Exponential and Trigonometric Brownian Examples

Three Functions, Three Ito Decompositions

Applying the ordinary chain rule to a smooth function of a differentiable path produces only a first-derivative term. Brownian motion has nonzero quadratic variation, so Ito’s formula adds a second-derivative correction. For the exponential, sine, and cosine, that correction is especially transparent: the exponential’s second derivative has the same sign as the function, while the sine and cosine each acquire a negative second derivative. Tracking these signs gives three frequently used decompositions.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$ relative to $(\mathcal{F}_t)$. In particular, for $0\le s<t$, $B_t-B_s$ is independent of $\mathcal{F}_s$ and has distribution $N(0,t-s)$. Fix a finite horizon $T>0$. An Ito integral $\int_0^t H_s\,dB_s$, $0\le t\le T$, below is taken with a progressively measurable integrand $H$ in $\mathcal{L}^2$, meaning $\mathbb{E}\int_0^T H_s^2\,ds<\infty$.

Definition: For a twice continuously differentiable function $f:\mathbb{R}\to\mathbb{R}$, Ito’s formula for Brownian motion is $f(B_t)=f(B_0)+\int_0^t f'(B_s)\,dB_s+\tfrac12\int_0^t f''(B_s)\,ds$. The integral is an Ito integral, and the second integral is an ordinary time integral. The correction comes from $dB_t\,dB_t=dt$, while $dB_t\,dt=0$ and $dt\,dt=0$. We use this formula, established in the earlier material on Ito’s formula for Brownian motion.

For each of the three functions, the task is to identify both derivatives before substituting them. The coefficient of $dB_t$ is the first derivative; the coefficient of $dt$ is one half of the second derivative. The integrability checks ensure that the stochastic integrals are well defined and square-integrable on every finite time interval.

The Exponential Function

Worked Example: The Exponential of Brownian Motion

Take $f(x)=e^x$. Its derivatives are $f'(x)=e^x$ and $f''(x)=e^x$. Thus the first-derivative term is $e^{B_t}\,dB_t$, and the explicit second-derivative correction is $\tfrac12 e^{B_t}\,dt$. Since $B_0=0$, Ito’s formula gives, for $0\le t\le T$,

e^{B_t}=1+\int_0^t e^{B_s}\,dB_s+\frac12\int_0^t e^{B_s}\,ds, \qquad de^{B_t}=e^{B_t}\,dB_t+\frac12e^{B_t}\,dt.

To check the integrability of the Ito integrand, the Gaussian moment-generating function gives $\mathbb{E}[e^{2B_s}]=e^{2s}$. Therefore

\mathbb{E}\int_0^T(e^{B_s})^2\,ds =\int_0^T e^{2s}\,ds =\frac{e^{2T}-1}{2}<\infty.

So $e^{B_s}$ is an admissible $\mathcal{L}^2$ integrand. The Ito integral is a square-integrable martingale: its integrability follows from the Ito isometry, and its conditional-expectation property follows first for simple adapted integrands from the zero conditional mean of Brownian increments, then for $\mathcal{L}^2$ integrands by the Ito-isometry limit.

Check by expectation: the Gaussian moment-generating function also gives $\mathbb{E}[e^{B_T}]=e^{T/2}$. The stochastic integral has mean zero. Moreover, $\mathbb{E}[e^{B_s}]=e^{s/2}$, so the expected right-hand side of the integral decomposition is

1+\frac12\int_0^T e^{s/2}\,ds =1+\frac12\cdot 2(e^{T/2}-1) =e^{T/2}.

This agrees with the direct Gaussian calculation. The check confirms the positive sign and the factor $\tfrac12$ in the drift correction.

Sine and Cosine

Worked Example: The Sine of Brownian Motion

Take $f(x)=\sin x$. Then $f'(x)=\cos x$ and $f''(x)=-\sin x$. The first-derivative term is $\cos(B_t)\,dB_t$; the correction is $\tfrac12(-\sin(B_t))\,dt$, so it has a minus sign. Since $\sin(B_0)=0$,

\sin(B_t)=\int_0^t\cos(B_s)\,dB_s-\frac12\int_0^t\sin(B_s)\,ds, \qquad d\sin(B_t)=\cos(B_t)\,dB_t-\frac12\sin(B_t)\,dt.

The integrand is progressively measurable and bounded in absolute value by $1$. Hence $\mathbb{E}\int_0^T\cos^2(B_s)\,ds\le T<\infty$, and its Ito integral is a square-integrable martingale with zero expectation.

Check by expectation: the centered normal distribution of $B_T$ is symmetric, so $\mathbb{E}[\sin(B_T)]=0$. Also, $\mathbb{E}[\sin(B_s)]=0$ for every $s$, and the stochastic integral has mean zero. Taking expectations in the displayed decomposition therefore gives $0=0-\tfrac12\int_0^T0\,ds$, as required. This expectation check is consistent with the negative correction; the derivative calculation itself fixes its sign.

Worked Example: The Cosine of Brownian Motion

Take $f(x)=\cos x$. Its derivatives are $f'(x)=-\sin x$ and $f''(x)=-\cos x$. Thus the stochastic coefficient is $-\sin(B_t)$, and the correction is $-\tfrac12\cos(B_t)\,dt$. Since $\cos(B_0)=1$, Ito’s formula gives

\cos(B_t)=1-\int_0^t\sin(B_s)\,dB_s-\frac12\int_0^t\cos(B_s)\,ds, \qquad d\cos(B_t)=-\sin(B_t)\,dB_t-\frac12\cos(B_t)\,dt.

Here $|\sin(B_s)|\le1$, so $\mathbb{E}\int_0^T\sin^2(B_s)\,ds\le T<\infty$. The Ito integral is consequently a square-integrable martingale and has mean zero.

Check by expectation: the characteristic function of $N(0,s)$ gives $\mathbb{E}[\cos(B_s)]=e^{-s/2}$. The expectation of the right-hand side at time $T$ is therefore

1-\frac12\int_0^T e^{-s/2}\,ds =1-\frac12\cdot 2(1-e^{-T/2}) =e^{-T/2}.

Directly, $\mathbb{E}[\cos(B_T)]=e^{-T/2}$, so the two calculations agree. In particular, the drift is negative, not positive.

The Decompositions and Their Integrability

Theorem: On every finite interval $[0,T]$, the following identities hold almost surely for all $t\in[0,T]$: $$ \begin{aligned} e^{B_t}&=1+\int_0^t e^{B_s}\,dB_s+\frac12\int_0^t e^{B_s}\,ds,\\ \sin(B_t)&=\int_0^t\cos(B_s)\,dB_s-\frac12\int_0^t\sin(B_s)\,ds,\\ \cos(B_t)&=1-\int_0^t\sin(B_s)\,dB_s-\frac12\int_0^t\cos(B_s)\,ds. \end{aligned} $$ Each Ito integral uses its displayed integrand, which is progressively measurable and in $\mathcal{L}^2$ on $[0,T]$.

Proof. Apply Ito’s formula to $f(x)=e^x$, whose first and second derivatives are both $e^x$, and use $e^{B_0}=1$. The resulting Ito integrand satisfies $\mathbb{E}\int_0^T e^{2B_s}\,ds=(e^{2T}-1)/2<\infty$, as computed above. Apply the same formula to $f(x)=\sin x$, with $f'(x)=\cos x$ and $f''(x)=-\sin x$; its Ito integrand satisfies $\mathbb{E}\int_0^T\cos^2(B_s)\,ds\le T$. Finally, for $f(x)=\cos x$, $f'(x)=-\sin x$ and $f''(x)=-\cos x$, and $\mathbb{E}\int_0^T\sin^2(B_s)\,ds\le T$. These bounds establish the required $\mathcal{L}^2$ conditions. Substitution in Ito’s formula gives exactly the three stated identities, including their initial values and correction terms. $\square$

The theorem’s stochastic integrals are true square-integrable martingales, not merely local martingales. For each, the $\mathcal{L}^2$ bound gives finite second moment by the Ito isometry; the martingale conditional-expectation property follows from the Ito-integral construction, as described in the exponential example. The processes on the left need not themselves be martingales: their decompositions also contain time-integral terms.

A Normalized Exponential Martingale

The positive correction in the exponential formula explains why $e^{B_t}$ is not a martingale. Removing that deterministic growth produces a martingale, a useful distinction when handling exponential Brownian expressions.

Proposition: The process $M_t=e^{B_t-t/2}$ is a true martingale on every finite time interval. In contrast, $e^{B_t}$ is not a martingale.

Proof. The Gaussian moment-generating function gives $\mathbb{E}[M_t]=e^{-t/2}e^{t/2}=1$, so $M_t$ is integrable. For $0\le s\le t$, write $B_t=B_s+(B_t-B_s)$. Since the increment is independent of $\mathcal{F}_s$,

\mathbb{E}[M_t\mid\mathcal{F}_s] =e^{B_s-t/2}\mathbb{E}[e^{B_t-B_s}] =e^{B_s-t/2}e^{(t-s)/2} =e^{B_s-s/2} =M_s.

This proves the martingale property, including the case $s=t$. For $t>s$, the same conditional-expectation calculation without the normalizing factor gives $\mathbb{E}[e^{B_t}\mid\mathcal{F}_s]=e^{B_s}e^{(t-s)/2}$, which differs from $e^{B_s}$. Thus $e^{B_t}$ is not a martingale. $\square$

Common Sign and Martingale Pitfalls

A frequent error is to differentiate $\cos x$ as $+\sin x$, or to forget that its second derivative is negative. That incorrect calculation would give $d\cos(B_t)=\sin(B_t)\,dB_t+\tfrac12\cos(B_t)\,dt$. The correction is concrete: at $x=0$, the actual derivatives are $f'(0)=0$ and $f''(0)=-1$, so Ito’s formula gives a zero diffusion coefficient and drift $-\tfrac12\,dt$ there. The erroneous expression gives drift $+\tfrac12\,dt$, contradicting the second derivative.

Another error is to label $e^{B_t}$ a martingale because it is an exponential of Brownian motion. Its conditional expectation is $e^{B_s}e^{(t-s)/2}$, not $e^{B_s}$. It is the normalized process $e^{B_t-t/2}$ that has the martingale property. Finally, the Ito correction is not obtained by applying the ordinary chain rule: it is the term $\tfrac12 f''(B_t)\,dt$ required by Brownian quadratic variation.

Check Your Understanding

For each question, track the first derivative separately from the second-derivative correction.

Write the Ito decomposition of $e^{2B_t}$, displaying both derivatives and the second-derivative correction.
What are the diffusion coefficient and drift correction in the decomposition of $\sin(B_t)$?
For the cosine decomposition, identify the sign of each term and state the values of $f'(0)$ and $f''(0)$.
Why is $\int_0^t e^{B_s}\,dB_s$ a square-integrable martingale on a finite interval?
For $t>s$, calculate $\mathbb{E}[e^{B_t}\mid\mathcal{F}_s]$ and explain why it rules out the martingale property for $e^{B_t}$.