What Quadratic Variation Records for an Ito Process
For a smooth path, squared increments disappear as a partition is refined. For a Brownian-driven process, they do not: the Brownian part accumulates nonzero quadratic variation, while an integrable drift has finite variation and contributes none. Covariation records the corresponding interaction between the noisy parts of two processes. These quantities depend on the diffusion coefficients and on which Brownian drivers the processes share.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying the Brownian motions used below. For a fixed \(T>0\), every stochastic integral in this tutorial is an Ito integral with a progressively measurable integrand \(H\) satisfying \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\), that is, \(H\in\mathcal{L}^2\). Such an integral is a square-integrable martingale: its second moment is finite by the Ito isometry, and its conditional mean increment is zero by the martingale property of the Ito integral. Drift coefficients will be progressively measurable with \(\mathbb{E}\int_0^T|\mu_s|\,ds<\infty\), so their time integrals have finite variation almost surely.
For continuous semimartingales, covariation also satisfies the polarization identity \([X,Y]=\tfrac14([X+Y]-[X-Y])\). Indeed, on each partition, \((\Delta_iX+\Delta_iY)^2-(\Delta_iX-\Delta_iY)^2=4\Delta_iX\Delta_iY\); taking the limits gives the identity. In particular, \([B]_t=t\) for Brownian motion. This is a limit of squared increments, not the ordinary chain rule for differentiable paths.
The Formulas for One and Two Ito Processes
Let \(B\) be a Brownian motion and let \(X_t=X_0+\int_0^t\mu_s\,ds+\int_0^t\sigma_s\,dB_s\), with the integrability conditions above. The first term after \(X_0\) has finite variation; the second is the stochastic part. For two processes driven by the same Brownian motion, write \(Y_t=Y_0+\int_0^t\nu_s\,ds+\int_0^t\eta_s\,dB_s\).
Proof. First consider stochastic integrals with bounded, adapted step-process integrands. On an interval where the integrand is the \(\mathcal{F}_{t_{j-1}}\)-measurable value \(h_j\), the integral increment is \(h_j\Delta B_i\). The Brownian quadratic variation sums converge in probability to the interval length; multiplying by the bounded \(h_j^2\) gives the limit \(h_j^2\) times that length. There are finitely many step intervals, and partition increments straddling their endpoints vanish in probability by continuity of Brownian paths and boundedness of the integrands. Thus the quadratic variation limit for a step integrand \(H\) is \(\int_0^tH_s^2\,ds\).
For a general \(H\in\mathcal{L}^2\), choose bounded adapted step processes \(H^{(m)}\) with \(\mathbb{E}\int_0^T|H_s-H_s^{(m)}|^2ds\to0\); this is the usual step-process density in the Ito integral construction. Write \(M=\int H\,dB\) and \(M^{(m)}=\int H^{(m)}\,dB\). For any deterministic partition, the martingale increments are orthogonal in \(L^2\), so the Ito isometry gives $$ \mathbb{E}\sum_i\big(\Delta_i(M-M^{(m)})\big)^2 =\mathbb{E}\int_0^T|H_s-H_s^{(m)}|^2\,ds. $$ Also, by Cauchy-Schwarz for finite sums, $$ \left|\sum_i(\Delta_iM)^2-\sum_i(\Delta_iM^{(m)})^2\right| \le \left(\sum_i\big(\Delta_i(M-M^{(m)})\big)^2\right)^{1/2} \left(\sum_i\big(|\Delta_iM|+|\Delta_iM^{(m)}|\big)^2\right)^{1/2}. $$ The first factor becomes small in probability uniformly over the partition as \(m\) increases, by the displayed expectation identity and Markov's inequality. The second factor is bounded in probability, since its squared expectation is bounded by a constant times \(\mathbb{E}\int_0^T(H_s^2+|H_s^{(m)}|^2)\,ds\). Finally, \(\int_0^t|H^{(m)}|^2ds\to\int_0^tH^2ds\) in \(L^1\), by Cauchy-Schwarz and the \(L^2\) approximation. These facts transfer the step-process quadratic variation limit to \(M\).
The drift \(A_t=\int_0^t\mu_sds\) has finite variation \(V_T=\int_0^T|\mu_s|ds<\infty\) almost surely. Its partition sum satisfies \(\sum_i(\Delta_iA)^2\le(\max_i|\Delta_iA|)V_T\to0\) by continuity of \(A\). The cross sum with \(M\) is bounded in absolute value by \((\sum_i(\Delta_iA)^2)^{1/2}(\sum_i(\Delta_iM)^2)^{1/2}\); the second factor is bounded in probability by the quadratic variation result just proved. Hence adding the drift does not change the limit. This proves \([X]_t=\int_0^t\sigma_s^2ds\).
For \(X\) and \(Y\) with the same driver, apply the quadratic variation formula to \(X+Y\) and \(X-Y\). Their diffusion coefficients are \(\sigma+\eta\) and \(\sigma-\eta\). Polarization gives $$ [X,Y]_t=\frac14\int_0^t\big((\sigma_s+\eta_s)^2-(\sigma_s-\eta_s)^2\big)\,ds =\int_0^t\sigma_s\eta_s\,ds. $$ The integrals are finite almost surely by Cauchy-Schwarz and the assumed square integrability. \(\square\)
Shared and Independent Brownian Drivers
For independent drivers, covariation reflects which noises occur in both processes. Let \(B\) and \(C\) be independent Brownian motions on the filtered space. Consider $$ X_t=X_0+\int_0^t\mu_sds+\int_0^ta_s\,dB_s+\int_0^tb_s\,dC_s,\qquad Y_t=Y_0+\int_0^t\nu_sds+\int_0^tc_s\,dB_s+\int_0^td_s\,dC_s, $$ where all four stochastic integrands are progressively measurable and in \(\mathcal{L}^2\). Applying the same quadratic-variation argument to a linear combination of the two independent Brownian integrals gives \([\,\int a\,dB+\int b\,dC\,]_t=\int_0^t(a_s^2+b_s^2)ds\). The cross term vanishes because independent Brownian increments have zero product mean; for step integrands, the sum of these products has variance bounded by a constant times \(\sum_i(\Delta t_i)^2\), which tends to zero. The step approximation and Ito-isometry argument used above extend the result to \(\mathcal{L}^2\) integrands. Polarization then yields the following formula.
Worked Example: Drift Does Not Change Quadratic Variation
Let \(X_t=1+2t+3B_t\) on \([0,2]\). Compute its quadratic variation and compare it with the finite-partition sum along \(t_i=2i/n\).
Here \(\mu=2\) and \(\sigma=3\), so the formula gives \([X]_t=\int_0^t9\,ds=9t\), and \([X]_2=18\). On the stated partition, \(\Delta t_i=2/n\), and $$ \Delta_iX=2\Delta t_i+3\Delta_iB=\frac4n+3\Delta_iB. $$ Thus $$ \sum_i(\Delta_iX)^2 =\frac{16}{n}+ \frac{24}{n}\sum_i\Delta_iB+9\sum_i(\Delta_iB)^2. $$ The first term tends to zero. The middle term equals \((24/n)B_2\), which tends to zero almost surely because \(B_2\) is finite almost surely. The last term tends to \(9\cdot2=18\) in probability, by Brownian quadratic variation.
Check: The expectation of the sum is \(n(2\Delta t_i)^2+9n\Delta t_i=16/n+18\). The cross term has expectation zero because each Brownian increment is centered. As \(n\to\infty\), this expectation tends to \(18\), agreeing with the limit; the drift contribution \(16/n\) disappears.
Worked Example: A Time-Varying Diffusion Coefficient
Let \(X_t=\int_0^t s\,dB_s\) on \([0,3]\). Find \([X]_t\) and \([X]_3\), then check the endpoint using the Ito isometry.
The drift is zero and the diffusion coefficient is \(\sigma_s=s\). Therefore $$ [X]_t=\int_0^t s^2\,ds=\frac{t^3}{3},\qquad [X]_3=\frac{3^3}{3}=9. $$ For an independent check, the Ito isometry gives $$ \mathbb{E}[X_3^2]=\mathbb{E}\int_0^3s^2\,ds=\frac{3^3}{3}=9. $$ This agrees with the quadratic variation at the endpoint. The equality of these two numbers is a check for this zero-mean martingale, not a definition of quadratic variation: quadratic variation is the limit of squared increments, whereas the isometry computes a second moment.
Worked Example: Shared and Independent Noise in Two Processes
Let \(B,C\) be independent Brownian motions and set \(X_t=t+2B_t+2C_t\), \(Y_t=3t+3B_t-C_t\). Compute their quadratic variations and covariation on \([0,2]\).
The diffusion coefficients of \(X\) are \(a=2,b=2\), and those of \(Y\) are \(c=3,d=-1\). Hence $$ [X]_t=\int_0^t(2^2+2^2)\,ds=8t,\qquad [Y]_t=\int_0^t(3^2+(-1)^2)\,ds=10t, $$ and $$ [X,Y]_t=\int_0^t(2\cdot3+2\cdot(-1))\,ds=4t. $$ At \(t=2\), these values are \(16\), \(20\), and \(8\), respectively. The \(B\)-driver contributes \(2\cdot3=6\) per unit time, and the \(C\)-driver contributes \(2\cdot(-1)=-2\); together they give \(4\) per unit time.
Check: The identity \([X+Y]_2=[X]_2+[Y]_2+2[X,Y]_2\) gives \(16+20+2\cdot8=52\). Directly, \(X+Y=4t+5B_t+C_t\), whose quadratic variation is \(\int_0^2(5^2+1^2)ds=26\cdot2=52\), as required.
Interpretation and a Common Mistake
Quadratic variation measures accumulated diffusion energy; covariation measures the signed overlap of diffusion coefficients along shared noise. The Brownian identities \(dB_t\,dB_t=dt\), \(dB_t\,dt=0\), and \(dt\,dt=0\) are shorthand for the corresponding limits of partition sums. They do not turn individual increments into deterministic quantities. In particular, the statement \([B]_t=t\) is about the limiting sum of squares, not the ordinary chain rule.
Common mistake: adding drift squares to quadratic variation. For \(X_t=2t+3B_t\), a wrong calculation might report \([X]_t=(2^2+3^2)t=13t\), treating drift and diffusion as if both accumulated like Brownian increments. The correction is that drift increments have size proportional to \(\Delta t\), so their squared sum vanishes; only the diffusion contributes, giving \([X]_t=9t\). Another error is assigning nonzero covariation to independent noises merely because both processes are random. In the formula, independent drivers are separate terms: only coefficients attached to the same driver multiply.
Check Your Understanding
Use the stated integrability assumptions and distinguish the Brownian drivers in each calculation.
- If \(X_t=\int_0^t(1+s)\,dB_s\), compute \([X]_t\) and \([X]_1\).
- For \(X_t=\int_0^t2\,dB_s\) and \(Y_t=\int_0^t(-3)\,dB_s\), compute \([X,Y]_t\).
- If \(B\) and \(C\) are independent and \(X_t=B_t+C_t\), \(Y_t=B_t-C_t\), compute \([X,Y]_t\).
- Explain why an absolutely continuous drift \(\int_0^t\mu_sds\) has zero quadratic variation under the assumptions in this tutorial.
- For two processes driven by independent \(B,C\), identify which coefficient products contribute to their covariation.