Shifting and Rescaling the Brownian Input
A constant shift changes where a function is evaluated, while a constant rescaling changes both the stochastic term and the size of the Ito correction. To keep these effects distinct, fix constants \(a,b\in\mathbb{R}\) and consider \(X_t=a+bB_t\). Since \(B_0=0\), the initial value is \(X_0=a\). The stochastic coefficient is \(b\), and the quadratic variation of \(X\) over time \(t\) is \(b^2t\): the sign of \(b\) affects increments of \(X\), but not their squared size.
Throughout, let \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carry a standard Brownian motion \(B_t\) relative to the filtration. An Ito integrand \(H_s\) used below is progressively measurable and belongs to \(\mathcal{L}^2\) on each stated finite time interval: \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\). The resulting Ito integral is a true square-integrable martingale, not merely a local martingale. The Ito multiplication rules are \(dB_t\,dB_t=dt\), \(dB_t\,dt=0\), and \(dt\,dt=0\); these encode quadratic variation and are not the ordinary chain rule.
Proof. Apply the standard Ito formula for a \(C^2\) function of Brownian motion to \(g(B_t)\), where \(g(x)=f(a+bx)\). Its derivatives are \(g'(x)=b f'(a+bx)\) and \(g''(x)=b^2 f''(a+bx)\). The formula gives \(g(B_t)=g(B_0)+\int_0^t g'(B_s)\,dB_s+\tfrac12\int_0^t g''(B_s)\,ds\). Substituting \(B_0=0\), \(g(0)=f(a)\), and the two derivatives gives the displayed identity. The assumed integrability makes the stochastic integral an Ito integral in \(\mathcal{L}^2\), and ensures that the correction integral is integrable. \(\square\)
The term \(b\) appears once in the stochastic integrand because it comes from differentiating \(f(a+bx)\). The correction has a factor \(b^2\) because it uses the second derivative of that composition. If \(b\) changes sign, the argument \(a+bB_s\) changes as well; it is therefore not generally correct to say that the entire stochastic term simply changes sign. Nor is the correction necessarily unchanged apart from its explicit \(b^2\): \(f''\) is evaluated at the changed argument.
Polynomial Checks and the Sign of the Scale
Worked Example: A Square with Positive and Negative Scales
Take \(f(x)=x^2\), \(a=3\), and compare \(b=-2\) with \(b=2\). Here \(f'(x)=2x\) and \(f''(x)=2\). For either value of \(b\), the stochastic integrand is \(b f'(3+bB_s)\), and the correction rate is \(\tfrac12b^2 f''(3+bB_s)=4\).
For \(b=-2\), the stochastic integrand is \(-4(3-2B_s)=-12+8B_s\). Thus \[ (3-2B_t)^2=9+\int_0^t(-12+8B_s)\,dB_s+4t. \] Using the already established square identity \(\int_0^t B_s\,dB_s=\tfrac12(B_t^2-t)\), the right-hand side becomes \(9-12B_t+8\cdot\tfrac12(B_t^2-t)+4t=9-12B_t+4B_t^2\). Direct expansion gives \((3-2B_t)^2=9-12B_t+4B_t^2\), as required.
For \(b=2\), the stochastic integrand is \(4(3+2B_s)=12+8B_s\), so \[ (3+2B_t)^2=9+\int_0^t(12+8B_s)\,dB_s+4t. \] Evaluating the integral gives \(12B_t+4(B_t^2-t)\); adding \(9+4t\) yields \(9+12B_t+4B_t^2=(3+2B_t)^2\).
Check: the two stochastic integrands, \(-12+8B_s\) and \(12+8B_s\), are not negatives of one another: their \(8B_s\) terms agree. The correction rate is \(4\) in both cases, but the arguments of \(f''\) happen to make no difference here because \(f''\) is constant. The direct expansions confirm both identities.
Worked Example: A Cubic and an Expectation Check
Let \(f(x)=x^3\), \(a=1\), and \(b=2\). The derivatives are \(f'(x)=3x^2\) and \(f''(x)=6x\). The formula gives \[ (1+2B_t)^3 =1+6\int_0^t(1+2B_s)^2\,dB_s +12\int_0^t(1+2B_s)\,ds. \] The stochastic integrand is square-integrable on each finite interval: its square is a polynomial in \(B_s\), whose expectation is finite because Brownian motion has finite moments of every order. The time integral is integrable for the same reason.
To check the scale and the correction, expand the last term: \(12\int_0^t(1+2B_s)\,ds=12t+24\int_0^tB_s\,ds\). The formula therefore has a correction contribution of \(12t\). Independently, because \(B_t\sim N(0,t)\), its first and third moments are zero and its second moment is \(t\). Direct expansion gives \[ \mathbb{E}[(1+2B_t)^3] =\mathbb{E}[1+6B_t+12B_t^2+8B_t^3] =1+12t. \] On the formula’s right-hand side, the Ito integral has expectation zero, so the expectation is \(1+12t+24\int_0^t\mathbb{E}[B_s]\,ds=1+12t\). The independent Gaussian-moment calculation agrees.
The zero expectation of the stochastic integral is justified here, rather than inferred from its appearance: its integrand is progressively measurable and square-integrable, so the Ito integral is a true square-integrable martingale starting at zero. In particular, its expectation at time \(t\) is zero.
Exponential Functions and the Correction
Worked Example: A Negative Scale in an Exponential
Set \(f(x)=e^x\), \(a=\ln 2\), and \(b=-1\). Since \(f'=f''=e^x\), the identity is \[ 2e^{-B_t}=2-\int_0^t e^{\ln 2-B_s}\,dB_s +\frac12\int_0^t e^{\ln 2-B_s}\,ds =2-2\int_0^t e^{-B_s}\,dB_s+\int_0^t e^{-B_s}\,ds. \] The Ito integrand \(-2e^{-B_s}\) is progressively measurable, and \(\mathbb{E}\int_0^T4e^{-2B_s}\,ds=4\int_0^T e^{2s}\,ds=2(e^{2T}-1)<\infty\). Here \(\mathbb{E}[e^{-2B_s}]=e^{2s}\), by the moment-generating function of \(N(0,s)\). Thus the stochastic integral is a true martingale, with expectation zero.
For an independent check, the same normal moment-generating function gives \(\mathbb{E}[2e^{-B_t}]=2e^{t/2}\). Taking expectations on the formula’s right-hand side gives \(2+\int_0^t\mathbb{E}[e^{-B_s}]\,ds =2+\int_0^t e^{s/2}\,ds =2+2(e^{t/2}-1)=2e^{t/2}\). The two expectations agree. The positive time-integral term is the Ito correction; it remains positive even though the scale \(b\) is negative.
Expectation Identity and Why the Formula Works
Proof. The integrand \(H_s=b f'(a+bB_s)\) is progressively measurable and satisfies \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\), so the Ito integral \(M_t\) is square-integrable. To recall why it is a martingale, first take an elementary adapted integrand, constant on each time interval with its value measurable at the interval’s left endpoint. For \(u\le v\), the corresponding increment is a sum of left-endpoint-measurable coefficients multiplied by Brownian increments after those endpoints. Conditioning successively on the past gives conditional mean zero, by the independent, mean-zero Brownian increments. General progressively measurable \(\mathcal{L}^2\) integrands are approximated in \(\mathbb{E}\int H_s^2ds\) by such elementary integrands; Ito isometry gives convergence of their integrals in \(L^2\), hence in \(L^1\). Conditional expectation is continuous in \(L^1\), so the limiting integral also satisfies \(\mathbb{E}[M_t-M_u\mid\mathcal{F}_u]=0\). Thus it is a true martingale and \(\mathbb{E}[M_t]=0\). Taking expectations in the theorem’s identity now proves the corollary; the assumed integrability justifies taking the expectation inside the time integral. \(\square\)
This conditional-expectation argument also clarifies the distinction between a true martingale and a local martingale. A locally defined Ito integral need not be a true martingale without suitable integrability. Here the \(\mathcal{L}^2\) hypothesis supplies that integrability, so the zero-mean check is valid. In the exponential example, for instance, it is the integral term that has mean zero; the exponential process itself has expectation \(2e^{t/2}\), not the constant initial value \(2\).
The formula matters whenever a change of scale is made inside a nonlinear function. The derivative creates the factor \(b\) in the Ito integral, while the quadratic variation of \(a+bB_t\) creates the factor \(b^2\) in the correction. An ordinary chain rule retains the first effect but misses the second. The error is not fixed by choosing \(b\) negative or by shifting the starting point.
Common Mistake: Applying the Ordinary Chain Rule
For a concrete wrong calculation, take \(f(x)=x^2\), \(a=1\), and \(b=2\). Treating Brownian motion as an ordinary differentiable path would suggest \((1+2B_t)^2=1+4\int_0^t(1+2B_s)\,dB_s\), with no correction. But the actual Ito formula has \(f'(x)=2x\), \(f''(x)=2\), stochastic integrand \(2\cdot2(1+2B_s)=4(1+2B_s)\), and correction \(\tfrac12\cdot 2^2\cdot2\,dt=4\,dt\). Therefore \[ (1+2B_t)^2=1+4\int_0^t(1+2B_s)\,dB_s+4t. \] At \(t=1\), the incorrect expression has expectation \(1\), because its Ito integral is square-integrable and mean zero. The left-hand side has expectation \(1+4\mathbb{E}[B_1]+\!4\mathbb{E}[B_1^2]=5\). The omitted \(4t\) is exactly the missing expectation. This check pinpoints why the ordinary chain rule fails.
Check Your Understanding
In each question, keep the scale in the argument as well as in the explicit factors.
- For general \(a,b\), write the Ito formula for \(f(a+bB_t)\) and identify the stochastic integrand and correction rate.
- For \(f(x)=x^2\), \(a=1\), and \(b=-3\), compute both terms on the right-hand side and check the identity by expansion.
- For \(f(x)=e^x\), \(a=0\), and \(b=2\), state the Ito correction rate and verify the expectation of the formula using the normal moment-generating function.
- Why does changing \(b\) to \(-b\) not generally make the full stochastic integrand its negative?
- Which integrability property ensures that the Ito integral in the theorem is a true martingale, allowing its expectation to be set equal to zero?