Numerically Testing Ito's Formula for the Square Function

What the Numerical Test Measures

Ito’s formula for the square function says that Brownian motion accumulates a correction from its quadratic variation. A grid-based test can make that correction visible: from simulated increments, compare the terminal square \(B_T^2\) with the left-endpoint sum \(2\sum_k B_{t_k}\Delta B_k\), including the time correction \(T\). The difference between these quantities is not numerical roundoff alone. On a finite grid it is exactly the difference between the grid’s quadratic-variation sum and \(T\).

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\) relative to \((\mathcal{F}_t)\). Fix \(T>0\) and a deterministic partition \(0=t_0<t_1<\cdots<t_n=T\). Write \(\Delta t_k=t_{k+1}-t_k\) and \(\Delta B_k=B_{t_{k+1}}-B_{t_k}\), for \(k=0,\ldots,n-1\). In a simulation on an equal grid, one generates independent increments with distribution \(N(0,T/n)\) and obtains path values recursively from \(B_{t_0}=0\) and \(B_{t_{k+1}}=B_{t_k}+\Delta B_k\).

The stochastic integral corresponding to \(S_n\) is the Ito integral \(\int_0^T B_s\,dB_s\). Its integrand \(H_s=B_s\) is adapted and progressively measurable, and it belongs to \(\mathcal{L}^2\) on \([0,T]\), since \(\mathbb{E}\int_0^T B_s^2\,ds=\int_0^T s\,ds=T^2/2<\infty\). Ito’s formula, applied to \(f(x)=x^2\), has \(f'(x)=2x\), \(f''(x)=2\), and correction \(\tfrac12 f''(B_t)\,d[B]_t=dt\), because \([B]_t=t\). Thus \(B_T^2=2\int_0^T B_s\,dB_s+T\). This uses the Ito integral, not the ordinary chain rule or the Stratonovich integral: the Stratonovich chain rule has no \(T\) correction for this function.

The Exact Grid Identity

Proof. For each grid interval, expand the difference of consecutive squares using \(B_{t_{k+1}}=B_{t_k}+\Delta B_k\):

Summing over \(k=0,\ldots,n-1\), the left side telescopes to \(B_T^2-B_0^2=B_T^2\). The two terms on the right sum to \(2S_n+Q_n\). Therefore \(B_T^2=2S_n+Q_n\), and subtracting \(2S_n+T\) from both sides gives \(R_n=Q_n-T\). This also shows why the identity is exact on the grid, even though its stochastic-integral counterpart is a limiting statement. \(\square\)

For the equal grid, the distribution of the residual gives a quantitative benchmark for a simulation. The increments are independent \(N(0,T/n)\) random variables. A centered normal variable of variance \(h\) has second moment \(h\) and fourth moment \(3h^2\). Consequently, each squared increment has mean \(T/n\) and variance \(3(T/n)^2-(T/n)^2=2(T/n)^2\). Independence lets us add these variances.

Proof. By the proposition, \(R_n=Q_n-T\). Put \(h=T/n\). Each \(\Delta B_k\) has distribution \(N(0,h)\), so \(\mathbb{E}[(\Delta B_k)^2]=h\) and \(\operatorname{Var}((\Delta B_k)^2)=3h^2-h^2=2h^2\). Adding the \(n\) means gives \(\mathbb{E}[Q_n]=nh=T\); adding the variances of the independent squared increments gives \(\operatorname{Var}(Q_n)=2nh^2=2T^2/n\). Hence \(R_n\) has mean zero and variance \(2T^2/n\). Since its mean is zero, \(\mathbb{E}[R_n^2]=\operatorname{Var}(R_n)\), and taking the nonnegative square root gives the stated root-mean-square size. This tends to zero, which is precisely \(L^2\) convergence of \(R_n\) to zero. \(\square\)

This theorem predicts the scale of a typical residual, not its value on every simulated path. Refining the grid reduces the root-mean-square residual across repeated simulations. It does not require the absolute residual on one particular path to decrease at every refinement.

Worked Grid Calculations

A Practical Simulation Procedure

For each chosen grid size \(n\), compute one path and its residual using the same increments throughout. In particular, generate the path values before forming the left-endpoint sum; using \(B_{t_{k+1}}\) in place of \(B_{t_k}\) changes the discretization. To compare grid sizes reliably, repeat the simulation over many independent paths and summarize the residuals at each fixed \(n\).

For many independent paths, the sample mean of \(R_n\) should fluctuate around zero, while the sample mean of \(R_n^2\) should approach \(2T^2/n\) as the number of paths grows. A finite Monte Carlo sample will not match the theoretical values exactly. Grid refinement and increasing the number of simulated paths address different sources of discrepancy: refinement reduces the theoretical residual variance, while more paths improve the precision of estimated averages.

Common Mistake: Dropping the Quadratic Correction

A tempting but incorrect calculation is to use \(B_T^2\approx2S_n\), omitting \(T\). For the four-increment path in the zero-residual example, this gives \(2S_4=-1\), even though \(B_1^2=0\). The resulting error \(B_1^2-2S_4=1\) is not the residual defined for the Ito test. The correction is to include \(T\): \(2S_4+T=-1+1=0\), which matches the terminal square. In general, the exact grid identity \(B_T^2=2S_n+Q_n\) shows the source of the correction; the limit \(Q_n\to T\) explains why Ito’s formula uses \(T\).

Another mistake is to demand that a single path’s absolute residual shrink at every refinement. The residual is \(Q_n-T\), and the quadratic-variation sum fluctuates. Its \(L^2\) convergence and decreasing variance are statements about the sequence of random variables, not a pathwise monotonicity guarantee. The useful numerical conclusion is statistical: at equal grid size \(n\), compare residuals across paths at that same \(n\), and compare their empirical mean-square size with \(2T^2/n\).