Itô’s Lemma and the Meaning of Brownian Stochastic Integrals
Summary
The document derives the Itô differential of squared Brownian motion, obtaining a stochastic term and a time correction from the quadratic variation. Integrating that identity relates the squared process to an Itô integral and elapsed time. This illustrates why ordinary differential intuition needs adjustment when working with Brownian paths.
The answer clarifies that an expression such as the current Brownian value multiplied by its differential is informal on its own; the corresponding integral is the mathematically defined object. It sketches the integral’s construction as a limit of sums using Brownian values at the left endpoints of time slices. The discussion is introductory and does not develop convergence conditions or the full definition of the Itô integral. The displayed identity is about the integral over an interval, rather than a pointwise value for the informal differential product.
Key ideas
- Itô’s lemma adds a time correction when applied to squared Brownian motion.
- The quadratic variation of Brownian motion contributes the time increment in the differential calculation.
- The integral of the Brownian process against its own increments is defined as an Itô integral.
- The Itô integral can be motivated by limits of sums that use left-endpoint process values.
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# Question on Ito's lemma involving $\mathrm{d}W(t)$
# Question on Ito's lemma involving $\mathrm{d}W(t)$
I am new to Ito-calculus, so please forgive me if the question is stupid.
Let $W(t)$ be a Brownian-Motion and $f(W(t))=W(t)^2$. If I want to calculate the differential $\mathrm{d}f(W(t))$, Ito's lemma yields: \begin{align*} \mathrm{d}f(W(t))&=\frac{\mathrm{d} f(W(t))}{\mathrm{d} W(t)} \mathrm{d}W(t)+\frac{1}{2}\frac{\mathrm{d}^2 f(W(t))}{\mathrm{d} W(t)^2} \mathrm{d}W(t)^2 \\ &=2W(t)\mathrm{d}W(t)+\frac{1}{2}\cdot 2 \cdot \underbrace{\mathrm{d}W(t)^2}_{=\mathrm{d}t} \\ &=2W(t)\mathrm{d}W(t)+\mathrm{d}t \end{align*} Using the integral notation instead of the differential notation I get: \begin{align} W(t)^2-W(0)^2&=2\int_0^tW(u)\mathrm{d}W(u)+\int_0^t\mathrm{d}t \\ W(t)^2&=2\int_0^tW(u)\mathrm{d}W(u)+t \end{align} I assume that my calculations are correct up to this point. However, I wonder how we deal with functions that depend on $\mathrm{d}W(t)$. For instance, $$ W(t)\mathrm{d}W(t) $$ From my understanding, Ito's lemma only holds for functions involving $W(t)$ and not $\mathrm{d}W(t)$. From the calculations above I conclude that: $$ W(t)\mathrm{d}W(t)=\frac{1}{2}W(t)^2-\frac{t}{2} $$ But how do we to calculate $W(t)\mathrm{d}W(t)$ directly ? Did I misunderstood something or is there another way on how to approach this type of functions ?
Thanks in advance.
## Answer by Arshdeep (score 3, accepted)
https://quant.stackexchange.com/a/66150
Ito's lemma is for twice differentiable functions of the form $f(t,W(t))$. You speak of $W(t)dW(t)$ - this is informal notation and doesn't have a mathematical meaning. Although once you put the integral sign, it becomes mathematically precise. So there's nothing known as $W(t)dW(t)$, but $I(t)=∫_0^{t}W(u)dW(u)$ is well defined via the definition of an Ito integral.
The more fundamental way to calculate this integral is to break up the time axis into slices and sum up the pieces, :
$W(t_j)[W(t_{j+1})-W(t_j)]$ across all $j$, and taking the limit as number of slices become infinite. This is a standard proof you can look up.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.