Ito’s Integral

Recall the stochastic differential equation (SDE) form:

\[dX_t = \mu(X_t, t)\, dt + \sigma(X_t, t)\, dW_t, \quad X_0 = x_0.\]

For simplicity, we’ll focus on the driving noise term and replace $dW_t$ with $dB_t$, where $B_t$ is a standard Brownian motion (Wiener process).

Integrating both sides gives:

\[X_T - X_0 = \int_0^T \mu(X_t, t)\, dt + \int_0^T \sigma(X_t, t)\, dB_t.\]

The first integral is a standard Riemann integral—familiar and deterministic. But the second term is trickier: it’s a stochastic integral with respect to the highly irregular paths of Brownian motion. We haven’t yet defined what $\int f(t) \, dB_t$ even means.

This note builds intuition for why stochastic integration works by exploring two key properties of Brownian motion paths: total variation and quadratic variation.

Total Variation

Consider a partition of the interval $[0, T]$ into $n$ equal parts: $t_k = k \cdot \Delta t$, where $\Delta t = T/n$ and $k = 0, 1, \dots, n-1$.

The total variation along this partition is the sum of absolute increments:

\[V_n = \sum_{k=0}^{n-1} |\Delta B_{t_k}| = \sum_{k=0}^{n-1} |B_{t_{k+1}} - B_{t_k}|,\]

where each $\Delta B_{t_k} \sim \mathcal{N}(0, \Delta t)$ independently.

Why use absolute values? They measure the “total distance traveled” by the path, regardless of direction.

Now compute the expectation:

Let $Z \sim \mathcal{N}(0, \Delta t)$. Then $$\mathbb{E}[

] = \sqrt{\Delta t} \cdot \sqrt{2/\pi}$$.

Thus,

\[\mathbb{E}[V_n] = n \cdot \sqrt{\Delta t} \cdot \sqrt{2/\pi} = \sqrt{n} \cdot \sqrt{T} \cdot \sqrt{2/\pi} = \sqrt{n T} \cdot \sqrt{2/\pi}.\]

As $n \to \infty$, $\mathbb{E}[V_n] \to \infty$.

This shows that the total variation of a Brownian path over $[0, T]$ is infinite. The path is continuous but has infinite length—like a fractal curve that wiggles infinitely much at every scale. This is why Brownian paths are nowhere differentiable.

Quadratic Variation

Now consider the sum of squared increments :

\[Q_n = \sum_{k=0}^{n-1} (\Delta B_{t_k})^2.\]

This is the quadratic variation along the partition.

Expectation:

\[\mathbb{E}[Q_n] = \sum_{k=0}^{n-1} \mathbb{E}[(\Delta B_{t_k})^2] = \sum_{k=0}^{n-1} \Delta t = T.\]

Variance:

For $Z \sim \mathcal{N}(0, \Delta t)$, $Z^2$ has variance $2 (\Delta t)^2$, so

\[\mathrm{Var}(Q_n) = n \cdot 2 (\Delta t)^2 = 2 T^2 / n \to 0 \quad \text{as } n \to \infty.\]

Thus, the quadratic variation of Brownian motion over $[0, T]$ is finite and equals 0.

Observe that if $n \to \infty$, $\mathbb{E}[(\Delta B_t)^2]=\Delta t \\ \mathrm{Var}((\Delta B_t)^2)=0$, we can somehow treat $(\Delta B_t)^2$ as a number instead of random variable, which leads to the formula that $dB_t \cdot dB_t=dt$