Use the nested dyadic grids of \([0,T]\), and let \(\mathcal{G}_n\) be the sigma-field generated by the Brownian values on the \(n\)th grid. The union of these grids is dense in time. By continuity of Brownian paths, the values on this countable dense set determine the path, so \(\sigma(\bigcup_n\mathcal{G}_n)=\mathcal{F}_T\) up to null sets. The standard \(L^2\) martingale convergence theorem for conditional expectations gives \(\mathbb{E}[X\mid\mathcal{G}_n]\to X\) in \(L^2(\mathbb{P})\).

Choose \(n\) large enough that \(\|X-\mathbb{E}[X\mid\mathcal{G}_n]\|_2<\varepsilon/2\). The conditional expectation is square-integrable and is a measurable function of the finitely many independent normalized increments on that grid. The finite-dimensional Gaussian density theorem therefore supplies a finite linear combination \(Y\) of exponentials with grid-constant deterministic integrands such that \(\|\mathbb{E}[X\mid\mathcal{G}_n]-Y\|_2<\varepsilon/2\). Each such exponential is \(Z_T(h)\) for a deterministic step function \(h\), by the coefficient-to-integrand calculation above. The triangle inequality gives \(\|X-Y\|_2<\varepsilon\). This proves the assertion. \(\square\)

The assumption on the filtration matters. If \(\mathcal{F}_T\) also contains randomness not generated by \(B\), Brownian exponentials need not be dense in the larger space \(L^2(\mathcal{F}_T)\). The theorem concerns terminal variables measurable with respect to the Brownian path, up to null sets.

Common Pitfall and Why Density Matters

A common mistake is to say that the collection of exponentials itself is dense. The result concerns finite linear combinations. Each \(Z_T(h)\) is positive, while a general square-integrable variable may take both positive and negative values; linear combinations allow cancellations and are essential to the density statement.

Another mistake is to infer density merely from the fact that each exponential has expectation one. A shared mean does not establish approximation of arbitrary random variables. The proof uses two distinct ingredients: finite-dimensional Gaussian exponential density, and approximation of Brownian terminal variables by conditioning on increasingly fine grids.

This density result supplies a useful test family for Brownian \(L^2\) arguments. If a continuous linear functional vanishes on every deterministic-step exponential, then it vanishes on their linear span and hence, by density, on every Brownian terminal variable in \(L^2\). This is one reason the exponential family is effective in proofs involving Brownian martingales and representation.