Deterministic Brownian Exponentials and Gaussian Tilting

From Brownian Integrals to Gaussian Tilting

A deterministic Brownian integral is Gaussian, so its exponential moments can be calculated exactly. The resulting stochastic exponential has mean one, and weighting probabilities by it shifts the means of Gaussian quantities without changing their covariances. These facts give a concrete model for exponential tilting before considering general adapted integrands or general changes of measure.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$ relative to $(\mathcal{F}_t)$. Fix a finite horizon $T$. Every stochastic integral below is an Ito integral: its integrand is progressively measurable and belongs to $\mathcal{L}^2$, meaning $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. In particular, a deterministic measurable function $h$ with $\int_0^T h(s)^2\,ds<\infty$ is an admissible integrand.

Definition: For such a deterministic $h$, set $M_t=\int_0^t h(s)\,dB_s$, $q_t=\int_0^t h(s)^2\,ds$, and $Z_t=\exp(M_t-\tfrac12q_t)$, for $0\le t\le T$. The process $Z_t$ is the stochastic exponential of $M_t$. The term $-\tfrac12q_t$ is its quadratic-variation correction, since $[M]_t=q_t$.

The deterministic Brownian integral $M_t$ is centered normal with variance $q_t$. To see why, first take $h$ to be a deterministic step function. Then $M_t$ is a sum of constant multiples of Brownian increments on disjoint intervals; those increments are independent centered normal variables. Its variance is the sum of the increment variances multiplied by the squared constants, namely $q_t$. For general deterministic $h\in L^2([0,T])$, approximate $h$ in $L^2$ by step functions. The Ito isometry gives convergence of their integrals in $L^2(\mathbb{P})$, and hence convergence of characteristic functions, since $|e^{ix}-e^{iy}|\le |x-y|$. The limiting characteristic function is that of $N(0,q_t)$. This also covers $q_t=0$, when $M_t=0$ almost surely.

Mean, Variance, and the Martingale Property

Theorem: For every deterministic $h\in L^2([0,T])$ and $0\le t\le T$, $$ \mathbb{E}[Z_t]=1,\qquad \mathbb{E}[Z_t^2]=e^{q_t},\qquad \operatorname{Var}(Z_t)=e^{q_t}-1. $$ Moreover, $Z_t$ is a martingale with respect to $(\mathcal{F}_t)$.

Proof. If $G\sim N(0,q_t)$, its moment-generating function is $\mathbb{E}[e^{aG}]=e^{a^2q_t/2}$ for every real $a$; this formula remains valid when $q_t=0$. Since $M_t$ has this law,

\mathbb{E}[Z_t] =e^{-q_t/2}\mathbb{E}[e^{M_t}] =e^{-q_t/2}e^{q_t/2}=1, \qquad \mathbb{E}[Z_t^2] =e^{-q_t}\mathbb{E}[e^{2M_t}] =e^{-q_t}e^{2q_t}=e^{q_t}.

Thus $Z_t$ is integrable and $\operatorname{Var}(Z_t)=\mathbb{E}[Z_t^2]-(\mathbb{E}[Z_t])^2=e^{q_t}-1$. For $0\le s\le t$, write $M_t=M_s+(M_t-M_s)$ and $q_t=q_s+(q_t-q_s)$. The Brownian increments after $s$ are independent of $\mathcal{F}_s$, so $M_t-M_s$ is independent of $\mathcal{F}_s$ and is normal with mean zero and variance $q_t-q_s$. Consequently,

\mathbb{E}[Z_t\mid\mathcal{F}_s] =Z_s\,\mathbb{E}\!\left[ e^{M_t-M_s-(q_t-q_s)/2}\mid\mathcal{F}_s \right] =Z_s e^{-(q_t-q_s)/2}e^{(q_t-q_s)/2} =Z_s.

This proves the martingale property, including the case $q_t-q_s=0$. In general, a local martingale need not be a true martingale; here the conditional-expectation calculation and integrability verify the stronger property directly. $\square$

The same process can be related to Ito’s formula. With $Z_t=f(t,M_t)$, take $f(t,x)=\exp(x-q_t/2)$. Its derivatives are $f_t(t,x)=-\tfrac12 h(t)^2f(t,x)$, $f_x(t,x)=f(t,x)$, and $f_{xx}(t,x)=f(t,x)$, for almost every $t$. Since $d[M]_t=h(t)^2dt$, Ito’s formula gives

dZ_t=f_t\,dt+f_x\,dM_t+\tfrac12f_{xx}\,d[M]_t =-\tfrac12h(t)^2Z_t\,dt+Z_th(t)\,dB_t +\tfrac12h(t)^2Z_t\,dt =Z_th(t)\,dB_t.

The cancellation uses $dB_t\,dB_t=dt$, $dB_t\,dt=0$, and $dt\,dt=0$. This differential identity alone identifies a local-martingale form; the conditional-expectation proof above is what verifies the true martingale property here.

Worked Example: Mean and Variance for a Constant Integrand

Let $T=2$ and $h(s)=\tfrac12$ on $[0,2]$. This deterministic function is square-integrable, and its Ito integral is in the class specified above. Compute the mean and variance of $Z_2$.

First calculate $q_2=\int_0^2(\tfrac12)^2\,ds=2\cdot\tfrac14=\tfrac12$. Therefore $Z_2=\exp(\tfrac12B_2-\tfrac14)$. The theorem gives $\mathbb{E}[Z_2]=1$, $\mathbb{E}[Z_2^2]=e^{1/2}$, and $\operatorname{Var}(Z_2)=e^{1/2}-1$.

Check: $B_2\sim N(0,2)$, so $\tfrac12B_2$ is centered normal with variance $\tfrac14\cdot2=\tfrac12$. Thus $\mathbb{E}[e^{\frac12B_2-\frac14}]=e^{-1/4}e^{(1/2)/2}=1$, and $\mathbb{E}[e^{B_2-\frac12}]=e^{-1/2}e^{2/2}=e^{1/2}$. Subtracting the squared mean gives $e^{1/2}-1$, as claimed.

The Gaussian Mean-Shift Formula

Let $g_1,\ldots,g_n$ be deterministic square-integrable functions on $[0,T]$, and define $Y_i=\int_0^T g_i(s)\,dB_s$. These Ito integrals use progressively measurable integrands with $\mathbb{E}\int_0^T g_i(s)^2\,ds<\infty$. For deterministic integrands, their joint Gaussian law has means zero and covariances $\mathbb{E}[Y_iY_j]=\int_0^Tg_i(s)g_j(s)\,ds$, by the Ito isometry and the joint Gaussian property of Brownian integrals.

Theorem: Define a probability measure $\mathbb{P}^h$ on $\mathcal{F}_T$ by $\mathbb{P}^h(A)=\mathbb{E}[\mathbf{1}_A Z_T]$, and write $\mathbb{E}^h[F]=\mathbb{E}[Z_TF]$ for integrable quantities under this weighting. Then $(Y_1,\ldots,Y_n)$ under $\mathbb{P}^h$ is Gaussian with means $c_i=\int_0^T h(s)g_i(s)\,ds$ and unchanged covariances $\int_0^Tg_i(s)g_j(s)\,ds$. Degenerate covariance matrices are allowed.

Proof. The preceding theorem gives $Z_T\ge0$ and $\mathbb{E}[Z_T]=1$, so the displayed formula defines a probability measure. Put $X=M_T=\int_0^T h(s)\,dB_s$. For any $a=(a_1,\ldots,a_n)\in\mathbb{R}^n$, let $Y_a=\sum_i a_iY_i=\int_0^T(\sum_i a_ig_i(s))\,dB_s$. The pair $(X,Y_a)$ is jointly Gaussian: every linear combination is a deterministic Brownian integral and is normal, by the Gaussian-integral argument above. Its variances and covariance are

\operatorname{Var}(X)=q_T,\qquad \operatorname{Var}(Y_a)=\int_0^T\left(\sum_i a_ig_i(s)\right)^2ds,\qquad \operatorname{Cov}(X,Y_a)=\int_0^T h(s)\sum_i a_ig_i(s)\,ds.

The moment-generating function of a centered jointly Gaussian pair gives, for every real $r$,

\mathbb{E}^h[e^{rY_a}] =e^{-q_T/2}\mathbb{E}[e^{X+rY_a}] =\exp\!\left(\tfrac12q_T+r\operatorname{Cov}(X,Y_a) +\tfrac12r^2\operatorname{Var}(Y_a)-\tfrac12q_T\right).

Thus $\mathbb{E}^h[e^{rY_a}]=\exp(r\operatorname{Cov}(X,Y_a)+\tfrac12r^2\operatorname{Var}(Y_a))$, the moment-generating function of a normal variable with mean $\operatorname{Cov}(X,Y_a)$ and variance $\operatorname{Var}(Y_a)$. This holds for every $a$, so the vector is Gaussian with mean entries $c_i$ and the original covariance entries. The conclusion also holds when some linear combinations have zero variance: their moment-generating functions then describe constants. $\square$

This construction is a Gaussian tilt on $\mathcal{F}_T$, not an assertion of a risk-neutral measure $\mathbb{Q}$ or a general Girsanov theorem. Since $Z_T$ is strictly positive, the weighting also assigns zero probability to exactly the $\mathbb{P}$-null events; the mean-shift conclusion itself follows from the calculation above.

Worked Example: Tilting Two Brownian Values

On $[0,1]$, take $h(s)=1$, and consider $Y_1=B_{1/2}$ and $Y_2=B_1$. Here $Z_1=e^{B_1-1/2}$, and $h$ and the indicator integrands for both Brownian values are deterministic square-integrable Ito integrands.

Use $g_1=\mathbf{1}_{[0,1/2]}$ and $g_2=\mathbf{1}_{[0,1]}$. The tilted means are $c_1=\int_0^{1/2}1\,ds=\tfrac12$ and $c_2=\int_0^1 1\,ds=1$. The covariance entries remain $\int g_1^2ds=\tfrac12$, $\int g_1g_2ds=\tfrac12$, and $\int g_2^2ds=1$. Therefore, under $\mathbb{P}^h$, the mean vector is $(\tfrac12,1)$, and the covariance matrix is

\begin{pmatrix} \tfrac12 & \tfrac12\\ \tfrac12 & 1 \end{pmatrix}.

Check: $X=B_1$ has variance $1$, and its covariance with $B_1$ is $1$. The joint-Gaussian moment calculation therefore gives $\mathbb{E}^h[B_1]=1$. Directly, the weighted marginal density of $B_1=x$ is proportional to $e^{x-1/2}e^{-x^2/2}$, which equals $e^{-(x-1)^2/2}$ up to the constant normalizing factor; this is the density of $N(1,1)$. The shifted mean and unchanged variance agree.

Worked Example: A Time-Dependent Integrand

On $[0,1]$, let $h(s)=2s$, and tilt the vector $(B_{1/2},B_1)$. The function $2s$ and the Brownian-value indicator integrands are deterministic and square-integrable, so the theorem applies. The normalizing quadratic variation is $q_1=\int_0^1 4s^2\,ds=\tfrac43$, giving $Z_1=\exp(\int_0^1 2s\,dB_s-\tfrac23)$.

The means are $c_1=\int_0^{1/2}2s\,ds=\tfrac14$ and $c_2=\int_0^1 2s\,ds=1$. The covariance matrix remains the one computed from the indicator integrands, namely

\operatorname{Cov}^{h}\!\begin{pmatrix}B_{1/2}\\B_1\end{pmatrix} = \begin{pmatrix} \tfrac12 & \tfrac12\\ \tfrac12 & 1 \end{pmatrix}.

Check: the stochastic integral $X=\int_0^1 2s\,dB_s$ is centered normal with variance $\tfrac43$, so $\mathbb{E}[Z_1]=e^{-2/3}e^{(4/3)/2}=1$. Also, its covariance with $B_1$ is $\int_0^1 2s\,ds=1$, confirming the stated tilted mean of $B_1$. The covariance of $B_{1/2}$ and $B_1$ stays $\tfrac12$, since exponential tilting changes the mean vector but not the covariance matrix.

Why the Correction Term Matters

The factor $-\tfrac12q_T$ is essential: it normalizes the exponential to have mean one. A common error is to use $e^{M_T}$ as if it were already a probability weight. For example, take $h(s)=1$ on $[0,1]$, so $M_1=B_1$ and $q_1=1$. The incorrect weight $e^{B_1}$ has expectation $e^{1/2}$, not one; weighting by it does not define a probability measure. The corrected weight $e^{B_1-1/2}$ has expectation $e^{-1/2}e^{1/2}=1$. Its variance is $e^1-1$, by the theorem. The correction is therefore not cosmetic: it fixes the total mass of the tilted distribution.

For Brownian values, choose $g_i=\mathbf{1}_{[0,t_i]}$ in the Gaussian mean-shift theorem. It follows that their tilted means become $\int_0^{t_i}h(s)\,ds$, while their covariances remain $\min(t_i,t_j)$. This finite-dimensional calculation explains the deterministic drift shift that appears in more general change-of-measure results, without assuming those later results. It also shows why deterministic integrands are a useful test case: their exponential moments and the resulting Gaussian laws can be checked exactly.

Check Your Understanding

Use the deterministic-integrand formulas and the Gaussian mean-shift theorem to answer the following.

For deterministic $h$, write the mean and variance of $Z_t$ in terms of $q_t$.
On $[0,1]$, take $h(s)=3$. Compute $q_1$, $\mathbb{E}[Z_1]$, and $\operatorname{Var}(Z_1)$.
If $Y=\int_0^T g(s)\,dB_s$, what are its mean and variance under the tilt $Z_T$?
For $h(s)=2s$ on $[0,1]$, compute the tilted mean of $B_{1/2}$.
Explain why the weight $e^{M_T}$, without the quadratic-variation correction, generally cannot define a probability measure by weighting $\mathbb{P}$.

Deterministic Brownian Exponentials and Gaussian Tilting

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