Conditional Expectations of Brownian Functionals

Conditioning on the Present

Knowing the current Brownian value does not determine the future, but it does determine the starting point of the future transition law. The remaining uncertainty is Gaussian: after time $s$, the increment $B_t-B_s$ is independent of the information available at time $s$, and has variance $t-s$. This lets us compute conditional expectations of functions of future values by averaging over Gaussian transition densities.

Throughout, let $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ be a filtered probability space carrying a standard Brownian motion $B_t$ with respect to this filtration. In particular, for $s<t$, $B_t-B_s$ is independent of $\mathcal{F}_s$ and has distribution $N(0,t-s)$. Conditional expectation identities below hold almost surely. All integrals in this tutorial are ordinary Lebesgue integrals against densities; no Ito or Stratonovich stochastic integral is used.

Definition: For $h>0$, the Brownian transition density is $p_h(x,y)=\dfrac{1}{\sqrt{2\pi h}}\exp\!\left(-\dfrac{(y-x)^2}{2h}\right)$. For a Borel function $f$ for which the expectation is finite, define $P_hf(x)=\int_{\mathbb{R}} f(y)p_h(x,y)\,dy$. Equivalently, $P_hf(x)=\mathbb{E}[f(x+Z)]$ when $Z\sim N(0,h)$.

The transition density is centered at $x$, not at zero. Conditional on $\mathcal{F}_s$, the future value $B_t$ is distributed as $B_s+Z$, where $Z\sim N(0,t-s)$ is independent of $\mathcal{F}_s$. Thus the present Brownian value remains inside the conditional formula.

One Future Value: The Gaussian Kernel Formula

Theorem: Let $0\le s<t$, and let $f:\mathbb{R}\to\mathbb{R}$ be Borel measurable with $\mathbb{E}|f(B_t)|<\infty$. Then $$ \mathbb{E}[f(B_t)\mid\mathcal{F}_s] =P_{t-s}f(B_s) =\int_{\mathbb{R}} f(y)p_{t-s}(B_s,y)\,dy $$ almost surely. The integral is finite for almost every outcome.

Proof. Write $Z=B_t-B_s$. Brownian independent increments give $Z\sim N(0,t-s)$ and make $Z$ independent of $\mathcal{F}_s$. Hence, conditional on $\mathcal{F}_s$, $B_t=B_s+Z$ has density $p_{t-s}(B_s,y)$ in the variable $y$. For bounded Borel $f$, integrating $f$ against this conditional density gives the stated conditional expectation. For an integrable $f(B_t)$, apply the bounded result to the truncations $f_n=\max(-n,\min(f,n))$. Since $f_n(B_t)\to f(B_t)$ in $L^1(\mathbb{P})$, their conditional expectations converge in $L^1$ to $\mathbb{E}[f(B_t)\mid\mathcal{F}_s]$; the density integrals converge to the corresponding integral for $f$ wherever that integral is finite. Integrability ensures finiteness almost surely. This proves the identity. $\square$

For example, if $f(y)=y^2$, then averaging over the future Gaussian increment gives $P_hf(x)=x^2+h$. For an indicator function, the same formula gives a conditional probability. The calculation is always an average over the transition law starting at the observed value $B_s$, rather than an average over the unconditional law of $B_t$.

Worked Example: A Conditional Second Moment

Compute $\mathbb{E}[B_2^2\mid\mathcal{F}_{1/2}]$. Set $x=B_{1/2}$ and $Z=B_2-B_{1/2}$. The increment $Z$ is independent of $\mathcal{F}_{1/2}$, has mean $0$, and has variance $2-\tfrac12=\tfrac32$. Expanding the square gives

\mathbb{E}[(x+Z)^2\mid\mathcal{F}_{1/2}] =x^2+2x\,\mathbb{E}[Z]+\mathbb{E}[Z^2] =x^2+\frac32.

Therefore $\mathbb{E}[B_2^2\mid\mathcal{F}_{1/2}]=B_{1/2}^2+\tfrac32$. Check: take expectations. Since $\mathbb{E}[B_{1/2}^2]=\tfrac12$, the result has expectation $\tfrac12+\tfrac32=2$, equal to $\mathbb{E}[B_2^2]$.

Several Future Values and Iterated Kernels

For several ordered future times, it is useful to work with successive increments. Fix $s<t_1<\cdots<t_n$, and set $t_0=s$. The increments $B_{t_i}-B_{t_{i-1}}$ are independent of one another and of $\mathcal{F}_s$; their respective variances are $t_i-t_{i-1}$. Thus the conditional joint density of the future values is a product of successive transition densities.

Theorem: With $s<t_1<\cdots<t_n$, let $g:\mathbb{R}^n\to\mathbb{R}$ be Borel measurable and suppose $\mathbb{E}|g(B_{t_1},\ldots,B_{t_n})|<\infty$. Then $$ \mathbb{E}[g(B_{t_1},\ldots,B_{t_n})\mid\mathcal{F}_s] = \int_{\mathbb{R}^n} g(x_1,\ldots,x_n) \prod_{i=1}^n p_{t_i-t_{i-1}}(x_{i-1},x_i)\,dx_1\cdots dx_n, \qquad x_0=B_s, $$ almost surely.

Proof. The vector of increments $(B_{t_1}-B_s,B_{t_2}-B_{t_1},\ldots,B_{t_n}-B_{t_{n-1}})$ consists of independent Gaussian variables, and the whole vector is independent of $\mathcal{F}_s$. Given $\mathcal{F}_s$, start at $x_0=B_s$. The first future value has density $p_{t_1-s}(x_0,x_1)$; conditional on that value, the next has density $p_{t_2-t_1}(x_1,x_2)$, and each subsequent transition has the corresponding density $p_{t_i-t_{i-1}}(x_{i-1},x_i)$. Multiplying these densities gives the conditional joint density. Integrating $g$ against it proves the formula for bounded Borel $g$. For an integrable $g$, truncation and $L^1$ convergence of conditional expectations give the same result, with the integral finite almost surely. $\square$

The same fact can be expressed as iterated transition operators. For a function of the last value alone, successive averaging uses the kernels in time order. The kernel composition rule is $P_a(P_bf)=P_{a+b}f$ when the expectations involved are defined. Indeed, adding two independent centered Gaussian increments of variances $a$ and $b$ gives a centered Gaussian increment of variance $a+b$. This is also the conditional-expectation tower property in transition-kernel form.

Worked Example: A Product at Two Future Times

Compute $\mathbb{E}[B_1B_2\mid\mathcal{F}_{1/2}]$. Let $x=B_{1/2}$, $U=B_1-B_{1/2}$, and $V=B_2-B_1$. Then $U$ and $V$ are independent of $\mathcal{F}_{1/2}$ and of each other, with means $0$, variances $\tfrac12$ and $1$, respectively. Since $B_1=x+U$ and $B_2=x+U+V$, expand:

$$ B_1B_2=x^2+2xU+xV+U^2+UV. $$

Conditionally, the terms linear in $U$ or $V$ have expectation zero; also $\mathbb{E}[U^2]=\tfrac12$ and $\mathbb{E}[UV]=\mathbb{E}[U]\mathbb{E}[V]=0$. Therefore $\mathbb{E}[B_1B_2\mid\mathcal{F}_{1/2}]=B_{1/2}^2+\tfrac12$.

Check: taking expectations gives $\tfrac12+\tfrac12=1$. The unconditional covariance formula for Brownian motion gives $\mathbb{E}[B_1B_2]=\min(1,2)=1$, in agreement.

Gaussian Conditioning for Linear Functionals

The increment representation also gives a direct rule for a linear combination of future values. For fixed $s<t_1<\cdots<t_n$ and real coefficients $a_1,\ldots,a_n$, the conditional mean and variance of $L=\sum_{i=1}^n a_iB_{t_i}$ are

m(B_s)=B_s\sum_{i=1}^n a_i,\qquad v=\sum_{i=1}^n\left(\sum_{j=i}^n a_j\right)^2(t_i-t_{i-1}).

In fact, the increment from $t_{i-1}$ to $t_i$ appears in $B_{t_i},B_{t_{i+1}},\ldots,B_{t_n}$, so its coefficient in $L$ is $\sum_{j=i}^n a_j$. These independent increments give the displayed variance.

Proposition: Given $\mathcal{F}_s$, $L$ is Gaussian with mean $m(B_s)$ and variance $v$. In particular, if $c\in\mathbb{R}$, then $\mathbb{E}[e^{cL}\mid\mathcal{F}_s]=\exp(cm(B_s)+\tfrac12c^2v)$. This includes the degenerate case $v=0$, when $L=m(B_s)$ conditionally.

Proof. Write each $B_{t_j}$ as $B_s+\sum_{i=1}^j(B_{t_i}-B_{t_{i-1}})$. Summing with coefficients $a_j$, the coefficient of $B_s$ is $\sum_j a_j$, and the coefficient of the $i$th increment is $\sum_{j=i}^n a_j$. Conditional on $\mathcal{F}_s$, these increments are independent centered normal variables with variances $t_i-t_{i-1}$. Their linear combination is therefore centered normal with variance $v$, added to the stated conditional mean. For a centered normal variable of variance $v>0$, completing the square in its density gives $\mathbb{E}[e^{cZ}]=e^{c^2v/2}$; for $v=0$, $Z=0$ almost surely and the same formula holds. Applying this identity conditionally proves the exponential formula. $\square$

Worked Example: An Exponential of Independent Future Increments

Compute $\mathbb{E}\!\left[\exp\!\left((B_1-B_{1/2})-(B_2-B_1)\right)\middle|\mathcal{F}_{1/2}\right]$. With $U=B_1-B_{1/2}$ and $V=B_2-B_1$, the exponent is $U-V$. These increments are independent of $\mathcal{F}_{1/2}$, independent of one another, and have variances $\tfrac12$ and $1$. The normal exponential formula, applied to each increment, gives

\mathbb{E}[e^{U-V}\mid\mathcal{F}_{1/2}] =e^{(1)^2(1/2)/2}e^{(-1)^2(1)/2} =e^{1/4}e^{1/2} =e^{3/4}.

Check: $U-V$ is centered normal with variance $\tfrac12+1=\tfrac32$, so its exponential moment is $e^{(3/2)/2}=e^{3/4}$, the same answer. The conditional expectation is constant because both increments are independent of the information at time $\tfrac12$.

Indicators, Iteration, and Common Mistakes

As another immediate application of the one-time formula, for $s<t$ and $a\in\mathbb{R}$, the conditional probability of exceeding $a$ is $\mathbb{P}(B_t>a\mid\mathcal{F}_s)=1-\Phi((a-B_s)/\sqrt{t-s})$, where $\Phi$ is the standard normal CDF. The strict inequality here corresponds to the event $B_t>a$; replacing it by $B_t\ge a$ gives the same conditional probability because a nondegenerate Gaussian distribution assigns probability zero to a single point.

For nested observations, condition first on the latest time and then on the earlier one. If $s<u<t$ and $f(B_t)$ is integrable, the tower property gives $\mathbb{E}[f(B_t)\mid\mathcal{F}_s]=\mathbb{E}[\mathbb{E}[f(B_t)\mid\mathcal{F}_u]\mid\mathcal{F}_s]$. The inner conditional expectation is $P_{t-u}f(B_u)$; averaging this function of $B_u$ over the transition from $B_s$ gives the same result as the single transition of duration $t-s$. This is useful when a functional is naturally evaluated through intermediate times.

Common mistake: forgetting the current value. A wrong calculation is $\mathbb{E}[B_2^2\mid\mathcal{F}_{1/2}]=\mathbb{E}[B_2^2]=2$, on the grounds that $B_2$ is “Gaussian.” It is Gaussian unconditionally, but it is not independent of $\mathcal{F}_{1/2}$. The corrected calculation writes $B_2=B_{1/2}+Z$, with $Z$ independent of $\mathcal{F}_{1/2}$, and gives $B_{1/2}^2+\tfrac32$. Its expectation is $2$, as required, but the conditional answer retains the information already observed.

Common mistake: multiplying marginal densities for dependent values. Future values $B_{t_1}$ and $B_{t_2}$, with $s<t_1<t_2$, are not conditionally independent given $\mathcal{F}_s$: both contain the increment from $s$ to $t_1$. The correct joint density uses the successive factors $p_{t_1-s}(B_s,x_1)p_{t_2-t_1}(x_1,x_2)$, not two independent densities each centered at $B_s$. The product example above exhibits the effect: the shared increment contributes the nonzero term $\tfrac12$.

Check Your Understanding

Use the transition-density or increment representation in each question, stating any integrability condition you need.

For $0\le s<t$, compute $\mathbb{E}[B_t\mid\mathcal{F}_s]$ and $\mathbb{E}[B_t^2\mid\mathcal{F}_s]$.
For $s<t$, express $\mathbb{P}(B_t\le a\mid\mathcal{F}_s)$ using $\Phi$.
For $s<u<t$, verify the kernel composition identity for $f(y)=y^2$.
Compute $\mathbb{E}[B_1B_3\mid\mathcal{F}_{1/2}]$ by writing both future values in terms of independent increments.
For $s<t$, find $\mathbb{E}[e^{c(B_t-B_s)}\mid\mathcal{F}_s]$ for real $c$, and explain why the result is nonrandom.