Time Dependence Adds a Derivative, Not a New Noise Term
When a function depends on both time and Brownian motion, the Itô formula has three contributions: a time derivative, a stochastic integral from the first spatial derivative, and a second-order correction from Brownian quadratic variation. The time derivative is an ordinary \(dt\) term; it does not replace or absorb the correction. The examples below compute each contribution explicitly and check the resulting identities using moments.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\) adapted to the filtration. For \(T>0\), \(\mathcal{L}^2\) denotes the progressively measurable processes \(H_s\) satisfying \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\). Every stochastic integral below is an Itô integral with an integrand in this class. Equalities between processes are understood almost surely.
The correction is \(\tfrac12 f_{xx}\,dt\), because Brownian quadratic variation satisfies \(dB_t\,dB_t=dt\); the other multiplication rules are \(dB_t\,dt=0\) and \(dt\,dt=0\). The integral here is Itô, not Stratonovich. The distinction matters: the formula above uses the Itô integral and its quadratic-variation correction.
Exponentials with Explicit Time Dependence
Proof. Use \(f(t,x)=e^{at+bx}\). Its derivatives are \(f_t=ae^{at+bx}\), \(f_x=be^{at+bx}\), and \(f_{xx}=b^2e^{at+bx}\). Since \(B_s\sim N(0,s)\), its normal moment-generating function gives \(\mathbb{E}[e^{cB_s}]=e^{c^2s/2}\) for every real \(c\). In particular, $$ \mathbb{E}\int_0^T f_x(s,B_s)^2\,ds =b^2\int_0^T e^{(2a+2b^2)s}\,ds<\infty. $$ This is finite for every finite \(T\), including when \(b=0\). The same moment calculation shows that the expected integrals of \(|f_t(s,B_s)|\) and \(|f_{xx}(s,B_s)|\) are finite. Thus the formula applies. Substitution of the three derivatives, with \(f(0,0)=1\), gives the claimed decomposition. By the Itô isometry the stochastic integral has finite second moment, and the square-integrable Itô integral is a martingale: for \(u\le t\), its conditional expectation at time \(u\) equals its value at \(u\). Hence it is a true martingale, not merely a local martingale. \(\square\)
Worked Example: An Exponential with Concrete Coefficients
Take \(Y_t=e^{t/2+2B_t}\) on \([0,1]\). Compute its Itô decomposition, verify that its stochastic-integral term is square-integrable, and check its expectation.
For \(f(t,x)=e^{t/2+2x}\), the partial derivatives are \(f_t=\tfrac12 f\), \(f_x=2f\), and \(f_{xx}=4f\). The time derivative contributes \(\tfrac12 f\,dt\), and the second-order term contributes \(\tfrac12(4f)\,dt=2f\,dt\). The total drift coefficient is therefore \(\tfrac52 f\). The formula gives
For the Itô integral, its integrand \(H_s=2e^{s/2+2B_s}\) is progressively measurable, and $$ \mathbb{E}\int_0^1 H_s^2\,ds =4\int_0^1 e^{s+8s}\,ds =4\int_0^1 e^{9s}\,ds =\frac{4(e^9-1)}{9}<\infty. $$ So the stochastic integral is a square-integrable martingale with expectation zero. Taking expectations in the decomposition at \(t=1\), or directly using the normal moment-generating function, gives \(\mathbb{E}[Y_1]=e^{1/2}e^{2}=e^{5/2}\). The decomposition gives the same value: its ordinary integral has expectation \(\int_0^1 e^{(5/2)s}\,ds=\tfrac25(e^{5/2}-1)\), and multiplying by \(\tfrac52\) and adding 1 yields \(e^{5/2}\).
Check: The expectation calculated from the formula agrees with the direct Gaussian calculation. The process \(Y_t\) is not claimed to be a martingale: its decomposition has the nonzero drift \(\tfrac52Y_t\,dt\).
The Product of Time and Brownian Motion
Proof. Take \(f(t,x)=tx\). Then \(f_t=x\), \(f_x=t\), and \(f_{xx}=0\). The time integrand \(B_s\) is integrable on finite intervals, since \(\mathbb{E}\int_0^T|B_s|\,ds\le\int_0^T\sqrt{s}\,ds<\infty\). The stochastic integrand \(s\) is deterministic and progressively measurable, with \(\mathbb{E}\int_0^T s^2\,ds=T^3/3<\infty\). Applying the time-dependent Itô formula, and noting \(f(0,0)=0\), gives the identity. The integral \(\int_0^t s\,dB_s\) is a square-integrable martingale: its second moment is \(\int_0^t s^2\,ds\), and its conditional expectation at an earlier time equals its value then. \(\square\)
Worked Example: Checking the Product Identity at Time Two
Use the identity to express \(2B_2\) as two integrals, then verify its mean and variance by computing the variances and covariance of those integrals.
At \(t=2\), the decomposition is $$ 2B_2=A+M,\qquad A=\int_0^2 B_s\,ds,\qquad M=\int_0^2 s\,dB_s. $$ The integrand \(s\) is in \(\mathcal{L}^2\), since \(\int_0^2s^2\,ds=8/3\). The process \(B_s\) is progressively measurable; its ordinary time integral is square-integrable because its covariance function is \(\mathbb{E}[B_sB_u]=\min(s,u)\), which has a finite integral on the square \([0,2]^2\).
Both terms have mean zero. For \(A\), its variance is $$ \operatorname{Var}(A)=\int_0^2\int_0^2\min(s,u)\,du\,ds =2\int_0^2\int_0^s u\,du\,ds =2\int_0^2\frac{s^2}{2}\,ds =\frac83. $$ By the Itô isometry, \(\operatorname{Var}(M)=\int_0^2s^2\,ds=8/3\). They are not independent and their covariance is not zero. The bilinear form of the Itô isometry gives \(\mathbb{E}[B_sM]=\int_0^s u\,du=s^2/2\). Therefore, using integrability to interchange expectation and the time integral, $$ \operatorname{Cov}(A,M)=\int_0^2\mathbb{E}[B_sM]\,ds =\int_0^2\frac{s^2}{2}\,ds =\frac43. $$ Consequently, \(\operatorname{Var}(A+M)=8/3+8/3+2(4/3)=8\). Directly, \(\operatorname{Var}(2B_2)=4\operatorname{Var}(B_2)=4\cdot2=8\). The means also agree: both are zero.
Check: The covariance term is essential. Adding only the two variances would give \(16/3\), which disagrees with the variance of \(2B_2\). The missing \(8/3\) is exactly twice the covariance \(4/3\).
A Time-Weighted Square and Its Correction
Worked Example: Itô’s Formula for \(tB_t^2\)
Find an integral decomposition of \(tB_t^2\), check the integrability of the stochastic integral on \([0,1]\), and verify the expectation at time 1.
Choose \(f(t,x)=tx^2\). Its partial derivatives are \(f_t=x^2\), \(f_x=2tx\), and \(f_{xx}=2t\). In particular, the second-order contribution is \(\tfrac12 f_{xx}\,dt=t\,dt\), which does not vanish. Since \(B_s\) is progressively measurable, \(H_s=2sB_s\) is also progressively measurable. Its square-integrability check is $$ \mathbb{E}\int_0^1 H_s^2\,ds =4\int_0^1s^2\mathbb{E}[B_s^2]\,ds =4\int_0^1s^3\,ds =1. $$ Also, \(\mathbb{E}\int_0^1B_s^2\,ds=\int_0^1s\,ds=1/2<\infty\), so the ordinary terms are integrable. Itô’s formula yields, for \(0\le t\le1\), $$ tB_t^2=\int_0^t B_s^2\,ds+\int_0^t2sB_s\,dB_s+\int_0^t s\,ds. $$ The middle term is a square-integrable martingale and has expectation zero. At \(t=1\), the left side has expectation \(\mathbb{E}[B_1^2]=1\). On the right, the first term has expectation \(\int_0^1\mathbb{E}[B_s^2]\,ds=\int_0^1s\,ds=1/2\), the stochastic integral has expectation zero, and the last term is \(\int_0^1s\,ds=1/2\). The total is \(1/2+0+1/2=1\), as required.
Check: At a general \(t\), the same check gives \(\mathbb{E}[tB_t^2]=t^2\), while the two ordinary terms have expectations \(\int_0^t s\,ds=t^2/2\) each. Their sum is \(t^2\).
A Common Error: Dropping the Second-Order Term
A frequent mistake is to apply the ordinary chain rule after differentiating with respect to time and space, while omitting the contribution from \(dB_t\,dB_t=dt\). For \(f(t,x)=tx^2\), that incorrect calculation would write $$ tB_t^2\overset{\text{wrong}}{=}\int_0^t B_s^2\,ds+\int_0^t2sB_s\,dB_s. $$ At \(t=1\), the right side has expectation \(1/2+0=1/2\), while the left side has expectation \(1\). The contradiction exposes the error. The correction is \(\tfrac12 f_{xx}(s,B_s)\,ds=s\,ds\); including it adds \(\int_0^1s\,ds=1/2\) and restores the correct expectation.
The examples also illustrate why integrability should be checked rather than inferred from smoothness alone. A smooth function can have derivatives that grow too rapidly for its spatial derivative along Brownian motion to lie in \(\mathcal{L}^2\). Here the chosen derivatives have finite second moments on every finite time interval, as shown explicitly. The Itô integral terms are therefore genuine square-integrable martingales, and expectation checks using their zero means are justified.
Check Your Understanding
For each question, state the assumptions needed for any Itô integral you use.
- For \(f(t,x)=e^{3t-x}\), compute \(f_t,f_x,f_{xx}\) and the drift coefficient in the Itô decomposition of \(f(t,B_t)\).
- For the product \(tB_t\), compute the variance of \(\int_0^T s\,dB_s\) for a finite \(T\).
- For \(tB_t^2\), identify the second-order correction and explain why it is nonzero.
- At \(t=1\), compute the expectations of the two ordinary integral terms in the decomposition of \(tB_t^2\).
- In the variance check for \(2B_2\), what covariance term must be included when adding the variances of the two integrals?