Random Coefficients from the Past
A simple stochastic integral need not have a deterministic coefficient. On an interval \((a,b]\), the coefficient may be a random variable such as \(B_a\), because its value is known at time \(a\). The Brownian increment \(B_b-B_a\), in contrast, is generated after that time. This separation between the information used to choose the coefficient and the increment being integrated is what makes the calculation work.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\) adapted to \((\mathcal{F}_t)\). Brownian increments after time \(a\) are independent of \(\mathcal{F}_a\), and \(B_b-B_a\) has distribution \(N(0,b-a)\) when \(a<b\). For the Ito integral, the relevant class \(\mathcal{L}^2\) consists of progressively measurable processes \(H_s\) satisfying \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\) on the time horizon in question.
The integral here is Ito: the coefficient is taken from the left endpoint and multiplied by the increment over the interval. A Stratonovich integral uses a different convention and may have a correction term when converted to Ito form. The computations below concern only simple Ito integrals. In particular, they rely on the fact that \(Y\) is known at time \(a\), not on an ordinary chain rule.
One Interval: Mean and Second Moment
Proof. Write \(\Delta B=B_b-B_a\) and \(h=b-a\). Since \(\Delta B\) is independent of \(\mathcal{F}_a\), it is independent of \(Y\), with \(\mathbb{E}[\Delta B]=0\) and \(\mathbb{E}[(\Delta B)^2]=h\). Independence gives \(\mathbb{E}[I^2]=\mathbb{E}[Y^2]\mathbb{E}[(\Delta B)^2]=h\mathbb{E}[Y^2]<\infty\), so \(I\) is integrable as well. Because \(Y\) is \(\mathcal{F}_a\)-measurable, conditioning on \(\mathcal{F}_a\) leaves \(Y\) fixed and the independent increment has mean zero and second moment \(h\). Thus \(\mathbb{E}[I\mid\mathcal{F}_a]=Y\mathbb{E}[\Delta B\mid\mathcal{F}_a]=0\) and \(\mathbb{E}[I^2\mid\mathcal{F}_a]=Y^2\mathbb{E}[(\Delta B)^2\mid\mathcal{F}_a]=hY^2\). Taking expectations of these conditional identities gives the stated unconditional formulas. \(\square\)
The conditional second moment retains the random size of the coefficient: once the past is known, the variance contributed by the interval is \((b-a)Y^2\). Averaging over the past produces the unconditional second moment. Since the mean is zero, this second moment is also the variance of \(I\).
Worked Example: Integrating the Coefficient \(B_1\)
For the integrand \(H_s=B_1\mathbf{1}_{(1,3]}(s)\), compute the expectation and second moment of \(I=\int_0^3 H_s\,dB_s\). The coefficient \(B_1\) is \(\mathcal{F}_1\)-measurable. Also, \(\mathbb{E}[B_1^2]=1\), so \(H\) is in \(\mathcal{L}^2\), with \(\mathbb{E}\int_0^3 H_s^2\,ds=(3-1)\mathbb{E}[B_1^2]=2\).
By the definition of this simple Ito integral, \(I=B_1(B_3-B_1)\). The interval length is \(3-1=2\), and the theorem gives
Check: Conditional on \(\mathcal{F}_1\), \(B_1\) is fixed and \(B_3-B_1\) has mean \(0\) and second moment \(2\). Hence \(\mathbb{E}[I^2\mid\mathcal{F}_1]=2B_1^2\); averaging gives \(2\mathbb{E}[B_1^2]=2\), as above.
Coefficients That Are Nonlinear Functions of Brownian Observations
A coefficient may be any square-integrable random variable measurable at the left endpoint; it need not be a linear function of the Brownian path. A useful calculation is therefore to find the coefficient's second moment first, and only then multiply by the interval length. If \(Y=g(B_a)\), for example, the required integrability condition is \(\mathbb{E}[g(B_a)^2]<\infty\).
Worked Example: A Centered Quadratic Coefficient
Set \(a=\tfrac12\), \(b=2\), and \(Y=B_{1/2}^2-\tfrac12\). Compute the mean and second moment of \(I=\int_0^2 Y\mathbf{1}_{(1/2,2]}(s)\,dB_s\), and check that the integrand is in \(\mathcal{L}^2\).
The coefficient is \(\mathcal{F}_{1/2}\)-measurable. Write \(X=B_{1/2}\), so \(X\sim N(0,\tfrac12)\). For a centered normal variable with variance \(a\), its second and fourth moments are \(a\) and \(3a^2\), respectively. Thus \(\mathbb{E}[X^2]=\tfrac12\) and \(\mathbb{E}[X^4]=3(\tfrac12)^2=\tfrac34\). Expanding the square gives
The interval length is \(2-\tfrac12=\tfrac32\). Consequently, the integrability check and the moment calculation are
Check: The coefficient is centered, since \(\mathbb{E}[Y]=\mathbb{E}[X^2]-\tfrac12=0\), but the integral's zero mean follows from the independent, mean-zero increment, not from that centering. Its conditional second moment is \(\tfrac32Y^2\); taking expectations gives \(\tfrac34\), matching the calculation.
Several Intervals
Now take a partition \(0=t_0<t_1<\cdots<t_n=T\). On each interval \((t_i,t_{i+1}]\), choose a square-integrable coefficient \(Y_i\) that is \(\mathcal{F}_{t_i}\)-measurable. The resulting integrand is \(H_s=\sum_{i=0}^{n-1}Y_i\mathbf{1}_{(t_i,t_{i+1}]}(s)\), and its integral is the finite sum \(\sum_{i=0}^{n-1}Y_i(B_{t_{i+1}}-B_{t_i})\). Its \(\mathcal{L}^2\) condition is \(\sum_{i=0}^{n-1}(t_{i+1}-t_i)\mathbb{E}[Y_i^2]<\infty\), which holds for a finite partition when each coefficient is square-integrable.
Proof. Define \(X_i=Y_i(B_{t_{i+1}}-B_{t_i})\). The one-interval theorem gives \(\mathbb{E}[X_i]=0\) and \(\mathbb{E}[X_i^2]=(t_{i+1}-t_i)\mathbb{E}[Y_i^2]\). For \(i<j\), \(X_i\) is \(\mathcal{F}_{t_j}\)-measurable because \(t_{i+1}\le t_j\). The one-interval conditional-mean identity, applied at \(t_j\), gives \(\mathbb{E}[X_j\mid\mathcal{F}_{t_j}]=0\). Therefore \(\mathbb{E}[X_iX_j]=\mathbb{E}[X_i\mathbb{E}[X_j\mid\mathcal{F}_{t_j}]]=0\). These expectations exist by the Cauchy–Schwarz inequality, since both variables are square-integrable. Expanding the square of the finite sum, the diagonal terms contribute \(\sum_i\mathbb{E}[X_i^2]\), and every cross term is zero. Linearity also gives \(\mathbb{E}[I]=\sum_i\mathbb{E}[X_i]=0\). Substituting the one-interval second moments proves the formula. \(\square\)
Worked Example: Two Successive Random Coefficients
On \([0,3]\), let \(H_s=0\) on \((0,1]\), \(H_s=B_1\) on \((1,2]\), and \(H_s=B_2\) on \((2,3]\). Compute the mean and second moment of \(I=\int_0^3H_s\,dB_s\).
The coefficients are measurable at their respective left endpoints. Their second moments are \(\mathbb{E}[B_1^2]=1\) and \(\mathbb{E}[B_2^2]=2\), so \(\mathbb{E}\int_0^3H_s^2\,ds=1\cdot1+1\cdot2=3<\infty\). The integral is
Both interval lengths are \(1\). The proposition gives \(\mathbb{E}[I]=0\) and
Check: The cross term has expectation zero: the first summand is \(\mathcal{F}_2\)-measurable, and the conditional expectation of \(B_2(B_3-B_2)\) given \(\mathcal{F}_2\) is \(B_2\mathbb{E}[B_3-B_2\mid\mathcal{F}_2]=0\). The two diagonal second moments are \(1\) and \(2\), so their sum is \(3\), as required.
A Common Mistake
Do not treat \(B_a\) and \(B_b\) as independent when \(a<b\). For example, a wrong calculation of \(\mathbb{E}[B_1(B_2-B_1)]\) might expand the product and set \(\mathbb{E}[B_1B_2]=0\), as if \(B_1\) and \(B_2\) were independent. That would give \(0-\mathbb{E}[B_1^2]=-1\), contradicting the single-interval theorem.
The correction is that \(B_2=B_1+(B_2-B_1)\), where the increment \(B_2-B_1\), not \(B_2\), is independent of \(\mathcal{F}_1\). Thus \(\mathbb{E}[B_1B_2]=\mathbb{E}[B_1^2]+\mathbb{E}[B_1(B_2-B_1)]=1+0=1\), and the expanded expression is \(1-1=0\). The safe procedure is to keep the integral in coefficient-times-increment form and apply independence to the increment over that interval.
For a simple Ito integral, the essential checks are therefore local to each interval: the coefficient must be measurable at the left endpoint, its second moment must be finite, and the increment must run from that endpoint to the right endpoint. With these conditions, the mean is zero and the second moment is the interval length times the coefficient's second moment; across several intervals, the cross terms vanish by conditioning on the past of each later increment.
Check Your Understanding
Use the left-endpoint measurability condition and the moment formulas to organize each calculation.
- For \(I=B_2(B_4-B_2)\), find \(\mathbb{E}[I]\) and \(\mathbb{E}[I^2]\).
- For \(Y=B_1^2-1\), find \(\mathbb{E}[Y^2]\) and then the second moment of \(Y(B_3-B_1)\).
- For coefficients \(Y_0,Y_1\) on two successive intervals, explain why the cross term in the squared integral has expectation zero.
- Explain why \(B_2\) is not an admissible coefficient on \((1,2]\) under the left-endpoint rule.
- State the \(\mathcal{L}^2\) condition for a coefficient \(Y\) used on a single interval \((a,b]\).