Explicit Representations for Polynomial and Exponential Payoffs

From a Terminal Payoff to Its Integrand

A Brownian terminal payoff $F(B_T)$ can often be represented as its mean plus an Ito integral. The useful route is to calculate its conditional expectation at each time $t$. This gives a process $M_t=\mathbb{E}[F(B_T)\mid\mathcal{F}_t]$; applying Ito's formula to $M_t$ then identifies the integrand. The process starts at the mean payoff and ends at the payoff itself.

Throughout, work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\geq0},\mathbb{P})$ carrying a standard Brownian motion $B_t$, with $B_0=0$. Assume that $(\mathcal{F}_t)$ is the completed, right-continuous natural filtration of $B$, restricted to the time interval $[0,T]$, where $T>0$. Thus future Brownian increments are independent of $\mathcal{F}_t$, and the Brownian martingale representation theorem applies. The course sometimes uses $W$ for Brownian motion; here the notation convention is $B$, so the terminal variable called $W_T$ in the focus is written $B_T$.

Definition: An admissible Ito integrand on $[0,T]$ is a progressively measurable process $H_s$ satisfying $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. Such an integrand belongs to $\mathcal{L}^2$, and $\int_0^t H_s\,dB_s$ is a square-integrable Ito integral. In this tutorial all stochastic integrals are Ito integrals under $\mathbb{P}$, not Stratonovich integrals.

For $F(B_T)\in L^2(\mathbb{P})$, the Brownian martingale representation theorem says that $F(B_T)=\mathbb{E}[F(B_T)]+\int_0^T H_s\,dB_s$ for a unique integrand in $\mathcal{L}^2$, up to equality almost everywhere with respect to $ds\,d\mathbb{P}$. That theorem is a standard result established in the earlier material on Brownian martingale representation. Here we compute the integrands directly and verify the representations.

Three Explicit Representations

Proposition: For $T>0$ and $a\in\mathbb{R}$, the following identities hold almost surely: $$ B_T=\int_0^T 1\,dB_s,\qquad B_T^2=T+\int_0^T 2B_s\,dB_s,\qquad e^{aB_T}=e^{a^2T/2}+\int_0^T a e^{aB_s+\frac12a^2(T-s)}\,dB_s. $$ Each displayed integrand belongs to $\mathcal{L}^2$. If $a=0$, the exponential representation reduces to the constant identity $1=1$ with zero integrand.

Proof. For the first identity, apply Ito's formula to $f(t,x)=x$ along $B_t$. Its derivatives are $f_t=0$, $f_x=1$, and $f_{xx}=0$, so the Ito correction $\tfrac12 f_{xx}\,dt$ is zero and $dB_t=1\,dB_t$. Integrating from $0$ to $T$ and using $B_0=0$ gives the identity. The integrand is progressively measurable and has squared norm $\mathbb{E}\int_0^T1\,ds=T<\infty$.

For the square, take $f(t,x)=x^2+T-t$. Its derivatives are $f_t=-1$, $f_x=2x$, and $f_{xx}=2$. Ito's formula, including its second-derivative correction, gives $$ df(t,B_t)=\left(-1+\tfrac12\cdot2\right)dt+2B_t\,dB_t=2B_t\,dB_t. $$ Consequently $f(T,B_T)-f(0,B_0)=B_T^2-T$. The integrand is progressively measurable and $$ \mathbb{E}\int_0^T(2B_s)^2\,ds=4\int_0^T\mathbb{E}[B_s^2]\,ds =4\int_0^T s\,ds=2T^2<\infty. $$

For the exponential, define $u(t,x)=e^{ax+\frac12a^2(T-t)}$. Its derivatives are $u_t=-\tfrac12a^2u$, $u_x=au$, and $u_{xx}=a^2u$. Therefore Ito's formula gives $$ du(t,B_t)=\left(-\tfrac12a^2u+\tfrac12a^2u\right)dt+au\,dB_t = a e^{aB_t+\frac12a^2(T-t)}\,dB_t. $$ At the endpoints, $u(T,B_T)=e^{aB_T}$ and $u(0,B_0)=e^{a^2T/2}$, proving the stated identity once the integrability of the integrand is checked. For every real $a$, $$ \mathbb{E}\int_0^T a^2u(s,B_s)^2\,ds =a^2\int_0^T e^{a^2T+a^2s}\,ds =e^{2a^2T}-e^{a^2T} $$ when $a\ne0$, which is finite. If $a=0$, the integrand is identically zero and the integral is zero. In each case, Ito's formula is applied to a smooth function, the integrand is in $\mathcal{L}^2$, and the equality is an equality of square-integrable random variables. $\square$

These are true martingale representations, not merely local martingale identities. Indeed, the Ito isometry and the $\mathcal{L}^2$ bounds just computed make each stochastic integral square-integrable. For any square-integrable Ito integral $N_t=\int_0^t H_s\,dB_s$, the defining properties of the Ito integral give $\mathbb{E}[N_t\mid\mathcal{F}_s]=N_s$ for $s\leq t$. A local martingale need not be a true martingale; here square-integrability supplies the required integrability and conditional-expectation property.

Worked Representations and Checks

Worked Example: The Terminal Brownian Value

Represent $B_2$ as its mean plus an Ito integral, and verify the formula.

The Brownian motion starts at zero and has mean zero, so $\mathbb{E}[B_2]=0$. From the first identity in the proposition, the integrand is $H_s=1$, and $$ B_2=0+\int_0^2 1\,dB_s. $$ The norm check is $\mathbb{E}\int_0^2 1^2\,ds=2$, so this is an admissible integrand.

Check: Applying Ito's formula to $f(t,x)=x$, with $f_t=0$, $f_x=1$, and $f_{xx}=0$, gives $dB_t=1\,dB_t+\tfrac12\cdot0\,dt$. Integration from $0$ to $2$ recovers $B_2-B_0=B_2$. Also, the integral has mean zero and variance $2$, matching $\mathbb{E}[B_2]=0$ and $\mathbb{E}[B_2^2]=2$.

Worked Example: The Square of the Terminal Value

Represent $B_2^2$ as its mean plus an Ito integral, and check both the terminal value and the variance.

Since $B_2$ is normal with mean zero and variance $2$, its second moment is $\mathbb{E}[B_2^2]=2$. The representation is $$ B_2^2=2+\int_0^2 2B_s\,dB_s. $$ The integrand is in $\mathcal{L}^2$, since $\mathbb{E}\int_0^2(2B_s)^2\,ds=4\int_0^2s\,ds=8$.

Check: For $f(t,x)=x^2+2-t$, the derivatives are $f_t=-1$, $f_x=2x$, and $f_{xx}=2$. Thus $$ df(t,B_t)=\left(-1+\tfrac12\cdot2\right)dt+2B_t\,dB_t =2B_t\,dB_t. $$ At $t=0$, $f(0,0)=2$; at $t=2$, $f(2,B_2)=B_2^2$, as required. Independently, the Ito isometry gives variance $8$ for the integral. A centered normal variable of variance $2$ has fourth moment $3\cdot2^2=12$, so $\operatorname{Var}(B_2^2)=12-2^2=8$, in agreement.

Worked Example: An Exponential Terminal Payoff

For $T=1$ and $a=2$, represent $e^{2B_1}$ as its mean plus an Ito integral.

The normal moment-generating formula gives $\mathbb{E}[e^{2B_1}]=e^{2}$. The conditional-mean process is $u(t,B_t)=e^{2B_t+2(1-t)}$, and its Ito differential has integrand $2u(t,B_t)$. Therefore $$ e^{2B_1}=e^2+\int_0^1 2e^{2B_s+2(1-s)}\,dB_s. $$ The squared-integrand check is $$ \mathbb{E}\int_0^1\left(2e^{2B_s+2(1-s)}\right)^2ds =4\int_0^1e^{4+4s}\,ds=e^8-e^4<\infty. $$

Check: For $u(t,x)=e^{2x+2(1-t)}$, the derivatives are $u_t=-2u$, $u_x=2u$, and $u_{xx}=4u$. Hence $$ du(t,B_t)=\left(-2u+\tfrac12\cdot4u\right)dt+2u\,dB_t =2u\,dB_t. $$ The initial value is $u(0,0)=e^2$, and the terminal value is $u(1,B_1)=e^{2B_1}$. As a second check, the integral's variance is $e^8-e^4$. The payoff's variance is $\mathbb{E}[e^{4B_1}]-(\mathbb{E}[e^{2B_1}])^2=e^8-(e^2)^2=e^8-e^4$.

Why the Time Correction Matters

For the exponential payoff, the representation integrand is not obtained just by differentiating $e^{aB_t}$. The martingale being differentiated is the conditional expectation of the terminal payoff: $$ \mathbb{E}[e^{aB_T}\mid\mathcal{F}_t] =e^{aB_t}\mathbb{E}[e^{a(B_T-B_t)}] =e^{aB_t+\frac12a^2(T-t)}. $$ The second equality uses independence of $B_T-B_t$ from $\mathcal{F}_t$ and the normal moment-generating function. The time factor makes the drift from Ito's second-derivative term cancel the time derivative.

A common error is to use $e^{aB_t-\frac12a^2t}$ as the conditional expectation of $e^{aB_T}$. That process is a different exponential, with a different initial and terminal value. For example, at $t=T$ it equals $e^{aB_T-\frac12a^2T}$, not $e^{aB_T}$, unless $a=0$ or $T=0$. Its martingale property does not make it the representation process for the unscaled terminal payoff. The correct process has the plus sign in its exponent and the remaining time $T-t$.

The distinction between drift and diffusion is visible in the Ito calculation: the exponential's time derivative contributes drift $-\tfrac12a^2u\,dt$, while its quadratic-variation correction contributes $+\tfrac12a^2u\,dt$. They cancel, leaving only the diffusion term. This uses $dB_t\,dB_t=dt$, $dB_t\,dt=0$, and $dt\,dt=0$; the ordinary chain rule would omit the correction. No change from $\mathbb{P}$ to a risk-neutral measure $\mathbb{Q}$ is involved.

For $B_T^2$, the same cancellation is encoded by subtracting elapsed time in the martingale $B_t^2+T-t$. The payoff mean is $T$, while the integral begins at zero in expectation. For $B_T$, there is no second-derivative correction because the function $x\mapsto x$ has second derivative zero. Together, these examples illustrate the general procedure: start from the conditional expectation of the terminal variable, apply Ito's formula with its correction term, and check that the resulting integrand is square-integrable.

Check Your Understanding

In each question, use the Ito-integral convention and assume $T>0$.

What is the integrand in the representation of $B_T$, and what is its $\mathcal{L}^2$ norm squared?
For $B_T^2$, identify the conditional-mean process and show how its time derivative cancels the Ito correction.
Write the exponential representation when $a=-1$, including its mean term and time-dependent integrand.
Verify directly that the exponential integrand has finite expected squared integral for every real $a$, including $a=0$.
Explain why $e^{aB_t-\frac12a^2t}$ does not in general have terminal value $e^{aB_T}$.

Explicit Representations for Polynomial and Exponential Payoffs

What You'll Learn