Itô's Lemma and Stochastic Calculus › Why Ordinary Calculus Fails

Before deriving Itô's Lemma, we need to understand why ordinary calculus cannot be applied directly to stochastic processes. Standard calculus fails because Brownian motion paths are irregular. Unlike smooth functions that we encounter in standard calculus, Brownian motion exhibits roughness at every scale. This roughness changes how we think about differentiation and integration.

Suppose we have a smooth function $f(x)$ and we want to know how $f$ changes when $x$ changes by a small amount $\Delta x$. The strategy in ordinary calculus is to approximate the function locally using its derivatives. The Taylor expansion gives us this approximation:

$$f(x + \Delta x) = f(x) + f'(x) \Delta x + \frac{1}{2} f''(x) (\Delta x)^2 + \frac{1}{6} f'''(x) (\Delta x)^3 + \cdots$$

where:

• $f(x)$: value of the function at $x$

• $f'(x)$: first derivative of $f$

• $f''(x)$: second derivative of $f$

• $\Delta x$: small change in $x$

In ordinary calculus, when we take the limit $\Delta x \to 0$, all terms beyond the first-order term become negligible. This is because higher powers of small quantities become vanishingly small compared to the first power. The resulting differential form is:

$$df = f'(x) \, dx$$

where:

• $df$: differential change in the function $f$
• $f'(x)$: first derivative of $f$ with respect to $x$
• $dx$: differential change in $x$

This works because $(\Delta x)^2$ is infinitesimally smaller than $\Delta x$. We have $\lim_{\Delta x \to 0} \frac{(\Delta x)^2}{\Delta x} = 0$. The second-order term vanishes in the limit, leaving only the familiar first-order derivative expression that forms the foundation of differential calculus.

Now consider what happens when $x$ is replaced by a Brownian motion $W_t$. Over a small time interval $\Delta t$, the change in Brownian motion is $\Delta W_t \sim \mathcal{N}(0, \Delta t)$. While the expected value of $\Delta W_t$ is zero, its variance is $\Delta t$. This variance property is the source of all the complications that follow.

Here's the critical observation: what is the expected value of $(\Delta W_t)^2$? To answer this, we use the fundamental relationship between moments:

$$\begin{aligned} \mathbb{E}[(\Delta W_t)^2] &= \text{Var}(\Delta W_t) + (\mathbb{E}[\Delta W_t])^2 && \text{(decomposition of second moment)} \\ &= \Delta t + 0 && \text{(properties of Brownian increment)} \\ &= \Delta t && \text{(simplify)} \end{aligned}$$

where:

• $\mathbb{E}[\cdot]$: expected value operator

• $\Delta W_t$: increment of Brownian motion over time $\Delta t$

• $\text{Var}(\cdot)$: variance operator

This result has important consequences: $(\Delta W_t)^2$ behaves like $\Delta t$, not like $(\Delta t)^2$. When we sum many such terms over an interval, the second-order terms don't vanish as they would in ordinary calculus. Instead, they accumulate to a finite, deterministic quantity. This accumulation is the fundamental reason why the ordinary chain rule fails for stochastic processes and why we need a modified version, namely Itô's Lemma.

Let's make the preceding observation mathematically precise. We want to understand what happens when we sum up squared increments of Brownian motion over an interval. Partition the interval $[0, T]$ into $n$ subintervals of length $\Delta t = T/n$. The quadratic variation of Brownian motion over $[0, T]$ is defined as the sum of squared increments:

$$\sum_{i=1}^{n} (W_{t_i} - W_{t_{i-1}})^2 = \sum_{i=1}^{n} (\Delta W_i)^2$$

where:

• $n$: number of subintervals in the partition

• $W_{t_i}$: value of Brownian motion at time $t_i$

• $\Delta W_i$: increment of Brownian motion over the $i$-th interval

Each squared increment $(\Delta W_i)^2$ has expected value $\Delta t$. The variance of each squared increment is $2(\Delta t)^2$. This variance result follows from the fact that if $Z$ is a standard normal random variable, then $Z^2$ follows a chi-squared distribution with 1 degree of freedom, which has variance 2.

The sum has expected value and variance given by:

$$\mathbb{E}\left[\sum_{i=1}^{n} (\Delta W_i)^2\right] = n \cdot \Delta t = T$$

where:

• $n$: number of subintervals

• $\Delta t$: length of each subinterval ($T/n$)

• $T$: total time horizon

The variance of the sum involves a key computation:

$$\text{Var}\left[\sum_{i=1}^{n} (\Delta W_i)^2\right] = n \cdot 2(\Delta t)^2 = \frac{2T^2}{n} \to 0 \text{ as } n \to \infty$$

where:

• $\text{Var}$: variance operator

• $2(\Delta t)^2$: variance of a squared Brownian increment

As the partition becomes finer (as $n$ increases), something important occurs. The variance goes to zero while the expected value stays fixed at $T$. This means the sum converges in a stronger sense than expectation alone. The sum converges in mean square to the constant $T$:

$$\sum_{i=1}^{n} (\Delta W_i)^2 \xrightarrow{L^2} T$$

where:

• $\xrightarrow{L^2}$: convergence in mean square ($L^2$ norm)

• $T$: deterministic limit of the quadratic variation

The symbolic result $(dW_t)^2 = dt$ distinguishes stochastic from ordinary calculus. In ordinary calculus, the quadratic variation of any smooth function is zero. In stochastic calculus, Brownian motion has positive, finite quadratic variation equal to the length of the time interval. This non-zero quadratic variation is what forces us to include second-order terms in our calculus.