Half of W Squared Is the Right Answer to the Wrong Sum

A particle starts at the origin and wanders at random for a time $T$. Chop the interval into $n$ equal pieces. For each piece, multiply the height the particle had when the piece began by the displacement it made during that piece, and add up all $n$ products. What number do you converge to as the pieces get small?

Almost everyone answers $\tfrac{1}{2}W(T)^2$ within a second, because the integral of $x\,dx$ is $\tfrac{1}{2}x^2$ and that fact is not in dispute. The limit is smaller by a constant:

The gap is $T/2$, it is deterministic, and it does not shrink with anything. A one-line argument rules the reflex answer out before any calculation: every term in the sum is a height that was fixed before its step, multiplied by a step of mean zero that is independent of that height, so every term has mean zero and so does the sum, for every partition rather than only in the limit. Meanwhile $\mathbb{E}\!\left[\tfrac{1}{2}W(T)^2\right] = T/2$, which is not zero. A quantity with mean $T/2$ cannot be the limit of quantities with mean zero.

One line of school algebra does the whole derivation

No stochastic machinery is needed to find the answer, only the observation that a product is half of a difference of squares. For any two numbers $a$ and $b$,

which you can read off a dissected square. Put $a = W(t_k)$ for the starting height and $b = W(t_{k+1}) - W(t_k)$ for the step, so that $a + b = W(t_{k+1})$, and sum over $k$. The $(a+b)^2$ of one piece is the $a^2$ of the next, so those two columns collapse to the single surviving term $W(T)^2$. Nothing collapses the third column.

Equation (3) is exact for every path and every partition, with no limit taken yet. So the reflex answer is the first term and the correction is the second, and the only remaining question is what the sum of squared steps does when the mesh is refined.

The squared steps refuse to disappear

In ordinary calculus that second sum is exactly the term you are trained to discard. If a function has a bounded derivative, a step over an interval of length $T/n$ is of size $T/n$, its square is of size $T^2/n^2$, and $n$ of those add to something of size $T/n$, which goes to zero. Brownian steps are bigger than that. A step over a window of length $T/n$ has variance $T/n$, so its typical size is $\sqrt{T/n}$, and its expected square is $T/n$ on the nose. Add $n$ of them:

Not small, not shrinking, and not even $n$-dependent. Refining the partition halves the size of each square and doubles their number, and the product is untouched. The variance of that sum is $2T^2/n$, so it converges in $L^2$ to the constant $T$: the correction term is not a leftover random variable, it is a number. Substituting $T$ into (3) gives (1).

The same answer without any sums

Once quadratic variation is on the table, the second-order term in a Taylor expansion stops being negligible, and that is the content of the chain rule for these paths. For a smooth $f$ applied to a Wiener process, $df(W) = f'(W)\,dW + \tfrac{1}{2}f''(W)\,dt$. Take $f(x) = x^2$, so $f' = 2x$ and $f'' = 2$:

Integrate both sides from $0$ to $T$. The left side is $W(T)^2$ because $W(0) = 0$, and the right side is $2\int_0^T W\,dW + T$. Rearranging is equation (1) again. The two derivations are the same computation seen from opposite ends: the sum shows you where the $dt$ in (5) comes from, and (5) saves you from ever writing the sum down.

The distribution follows for free. Since $W(T)/\sqrt{T}$ is standard normal, $W(T)^2/T$ is chi-squared on one degree of freedom, so the integral is an affine image of that law, shifted to have mean zero. Its variance is

using the fourth moment $\mathbb{E}[W(T)^4] = 3T^2$. Note the answer is bounded below by $-T/2$ and unbounded above, so it is visibly not normal, and anyone who reports a normal answer here has guessed rather than computed.

Where the analogy breaks, precisely

It is worth naming the exact property of the path that fails, because the failure is not about randomness. A Wiener path is continuous, so the usual pictures still apply. What it lacks is bounded variation: the total length of its ups and downs over any interval is infinite. Take a single path and measure it on nested meshes, each four times finer than the last. The total of the absolute steps doubles every time, which is the signature of $\sqrt{n}$ growth, while the total of the squared steps sits where it was.

That is the whole obstruction. The Riemann and Stieltjes theories buy their convergence with bounded variation, and once it is gone the value of a limit of sums starts to depend on where inside each interval you sample the integrand. For a smooth integrand that dependence washes out. Here it does not, and the choice becomes part of the definition rather than a matter of taste.

The reflex answer is a genuine integral, just not this one

Here is the part that makes the mistake worth taking seriously. Keep the same partition and the same path, and change only the height at which you evaluate the integrand: use the midpoint $\tfrac{1}{2}(W(t_k) + W(t_{k+1}))$ instead of the left end. The sum then collapses by the plain difference of squares and gives $\tfrac{1}{2}W(T)^2$ exactly, for every path and every $n$, with no limit and no correction. That is the Stratonovich integral:

On the simulated path drawn in Figure 2, the left-endpoint sum comes to $-0.2851$ and the midpoint sum over the same partition comes to $+0.2073$. The two conventions disagree about the sign of the answer, and their difference, $0.4923$, is half that path's accumulated squared steps.

So why is the left endpoint the standard? Because it is the only choice under which the integrand is known before the step it multiplies. That makes the sum a martingale, which is what gives it mean zero, and it is what makes the integral usable as the value of a strategy that cannot see the future. The midpoint rule quietly peeks at $W(t_{k+1})$ to decide how much of $dW$ to take. Stratonovich integrals obey the ordinary chain rule and are the right tool for pathwise and geometric questions, and they are useless for pricing anything. The $-T/2$ is the price of being unable to look ahead.

The general statement, and one honest caveat

Nothing above depended on equal steps. Equation (3) holds for any partition, and the expected total of the squared steps is $T$ for any partition too, so the answer is mesh-independent. The pattern also generalises: for $f(x) = x^m$ the chain rule gives $\int_0^T W^{m-1} dW = \frac{W(T)^m}{m} - \frac{m-1}{2}\int_0^T W^{m-2}\,dt$, so the correction is a genuine time integral rather than a constant as soon as $m > 2$, and only the case $m = 2$ is lucky enough to correct by a number.

The caveat is about convergence, not about the value. The sum of squared steps equals $T$ in the limit, in $L^2$ and almost surely along a refining sequence, and it does not equal $T$ at any finite $n$. Figure 2 shows 0.985 rather than 1, and Figure 3 shows 1.0645 at a coarse mesh. Anybody who claims exactness at finite $n$ is overclaiming; the identity that is exact at finite $n$ is (3), which is why it is worth writing down first.

Both identities here were checked symbolically and then against simulation. Over 80,000 paths of 1,000 steps each, the largest discrepancy in (3) was $2.3 \times 10^{-14}$, and (7) held to the same order. That is round-off rather than approximation error, which is the point: (3) and (7) are algebra, and only the value of the correction is probabilistic.

Sources and further reading

• Kiyosi Itô, “Stochastic Integral”, Proceedings of the Imperial Academy 20 (1944), 519–524. The four-page paper that defines the object in equation (1).
• Ruslan Stratonovich, “A New Representation for Stochastic Integrals and Equations”, SIAM Journal on Control4 (1966), 362–371, for the midpoint convention in equation (7).
• Wikipedia: Itô calculus, Quadratic variation and Wiener process.
• Wikipedia: Stratonovich integral, including the conversion formula between the two conventions.