The Area Under a Random Path Is Normal, With Variance T Cubed Over Three

Shade the region between a diffusing particle and the time axis over one second. The box is one wide and about one tall, so the eye guesses a variance of one, and the answer is one third because each increment counts only for the time remaining after it. No stochastic integration is needed to define the object, only continuity of the path, and the constant is pinned twice over: once by the weight (T minus t) and once by integrating the covariance min(s,t) across the square.

Watch a particle diffuse from the origin for a time $T$ and shade the region between its trajectory and the time axis, counting area below the axis as negative. That shaded quantity is a random variable. What is its distribution?

The answer is clean enough to be worth memorising, and the route to it is worth more than the answer:

\int_0^T W(t)\,dt \;\sim\; \mathcal{N}\!\left(0,\; \frac{T^3}{3}\right).

(1)

Two things in that line need justifying and they are usually conflated. Normality is one claim and the variance is another, and they have different proofs. Before either, there is a prior question that trips people who know too much stochastic analysis: what kind of integral is this?

It is an ordinary integral, and that matters

The instinct of anyone who has just learned Itô calculus is to reach for it here, and the instinct is wrong. There is no $dW$ anywhere in the expression. The integrand is $W(t)$ and the measure is $dt$ , and almost every Wiener path is continuous on $[0,T]$ , hence bounded and Riemann integrable there. So fix a path, integrate it the way you would integrate any continuous function, and you have a well-defined number. Repeat for each path and you have a random variable.

This is not a technicality. Stochastic integration theory exists because $\int f\,dW$ cannot be defined pathwise, the path having unbounded variation. Nothing forces that machinery on $\int W\,dt$ : it is only convenient later, as a device for computing the variance. Confusing the two costs candidates their footing on a problem that needs none.

What the object is

For each outcome, $A(T) = \int_0^T W(t)\,dt$ is the Riemann integral of a continuous function, defined path by path. Continuity holds for almost every path rather than every path, so $A(T)$ is defined almost surely, which is all a random variable needs.

The tempting number, and how far off it is

The eye estimates area as height times width. The window is $T$ wide and the path wanders to a height of order $\sqrt{T}$ , so the area is of order $T^{3/2}$ and its variance of order $T^3$ . Take the natural constant of one, at $T = 1$ , and you guess a variance of $1$ . The truth is $1/3$ , so the guess is three times too large.

The scaling half of that reasoning is exactly right and is worth isolating, because it means only one number is actually at stake. Brownian motion satisfies $W(ct) \overset{d}{=} \sqrt{c}\,W(t)$ . Substitute $t = Tu$ in the integral:

\int_0^T W(t)\,dt \;=\; T\int_0^1 W(Tu)\,du \;\overset{d}{=}\; T^{3/2}\int_0^1 W(u)\,du.

(2)

So the variance is $cT^3$ for some universal constant $c$ , forced by scaling alone, and the entire problem reduces to finding $c$ . The box argument guesses $c = 1$ . Getting the exponent right and the constant wrong is a good answer to give while you compute, and a bad one to stop at.

Fig. 1 — One path of 480 steps. The two lobes cancel in part, which is why the variance falls well short of the enclosing box.

Why the area is normal

Approximate the integral by a right-endpoint Riemann sum on $n$ equal steps of length $h = T/n$ :

A_n \;=\; h\sum_{k=1}^{n} W(kh) \;=\; h\sum_{k=1}^{n}\sum_{j=1}^{k} \Delta_j, \qquad \Delta_j = W(jh) - W((j-1)h).

(3)

Each $A_n$ is a finite linear combination of the increments $\Delta_j$ , which are independent normals, so each $A_n$ is exactly normal. The Gaussian family is closed under convergence in distribution, and $A_n \to A(T)$ almost surely by continuity of the path, so the limit is normal too. Mean zero comes from linearity, since every $W(t)$ has mean zero. No central limit theorem is needed and none should be invoked, because the approximating variables are already normal rather than merely asymptotically so.

A neater version of the same argument comes from integration by parts, which also hands over the variance. Applied to $d(tW)$ , Itô's product rule gives $d(tW) = W\,dt + t\,dW$ , so integrating from $0$ to $T$ and rearranging,

\int_0^T W(t)\,dt \;=\; T\,W(T) - \int_0^T t\,dW(t) \;=\; \int_0^T (T - t)\,dW(t).

(4)

The right-hand side is a Wiener integral of a deterministic function. Such an integral is normal with mean zero, and by the Itô isometry its variance is the plain $L^2$ norm of the weight.

Two routes to the constant

Equation (4) has a reading that requires no formalism. Each increment of the path shifts every later value of $W$ by the same amount, so an increment occurring at time $t$ contributes to the area for exactly the remaining time $T - t$ . The first wiggle counts for the whole horizon and the last counts for nothing. Squaring and averaging that weight is the answer:

\operatorname{Var}\left[\int_0^T W\,dt\right] \;=\; \int_0^T (T-t)^2\,dt \;=\; \left[-\frac{(T-t)^3}{3}\right]_0^T \;=\; \frac{T^3}{3}.

(5)

At $T = 1$ the weight declines linearly from one to zero and the mean of its square is $1/3$ , which is the whole of the missing constant. That number is familiar from elsewhere: the average of $u^2$ over the unit interval.

The second route uses no stochastic integration at all, only the covariance function. Write the variance of the integral as a double integral and use $\operatorname{Cov}(W(s), W(t)) = \min(s,t)$ :

\operatorname{Var}\left[\int_0^T W\,dt\right] = \int_0^T\!\!\int_0^T \min(s,t)\,ds\,dt = 2\int_0^T\!\!\int_0^t s\,ds\,dt = \int_0^T t^2\,dt = \frac{T^3}{3}.

(6)

The trick in the middle is to use the symmetry of $\min$ and integrate over the triangle below the diagonal only, then double. On the unit square that inner integral is $t^2/2$ , twice it is $t^2$ , and the outer integral of $t^2$ is one third. Two independent computations agreeing is worth saying out loud, because it is the only cheap protection against a slipped factor of two in exactly the place where one is easy to slip.

Fig. 2 — The covariance surface min(s, t) over the square. Its average height is one third, and that average is the variance.

What the discrete sum really equals

The Riemann sum of equation (3) has an exact variance at every $n$ , and it is not $T^3/3$ . Since $\operatorname{Cov}(W(ih), W(jh)) = h\min(i,j)$ , the variance of $A_n$ is $h^3 \sum_{i,j} \min(i,j)$ , and that lattice sum collapses to a sum of squares:

\operatorname{Var}[A_n] \;=\; \left(\frac{T}{n}\right)^{3}\frac{n(n+1)(2n+1)}{6} \;=\; T^3\,\frac{2n^2 + 3n + 1}{6n^2}.

(7)

At $T = 1$ that is $5/8$ for two steps, $15/32$ for four, and $6767/20000$ for a hundred. Every one of these exceeds one third, and the sequence decreases to it without ever reaching it. Even at ten thousand steps the exact value is $0.3333833\ldots$ , which is short of $1/3$ only in the fifth decimal place. The approach from above has a reason: a right-endpoint sum takes the height at the end of each step, which is on average further from the origin than any interior point of that step, so it overstates the spread.

Fig. 3 — The discrete variance is exact at each n and strictly above one third at each n.

Getting this distinction right is what separates a correct answer from a lucky one. The clean statement $T^3/3$ belongs to the limit and to nothing else, and any claim that a finite sum equals it is false.

Companion facts, and where the method stops

Two by-products fall out of equation (4) at no extra cost. The covariance of the area with the endpoint is $\operatorname{Cov}(A(T), W(T)) = \int_0^T (T-t)\,dt = T^2/2$ , so the two are strongly but not perfectly correlated, and the correlation coefficient works out to $\sqrt{3}/2 \approx 0.866$ . Scaling in (2) also gives every moment at once: at $T = 2$ the variance is $8/3$ , and the whole law is that of $2^{3/2}$ times the unit-horizon area.

Now the limits. The pathwise Riemann argument leans on continuity, which holds for almost every path and not for every path, so $A(T)$ is defined almost surely and no stronger statement is available. Replace the Wiener process by one with jumps, a Poisson process for instance, and the paths stop being continuous, though a Riemann integral survives for right-continuous paths of locally bounded variation, and the variance is no longer $T^3/3$ because the covariance is no longer $\min(s,t)$ .

The normality argument is the more fragile of the two. It uses only that the increments are jointly Gaussian, so it transfers unchanged to fractional Brownian motion, where the same reasoning gives a normal area with a different variance. It fails completely for geometric Brownian motion: the time average of an exponential of a Gaussian is not lognormal and has no closed-form law, which is precisely why Asian options are hard to price and why (1) should not be borrowed for them. The identity that survives is the one you can name the hypotheses of.

Every number above was checked twice, symbolically and by simulation. Two hundred thousand paths integrated with the trapezoid rule, which is exact for a piecewise linear path, gave a variance of $0.331$ at $T = 1$ against the predicted $0.3333$ , a sample kurtosis of $3.03$ against the normal value of $3$ , and a covariance with the endpoint of $0.4955$ against $0.5$ .

Sources and further reading

Norbert Wiener, “Differential Space”, Journal of Mathematics and Physics 2 (1923), 131–174, where the process and its covariance function are constructed.
Wikipedia: Wiener process (the covariance $\min(s,t)$ and the scaling property), Gaussian process and Itô isometry.
Wikipedia: Square pyramidal number for the identity $n(n+1)(2n+1)/6$ behind equation (7).
Wikipedia: Asian option, on why the same integral becomes intractable once the path is exponentiated.