Unit 5, lesson 35

Quadratic variation and stochastic limits

How sums of squared Brownian increments converge in mean square and in probability to elapsed time as the partition mesh shrinks, why finite variation paths give zero, and how quadratic covariation works for diffusions.

  • Video 35
  • Information, Brownian motion and quadratic variation

Realized variance adds up squared log returns over a period. In a constructed model whose log price moves as μt+0.2Wt\mu t + 0.2 W_t, with tt in years and WW a standard Brownian motion, each return shrinks when returns are sampled more often, so one might expect the sum of their squares to shrink to zero. It does not. Over one year it settles near 0.22=0.040.2^2 = 0.04, whatever the drift μ\mu. This lesson defines the partition limit behind that fact, proves it for Brownian motion in mean square from Gaussian fourth moments, shows why paths of finite variation give zero instead, and states the corresponding formulas for diffusions and their covariation.

Key terms

Partition mesh

Definition A partition of [0,T][0, T] is a finite set of times π={t0,t1,…,tn}\pi = \{t_0, t_1, \dots, t_n\} with 0=t0<t1<⋯<tn=T0 = t_0 < t_1 < \dots < t_n = T and interval lengths Δtk=tk+1−tk\Delta t_k = t_{k+1} - t_k. Its mesh is the length of its longest interval,

∣π∣=max⁡0≤k<nΔtk.\lvert \pi \rvert = \max_{0 \le k < n} \Delta t_k .

For a process XX, the quadratic sum over π\pi is

Qπ(X)=∑k=0n−1(Xtk+1−Xtk)2.\begin{aligned} &Q_\pi(X) \\[0.4em] &\quad = \sum_{k=0}^{n-1} \bigl(X_{t_{k+1}} - X_{t_k}\bigr)^2 . \end{aligned}

The partitions in this lesson are deterministic, fixed before any path is seen. A sequence of partitions with ∣πn∣→0\lvert \pi_n \rvert \to 0 may be nested or not, and its intervals need not be equal. The requirement is that the longest interval shrinks, so adding many points in one region while a long gap remains elsewhere does not make the mesh small.

In plain words The mesh measures how coarse a grid is at its coarsest place. Driving it to zero means sampling every part of the interval more and more finely, not just some parts.

Example The equal grid with n=100n = 100 intervals of [0,1][0, 1] has mesh 0.010.01. The partition with the points k/1000k/1000 for k=0,…,500k = 0, \dots, 500 followed by the single point 11 has 501501 intervals but mesh 0.50.5, because of the gap from 0.50.5 to 11. Refining only the left half never drives the mesh to zero.

Mean-square convergence

Definition Random variables XnX_n with E[Xn2]<∞\mathbb{E}[X_n^2] < \infty converge in mean square, or in L2L^2, to a random variable XX with E[X2]<∞\mathbb{E}[X^2] < \infty when

E[(Xn−X)2]→0.\mathbb{E}\bigl[(X_n - X)^2\bigr] \to 0 .

They converge in probability to XX when P(∣Xn−X∣>ε)→0\mathbb{P}(\lvert X_n - X \rvert > \varepsilon) \to 0 for every fixed ε>0\varepsilon > 0. Chebyshev’s inequality in the form

P(∣Xn−X∣>ε)≤E[(Xn−X)2]ε2\begin{aligned} &\mathbb{P}(\lvert X_n - X \rvert > \varepsilon) \\[0.4em] &\quad \le \frac{\mathbb{E}\bigl[(X_n - X)^2\bigr]}{\varepsilon^2} \end{aligned}

shows that mean-square convergence implies convergence in probability. The converse fails. Neither mode implies almost-sure convergence of the whole sequence, although convergence in probability gives almost-sure convergence along some subsequence. When the limit is a constant cc, the mean-square error splits as E[(Xn−c)2]=Var⁡(Xn)+(E[Xn]−c)2\mathbb{E}[(X_n - c)^2] = \operatorname{Var}(X_n) + (\mathbb{E}[X_n] - c)^2, variance plus squared bias.

In plain words Mean-square convergence makes the average squared error small. Convergence in probability only makes large errors rare, and a rare error may still be huge.

Example For the Brownian quadratic sum QnQ_n on the equal grid of [0,1][0, 1] with nn intervals, the next card shows that the mean is exactly 11 and the mean-square error is 2/n2/n. With ε=0.1\varepsilon = 0.1, Chebyshev gives P(∣Qn−1∣>0.1)≤200/n\mathbb{P}(\lvert Q_n - 1 \rvert > 0.1) \le 200/n. That bound is 22 for n=100n = 100 and says nothing, while for n=10000n = 10000 it is 0.020.02. For the converse, let Xn=nX_n = n on an event of probability 1/n1/n and Xn=0X_n = 0 otherwise. Then P(∣Xn∣>ε)≤1/n→0\mathbb{P}(\lvert X_n \rvert > \varepsilon) \le 1/n \to 0, so Xn→0X_n \to 0 in probability, while E[Xn2]=n→∞\mathbb{E}[X_n^2] = n \to \infty.

Quadratic variation

Definition A process XX has quadratic variation [X]T[X]_T on [0,T][0, T] when Qπn(X)→[X]TQ_{\pi_n}(X) \to [X]_T in probability for every sequence of partitions of [0,T][0, T] with ∣πn∣→0\lvert \pi_n \rvert \to 0. For a standard Brownian motion and every partition π\pi of [0,T][0, T],

E[Qπ(W)]=T,Var⁡(Qπ(W))=2∑k(Δtk)2≤2T∣π∣,\begin{aligned} \mathbb{E}[Q_\pi(W)] &= T, \\[0.4em] \operatorname{Var}(Q_\pi(W)) &= 2 \sum_{k} (\Delta t_k)^2 \\[0.4em] &\le 2T \lvert \pi \rvert , \end{aligned}

so Qπn(W)→TQ_{\pi_n}(W) \to T in mean square and in probability whenever ∣πn∣→0\lvert \pi_n \rvert \to 0, and [W]T=T[W]_T = T. To prove it, note that each increment ΔWk=Wtk+1−Wtk\Delta W_k = W_{t_{k+1}} - W_{t_k} is N(0,Δtk)N(0, \Delta t_k), so E[(ΔWk)2]=Δtk\mathbb{E}[(\Delta W_k)^2] = \Delta t_k, and these add to TT. Its fourth moment is 3(Δtk)23(\Delta t_k)^2, so Var⁡((ΔWk)2)=3(Δtk)2−(Δtk)2=2(Δtk)2\operatorname{Var}((\Delta W_k)^2) = 3(\Delta t_k)^2 - (\Delta t_k)^2 = 2(\Delta t_k)^2. Increments over disjoint intervals are independent, so the variances add, and (Δtk)2≤∣π∣ Δtk(\Delta t_k)^2 \le \lvert \pi \rvert \, \Delta t_k gives the bound. Since the mean is exactly TT, the mean-square error equals the variance and tends to zero, and Chebyshev gives

P(∣Qπ(W)−T∣>ε)≤2T∣π∣ε2.\begin{aligned} &\mathbb{P}\bigl(\lvert Q_\pi(W) - T \rvert > \varepsilon\bigr) \\[0.4em] &\quad \le \frac{2T \lvert \pi \rvert}{\varepsilon^2}. \end{aligned}

When ∑n∣πn∣<∞\sum_n \lvert \pi_n \rvert < \infty, for example for dyadic partitions, these bounds are summable and the Borel-Cantelli lemma upgrades the result to almost-sure convergence. The limit is taken along partitions fixed in advance and is not a supremum over partitions. The supremum of Qπ(W)Q_\pi(W) over all partitions of [0,T][0, T] is infinite almost surely, a path result stated here without proof.

In plain words A single squared increment is random, with mean equal to its interval length and a standard deviation 2\sqrt{2} times as large, so it has no tendency to equal that length. Over many independent intervals these errors average out and the sum concentrates at the elapsed time. Quadratic variation records how much squared movement a path accumulates, and for Brownian motion that amount is the clock itself.

Example On the constructed equal grid of [0,1][0, 1] with n=100n = 100, the quadratic sum has mean 11 and variance 2T2/n=0.022T^2/n = 0.02, a standard deviation of about 0.14140.1414. With n=10000n = 10000 the variance is 0.00020.0002 and the standard deviation about 0.014140.01414, ten times smaller. On an equal grid Qn=(T/n)∑kZk2Q_n = (T/n) \sum_k Z_k^2 with independent standard normal ZkZ_k, a scaled chi-square variable with nn degrees of freedom. For n=100n = 100 this gives P(∣Qn−1∣>0.5)≈0.0009\mathbb{P}(\lvert Q_n - 1 \rvert > 0.5) \approx 0.0009, far below the Chebyshev bound 0.080.08. The sum on any finite grid is random, and only its limit is TT.

Finite variation

Definition The total variation of a path AA on [0,T][0, T] is

TV⁡(A)=sup⁡π∑k∣Atk+1−Atk∣,\begin{aligned} &\operatorname{TV}(A) \\[0.4em] &\quad = \sup_{\pi} \sum_{k} \lvert A_{t_{k+1}} - A_{t_k} \rvert , \end{aligned}

the supremum over all partitions of [0,T][0, T], and AA has finite variation when TV⁡(A)<∞\operatorname{TV}(A) < \infty. Monotone paths have finite variation, continuously differentiable paths have TV⁡(A)=∫0T∣A′(t)∣ dt\operatorname{TV}(A) = \int_0^T \lvert A'(t) \rvert \, dt, and an integral At=∫0tμs dsA_t = \int_0^t \mu_s \, ds with ∫0T∣μs∣ ds<∞\int_0^T \lvert \mu_s \rvert \, ds < \infty is continuous with finite variation. If AA is continuous with finite variation, then

Qπ(A)≤max⁡k∣ΔAk∣TV⁡(A)⟶0(∣π∣→0),\begin{aligned} Q_\pi(A) &\le \max_{k} \lvert \Delta A_k \rvert \operatorname{TV}(A) \\[0.4em] &\longrightarrow 0 \quad (\lvert \pi \rvert \to 0), \end{aligned}

because AA is uniformly continuous on [0,T][0, T], so its largest increment over the partition tends to zero while the total variation stays fixed. The same reasoning gives ∣∑kΔAk ΔXk∣≤max⁡k∣ΔXk∣TV⁡(A)→0\bigl\lvert \sum_k \Delta A_k \, \Delta X_k \bigr\rvert \le \max_k \lvert \Delta X_k \rvert \operatorname{TV}(A) \to 0 for every continuous path XX. Continuity of AA is needed. A path that adds a single jump of size JJ to a continuous finite variation path has quadratic sums tending to J2J^2.

In plain words A path of finite variation travels a finite total distance, so as its moves get small, their squares get small faster and add up to nothing. A smooth or integrated drift therefore contributes no quadratic variation.

Example The constructed path At=3tA_t = 3t has TV⁡(A)=3T\operatorname{TV}(A) = 3T. On the equal grid each increment is 3T/n3T/n, so

Qn(A)=n(3Tn)2=9T2n,Q_n(A) = n \Bigl(\frac{3T}{n}\Bigr)^2 = \frac{9T^2}{n},

which is 0.090.09 for T=1T = 1 and n=100n = 100 and tends to zero. By contrast [3W]T=9[W]T=9T[3W]_T = 9 [W]_T = 9T, since every squared Brownian increment is multiplied by 99. The coefficient 33 is the same in both paths, but a smooth increment is of order hh and a Brownian increment of order h\sqrt{h}.

Brownian roughness

Definition Almost surely, a Brownian path has infinite total variation on every interval [s,t][s, t] with s<ts < t. For one interval [0,T][0, T], choose partitions with ∑n∣πn∣<∞\sum_n \lvert \pi_n \rvert < \infty, so that Qπn(W)→TQ_{\pi_n}(W) \to T almost surely. On a path with TV⁡(W)<∞\operatorname{TV}(W) < \infty the finite variation bound would force Qπn(W)→0Q_{\pi_n}(W) \to 0, since the path is continuous, which contradicts T>0T > 0. So P(TV⁡(W)<∞)=0\mathbb{P}(\operatorname{TV}(W) < \infty) = 0, and applying this to the countably many intervals with rational endpoints gives the statement for every interval. Consequently, almost every path is monotone on no interval, and the pathwise Stieltjes integral, which is built for integrators of finite variation, is not available for integrals against WW. That is why the Itô integral is constructed in The Ito integral and isometry. The stronger theorem that almost every Brownian path is nowhere differentiable is stated, not proved, in the course, and it does not follow from infinite variation alone.

In plain words A Brownian path wiggles at every scale. On finer and finer grids its squared moves still add up to the elapsed time, while its absolute moves add up to more and more without bound, so the path travels an infinite distance on every interval, however short.

Example On the equal grid of [0,1][0, 1] with nn intervals, each ∣ΔWk∣\lvert \Delta W_k \rvert has mean 2/(πn)\sqrt{2/(\pi n)}, so

E[∑k∣ΔWk∣]=2nπ,\mathbb{E}\Bigl[\sum_k \lvert \Delta W_k \rvert\Bigr] = \sqrt{\frac{2n}{\pi}},

about 7.987.98 for n=100n = 100 and 79.879.8 for n=10000n = 10000, while the variance of this sum is 1−2/π≈0.3631 - 2/\pi \approx 0.363 for every nn. The absolute sums grow like n\sqrt{n} while the quadratic sums settle at 11. Infinite variation and nowhere differentiability are different properties. The function f(x)=x2sin⁡(1/x2)f(x) = x^2 \sin(1/x^2) with f(0)=0f(0) = 0 is differentiable at every point of [0,1][0, 1] and still has infinite variation there.

Quadratic covariation

Definition For processes XX and YY on [0,T][0, T], the quadratic covariation [X,Y]T[X, Y]_T is the limit in probability of the cross sums

∑kΔXk ΔYkas ∣π∣→0,\sum_{k} \Delta X_k \, \Delta Y_k \quad \text{as } \lvert \pi \rvert \to 0,

when it exists. Since 4ΔXkΔYk=(ΔXk+ΔYk)2−(ΔXk−ΔYk)24 \Delta X_k \Delta Y_k = (\Delta X_k + \Delta Y_k)^2 - (\Delta X_k - \Delta Y_k)^2, it equals 14([X+Y]T−[X−Y]T)\tfrac14\bigl([X + Y]_T - [X - Y]_T\bigr) whenever both quadratic variations exist, and [X,X]T=[X]T[X, X]_T = [X]_T. The Cauchy-Schwarz inequality for the sums gives ∣[X,Y]T∣≤[X]T [Y]T\lvert [X, Y]_T \rvert \le \sqrt{[X]_T \, [Y]_T}. If WW and ZZ are independent standard Brownian motions and B=ρW+1−ρ2 ZB = \rho W + \sqrt{1 - \rho^2}\, Z with −1≤ρ≤1-1 \le \rho \le 1, then BB is a standard Brownian motion and [W,B]T=ρT[W, B]_T = \rho T. Each cross term ΔWkΔBk\Delta W_k \Delta B_k has mean ρΔtk\rho \Delta t_k and variance (1+ρ2)(Δtk)2(1 + \rho^2)(\Delta t_k)^2, so the cross sum has mean ρT\rho T and variance at most (1+ρ2)T∣π∣(1 + \rho^2) T \lvert \pi \rvert. For a continuous Itô process

Xt=X0+∫0tμs ds+∫0tσs dWs,\begin{aligned} X_t &= X_0 + \int_0^t \mu_s \, ds \\[0.4em] &\quad + \int_0^t \sigma_s \, dW_s , \end{aligned}

with an adapted, jointly measurable drift satisfying ∫0T∣μs∣ ds<∞\int_0^T \lvert \mu_s \rvert \, ds < \infty and a predictable coefficient satisfying ∫0Tσs2 ds<∞\int_0^T \sigma_s^2 \, ds < \infty almost surely, and YY built in the same way from a drift ν\nu, a coefficient η\eta and a Brownian motion BB for the same filtration with [W,B]t=∫0tρs ds[W, B]_t = \int_0^t \rho_s \, ds, the course states without proof

[X]T=∫0Tσs2 ds,[X,Y]T=∫0Tρsσsηs ds.\begin{gathered} [X]_T = \int_0^T \sigma_s^2 \, ds, \\[0.4em] [X, Y]_T = \int_0^T \rho_s \sigma_s \eta_s \, ds . \end{gathered}

For constant coefficients these follow from the cards above, because the drift terms have finite variation and drop out of every cross sum. The shorthand (dWt)2=dt(dW_t)^2 = dt, dWt dt=0dW_t \, dt = 0, (dt)2=0(dt)^2 = 0 and dWt dBt=ρt dtdW_t \, dB_t = \rho_t \, dt summarizes these partition limits and is not an identity between real numbers. For processes with jumps the squared jump sizes also enter. A Poisson process NN has [N]t=Nt[N]_t = N_t, the number of jumps up to time tt.

In plain words Covariation is the joint version of quadratic variation. Only the Brownian parts contribute, through the product of their coefficients and the correlation of their drivers, while drifts and other smooth terms drop out. Unlike a quadratic variation, a covariation can be negative.

Example For constructed drivers with ρ=0.5\rho = 0.5 on [0,1][0, 1] and n=100n = 100 equal intervals, the cross sum ∑kΔWkΔBk\sum_k \Delta W_k \Delta B_k has mean exactly 0.50.5 and variance (1+0.25)⋅100⋅0.012=0.0125(1 + 0.25) \cdot 100 \cdot 0.01^2 = 0.0125. For Xt=3Wt+3tX_t = 3 W_t + 3t on the same grid,

E[Qn(X)]=n(9h+9h2)=9.09\begin{aligned} \mathbb{E}[Q_n(X)] &= n\bigl(9h + 9h^2\bigr) \\[0.4em] &= 9.09 \end{aligned}

with h=0.01h = 0.01, which tends to [X]1=9[X]_1 = 9, since the drift contribution 9T2/n=0.099T^2/n = 0.09 and the cross terms vanish in the limit.

Check yourself

  1. On [0,2][0, 2] with n=200n = 200 equal intervals, find the mean, the variance and the standard deviation of the Brownian quadratic sum Qn(W)Q_n(W), and the Chebyshev bound for P(∣Qn(W)−2∣>0.5)\mathbb{P}(\lvert Q_n(W) - 2 \rvert > 0.5).
Show the answer

Each interval has length h=0.01h = 0.01. The mean is ∑kh=T=2\sum_k h = T = 2. The variance is

2∑kh2=2⋅200⋅0.0001=0.04,\begin{aligned} 2 \sum_k h^2 &= 2 \cdot 200 \cdot 0.0001 \\[0.4em] &= 0.04, \end{aligned}

so the standard deviation is 0.20.2. Chebyshev gives

P(∣Qn−2∣>0.5)≤0.040.25=0.16.\begin{aligned} &\mathbb{P}(\lvert Q_n - 2 \rvert > 0.5) \\[0.4em] &\quad \le \frac{0.04}{0.25} = 0.16 . \end{aligned}

The exact probability from the chi-square law with 200200 degrees of freedom is about 0.01270.0127, so the bound is crude, but it is enough to prove the limit because it tends to zero as the mesh shrinks.

  1. The path At=2tA_t = 2t for t<0.5t < 0.5 and At=2t+1A_t = 2t + 1 for t≥0.5t \ge 0.5 on [0,1][0, 1] has finite variation. A classmate concludes that its quadratic variation is zero. Compute Qn(A)Q_n(A) on the equal grid with nn intervals, find the limit, and name the hypothesis the classmate overlooked.
Show the answer

The total variation is 2+1=32 + 1 = 3, so the path does have finite variation. Every interval of the grid has the increment 2h2h with h=1/nh = 1/n, except the one interval (tk,tk+1](t_k, t_{k+1}] with tk<0.5≤tk+1t_k < 0.5 \le t_{k+1}, whose increment is 2h+12h + 1. Hence

Qn(A)=n(2h)2−(2h)2+(2h+1)2=1+8n⟶1.\begin{aligned} Q_n(A) &= n(2h)^2 - (2h)^2 \\[0.4em] &\quad + (2h + 1)^2 \\[0.4em] &= 1 + \frac{8}{n} \longrightarrow 1 . \end{aligned}

The zero limit holds only for continuous paths of finite variation. Here the jump of size 11 sits inside one interval of every partition, and its square survives in the limit. A realized variance computed from data with a large discrete move picks up that move in the same way.

  1. Let Xt=0.2Wt+0.05tX_t = 0.2 W_t + 0.05 t and Yt=0.3BtY_t = 0.3 B_t with B=−0.4W+0.84 ZB = -0.4 W + \sqrt{0.84}\, Z, where WW and ZZ are independent standard Brownian motions, and let T=0.5T = 0.5. Find [X]T[X]_T, [Y]T[Y]_T, [X,Y]T[X, Y]_T and the ratio [X,Y]T/[X]T[Y]T[X, Y]_T / \sqrt{[X]_T [Y]_T}. Why can [X,Y]T[X, Y]_T be negative when [X]T[X]_T cannot, and why does the term 0.05t0.05t not matter?
Show the answer

Since 0.84=1−0.42\sqrt{0.84} = \sqrt{1 - 0.4^2}, the process BB is a standard Brownian motion with [W,B]t=−0.4t[W, B]_t = -0.4t. Then

[X]T=0.22⋅0.5=0.02,[Y]T=0.32⋅0.5=0.045,[X,Y]T=0.06⋅(−0.4)⋅0.5=−0.012,\begin{aligned} [X]_T &= 0.2^2 \cdot 0.5 = 0.02, \\[0.4em] [Y]_T &= 0.3^2 \cdot 0.5 = 0.045, \\[0.4em] [X, Y]_T &= 0.06 \cdot (-0.4) \cdot 0.5 \\[0.4em] &= -0.012 , \end{aligned}

using 0.2⋅0.3=0.060.2 \cdot 0.3 = 0.06. The ratio is −0.012/0.02⋅0.045=−0.012/0.03=−0.4-0.012/\sqrt{0.02 \cdot 0.045} = -0.012/0.03 = -0.4, the correlation of the drivers. A quadratic variation is a limit of sums of squares, so it is never negative, while a cross sum ∑kΔXkΔYk\sum_k \Delta X_k \Delta Y_k has terms of either sign. The drift 0.05t0.05t is continuous with finite variation, so its quadratic sums and its cross sums with the continuous paths of WW and BB tend to zero. On any finite grid the cross sum even has mean exactly −0.012-0.012, since the drift multiplies increments of BB with mean zero.

How this lesson connects