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dt \cdot dt = 0

dt \cdot dW_t = 0

(dW_t)^2 = dt

Now, we're going to use this to build a new type of "summing" tool: the Itô Integral. This is the stochastic version of the simple integral we learned about in Lesson 1.5.

Part 1: The "Normal" Integral (A Quick Review)

In Lesson 1.5, we learned that a normal integral, $\int_a^b f(x) dx$ , is just a "perfect sum" of tiny rectangles.

Width: $\Delta t$ (which becomes $dt$ )
Height: $f(t)$
Area: $f(t) \cdot \Delta t$

To calculate the total sum, $\sum f(t_i) \Delta t$ , we have to choose a height $f(t)$ for each tiny slice. For the slice from $t_i$ to $t_{i+1}$ , we could pick:

The left point: $f(t_i)$
The right point: $f(t_{i+1})$
The midpoint: $f( (t_i + t_{i+1}) / 2 )$

In "normal" calculus, it doesn't matter. As the rectangles get infinitely skinny, all three choices give you the exact same answer.

Part 2: The "Random" Integral (The New Problem)

Now, we want to define a stochastic integral:

I = \int_0^T f(W_t) dW_t

This is a completely different and harder problem. We are trying to find the "total" of:

\sum (\text{Height}) \cdot (\text{Width}) \to \sum f(W_{t_i}) \cdot (\Delta W_i)

Our "width" is no longer a predictable $\Delta t$ . It's a random step $\Delta W_i$ .

This creates a huge problem: Does our choice of $t_i$ matter now?

Let's test it. Let's try to solve the simplest possible stochastic integral, $f(W_t) = W_t$ :

\int_0^T W_t dW_t

Our "normal" calculus brain says $\int x \, dx = \frac{1}{2}x^2$ , so the answer *must* be $\frac{1}{2}W_T^2$ . Let's see if that's true.

Part 3: The Proof - Why The "Choice of Point" Matters

We will use a simple algebra identity: $y(y-x) = \frac{1}{2}(y^2 - x^2) + \frac{1}{2}(y-x)^2$ . Let's define our step:

$x = W_{t_i}$ (the "left point")
$y = W_{t_{i+1}}$ (the "right point")
$y-x = \Delta W_i$ (the "step")

Plugging these in gives us two "magic" identities:

\begin{gathered} W_{t_i} \Delta W_i = \frac{1}{2}\left( W_{t_{i+1}}^2 - W_{t_i}^2 \right) - \frac{1}{2}(\Delta W_i)^2 \\ W_{t_{i+1}} \Delta W_i = \frac{1}{2}\left( W_{t_{i+1}}^2 - W_{t_i}^2 \right) + \frac{1}{2}(\Delta W_i)^2 \end{gathered}

Now, let's sum them up from i=1 to n.

Test Case A: The "Left-Point" (Itô) Integral

We sum the first identity:

\sum_{i=1}^n W_{t_i} \Delta W_i = \sum_{i=1}^n \left[ \frac{1}{2}\left( W_{t_{i+1}}^2 - W_{t_i}^2 \right) \right] - \sum_{i=1}^n \left[ \frac{1}{2}(\Delta W_i)^2 \right]

The first sum is a "telescoping sum," which simplifies to $\frac{1}{2}(W_T^2 - W_0^2) = \frac{1}{2}W_T^2$ .

The second sum is the Quadratic Variation, which simplifies to $\frac{1}{2}T$ .

The Itô Answer: $\int_0^T W_t dW_t = \frac{1}{2}W_T^2 - \frac{1}{2}T$

Test Case B: The "Right-Point" Integral

We sum the second identity:

\sum_{i=1}^n W_{t_{i+1}} \Delta W_i = \sum_{i=1}^n \left[ \frac{1}{2}\left( W_{t_{i+1}}^2 - W_{t_i}^2 \right) \right] + \sum_{i=1}^n \left[ \frac{1}{2}(\Delta W_i)^2 \right]

The only difference is the + sign. The result is: $\int_0^T W_t dW_t = \frac{1}{2}W_T^2 + \frac{1}{2}T$

We have just proven that in stochastic calculus, the "choice of point" does not just matter, it gives a completely different answer. The "normal" calculus answer,

\frac{1}{2}W_T^2

, is wrong in both cases.

Part 4: Itô's Choice (The 'Non-Anticipating' Rule)

Kiyosi Itô made the choice that is the foundation of all modern finance. He defined his integral using only the left-hand point.

The Itô Integral Definition:

\int_0^T f(t) dW_t = \lim_{n \to \infty} \sum_{i=1}^n f(t_i) \cdot (W_{t_{i+1}} - W_{t_i})

This is a subtle but profound choice. It says: "Your 'trading decision' (the amount you want to buy, $f(t_i)$ ) can only be based on information you have right now (at time $t_i$ ). You cannot base your decision on the future random event ( $W_{t_{i+1}} - W_{t_i}$ ) that hasn't happened yet." Itô's integral is the only mathematical framework that respects the arrow of time.

Lesson 2.4: The Itô Integral (Summing Up Randomness)

Part 1: The "Normal" Integral (A Quick Review)

Part 2: The "Random" Integral (The New Problem)

Part 3: The Proof - Why The "Choice of Point" Matters

Part 4: Itô's Choice (The 'Non-Anticipating' Rule)